Solutions to Exercises

Appendix D Solutions to Exercises

1 Curves
1.1 Derivatives, Velocity, Etc.

Exercises

1.1.1.

Solution.

(a) Since, on the specified part of the circle,

x = \sqrt{a^{2} - y^{2}}

and

y

runs from

0

a,

the parametrization is

r (y) = \sqrt{a^{2} - y^{2}} \hat{ı ı} + y \hat{ȷ ȷ},

0 \leq y \leq a .

(b) Let

θ

be the angle between

the radius vector from the origin to the point $(a \cos θ, a \sin θ)$ on the circle and
the positive $x$ -axis.

The tangent line to the circle at

(a \cos θ, a \sin θ)

is perpendicular to the radius vector and so makes angle

ϕ = \frac{π}{2} + θ

with the positive

x

axis. (See the figure on the left below.) As

θ = ϕ - \frac{π}{2},

the desired parametrization is

\begin{aligned} (x (ϕ), y (ϕ)) = (a \cos (ϕ - \frac{π}{2}), a \sin (ϕ - \frac{π}{2})) = (a \sin ϕ, - a \cos ϕ), \\ \frac{π}{2} \leq ϕ \leq π \end{aligned}

θ

be the angle between

the radius vector from the origin to the point $(a \cos θ, a \sin θ)$ on the circle and
the positive $x$ -axis.

The arc from

(0, a)

(a \cos θ, a \sin θ)

subtends an angle

\frac{π}{2} - θ

and so has length

s = a (\frac{π}{2} - θ) .

(See the figure on the right above.) Thus

θ = \frac{π}{2} - \frac{s}{a}

and the desired parametrization is

(x (s), y (s)) = (a \cos (\frac{π}{2} - \frac{s}{a}), a \sin (\frac{π}{2} - \frac{s}{a})), 0 \leq s \leq \frac{π}{2} a

1.1.2.

Solution.

We can find the time at which the curve hits a given point by considering the two equations that arise from the two coordinates. For the

y

-coordinate to be 0, we must have

(t - 5)^{2} = 0,

i.e.

t = 5 .

So, the point

(- 1 / \sqrt{2}, 0)

happens when

t = 5 .

Similarly, for the

y

-coordinate to be

25,

we need

(t - 5)^{2} = 25,

(t - 5) = \pm 5 .

When

t = 0,

the curve hits

(1, 25);

when

t = 10,

the curve hits

(0, 25) .

So, in order, the curve passes through the points

(1, 25),

(- 1 / \sqrt{2}, 0),

and

(0, 25) .

1.1.3.

Solution.

The curve “crosses itself” when the same coordinates occur for different values of

t,

say

t_{1}

and

t_{2} .

So, we want to know when

\sin t_{1} = \sin t_{2}

and also

t_{1}^{2} = t_{2}^{2} .

Since

t_{1}

and

t_{2}

should be different, the second equation tells us

t_{2} = - t_{1} .

Then the first equation tells us

\sin t_{1} = \sin t_{2} = \sin (- t_{1}) = - \sin t_{1} .

That is,

\sin t_{1} = - \sin t_{1},

\sin t_{1} = 0 .

That happens whenever

t_{1} = π n

for an integer

n .

So, the points at which the curve crosses itself are those points

(0, (π n)^{2})

where

n

is an integer. It passes such a point at times

t = π n

and

t = - π n .

So, the curve hits this point

2 π n

time units apart.

1.1.4.

Solution.

(a) Pretend that the circle is a spool of thread. As the circle rolls, it dispenses the thread along the ground. When the circle rolls

θ

radians, it dispenses the arc length

θ a

of thread and the circle advances a distance

θ a .

So the centre of the circle has moved

θ a

units to the right from its starting point,

x = a .

The centre of the circle always has

y

-coordinate

a .

So, after rolling

θ

radians, the centre of the circle is at position

c (θ) = (a + a θ, a) .

(b) Now, let’s consider the position of

P

on the circle, after the circle has rolled

θ

radians.

From the diagram, we see that

P

a \cos θ

units above the centre of the circle, and

a \sin θ

units to the right of it. So, the position of

P

(a + a θ + a \sin θ, a + a \cos θ) .

Remark: this type of curve is known as a cycloid.

1.1.5.

Solution.

We aren’t concerned with

x,

so we can eliminate it by solving for it in one equation, and plugging that into the other. Since

C

lies on the plane,

x = - y - z,

so:

\begin{aligned} 1 & = x^{2} - \frac{1}{4} y^{2} + 3 z^{2} = (- y - z)^{2} - \frac{1}{4} y^{2} + 3 z^{2} \\ = \frac{3}{4} y^{2} + 4 z^{2} + 2 y z \end{aligned}

Completing the square,

\begin{aligned} 1 & = \frac{1}{2} y^{2} + {(2 z + \frac{y}{2})}^{2} \\ 1 - \frac{y^{2}}{2} & = {(2 z + \frac{y}{2})}^{2} \end{aligned}

Since $y$ is small, the left hand is close to $1$ and the right hand side is close to $(2 z)^{2} .$ So $(2 z)^{2} \approx 1 .$ Since $z$ is negative, $z \approx - \frac{1}{2}$ and $2 z + \frac{y}{2} < 0 .$ Also, $1 - \frac{y^{2}}{2}$ is positive, so it has a real square root.

\begin{aligned} - \sqrt{1 - \frac{y^{2}}{2}} & = 2 z + \frac{y}{2} \\ - \frac{1}{2} \sqrt{1 - \frac{y^{2}}{2}} - \frac{y}{4} & = z \end{aligned}

1.1.6.

Solution.

To determine whether the particle is rising or falling, we only need to consider its

z

-coordinate:

z (t) = (t - 1)^{2} (t - 3)^{2} .

Its derivative with respect to time is

z^{'} (t) = 4 (t - 1) (t - 2) (t - 3) .

This is positive when

1 < t < 2

and when

3 < t,

so the particle is increasing on

(1, 2) \cup (3, \infty)

and decreasing on

(0, 1) \cup (2, 3) .

r (t)

is the position of the particle at time

t,

then its speed is

| r^{'} (t) | .

We differentiate:

r^{'} (t) = - e^{- t} \hat{ı ı} - \frac{1}{t^{2}} \hat{ȷ ȷ} + 4 (t - 1) (t - 2) (t - 3) \hat{k}

So,

r (1) = - \frac{1}{e} \hat{ı ı} - 1 \hat{ȷ ȷ}

and

r (3) = - \frac{1}{e^{3}} \hat{ı ı} - \frac{1}{9} \hat{ȷ ȷ} .

The absolute value of every component of

r (1)

is greater than or equal to that of the corresponding component of

r (3),

| r (1) | > | r (3) | .

That is, the particle is moving more swiftly at

t = 1

than at

t = 3 .

Note: We could also compute the sizes of both vectors directly:

| r^{'} (1) | = \sqrt{{(\frac{1}{e})}^{2} + (- 1)^{2}},

and

| r^{'} (3) | = \sqrt{{(\frac{1}{e^{3}})}^{2} + {(- \frac{1}{9})}^{2}} .

1.1.7.

Solution.

The red vector is

r (t + h) - r (t) .

The arclength of the segment indicated by the blue line is the (scalar)

s (t + h) - s (t) .

Remark: as

h

approaches 0, the curve (if it’s differentiable at

t

) starts to resemble a straight line, with the length of the vector

r (t + h) - r (t)

approaching the scalar

s (t + h) - s (t) .

This step is crucial to understanding Lemma 1.1.4.

1.1.8.

Solution.

Velocity is a vector-valued quantity, so it has both a magnitude and a direction. Speed is a scalar — the magnitude of the velocity. It does not include a direction.

1.1.9. (✳).

Solution.

By the product rule

\begin{aligned} \frac{d}{d t} [(r \times r^{'}) \cdot r^{″}] & = (r^{'} \times r^{'}) \cdot r^{″} + (r \times r^{″}) \cdot r^{″} + (r \times r^{'}) \cdot r^{‴} \end{aligned}

The first term vanishes because

r^{'} \times r^{'} = 0 .

The second term vanishes because

r \times r^{″}

is perpendicular to

r^{″} .

\frac{d}{d t} [(r \times r^{'}) \cdot r^{″}] = (r \times r^{'}) \cdot r^{‴}

which is (c).

1.1.10.

Solution.

we are told that

r (t) ⊥ r^{'} (t),

so that

r (t) \cdot r^{'} (t) = 0,

for all

t .

Consequently

\frac{d}{d t} | r (t) |^{2} = \frac{d}{d t} [r (t) \cdot r (t)] = 2 r (t) \cdot r^{'} (t) = 0

| r (t) |^{2}

is a constant, say

A,

independent of time and

r (t)

always lies on the sphere of radius

\sqrt{A}

centred on the origin.

1.1.11. (✳).

Solution.

We have

v (t) = r^{'} (t) = 5 \sqrt{2} \hat{ı ı} + 5 e^{5 t} \hat{ȷ ȷ} + 5 e^{- 5 t} \hat{k}

and hence

| v (t) | = | r^{'} (t) | = 5 | \sqrt{2} \hat{ı ı} + e^{5 t} \hat{ȷ ȷ} + e^{- 5 t} \hat{k} | = 5 \sqrt{2 + e^{10 t} + e^{- 10 t}}

Since

2 + e^{10 t} + e^{- 10 t} = (e^{5 t} + e^{- 5 t})^{2},

that’s (d).

1.1.12.

Solution.

We are told that

r (t) = a \cos t \hat{ı ı} + a \sin t \hat{ȷ ȷ} + c t \hat{k}

So, by definition,

\begin{aligned} velocity & = v (t) = r^{'} (t) = - a \sin t \hat{ı ı} + a \cos t \hat{ȷ ȷ} + c \hat{k} \\ speed & = \frac{d s}{d t} (t) = | r^{'} (t) | = \sqrt{a^{2} + c^{2}} \\ acceleration & = a (t) = r^{″} (t) = - a \cos t \hat{ı ı} - a \sin t \hat{ȷ ȷ} \end{aligned}

t

runs over an interval of length

2 π,

(x, y)

traces out a circle of radius

a

and

z

increases by

2 π c .

The path is a helix with radius

a

and with each turn having height

2 π c .

1.1.13. (✳).

Solution.

(a) Since

r^{'} (t) = (2 t, 0, t^{2}),

the specified unit tangent at

t = 1

\hat{T} (1) = \frac{(2, 0, 1)}{\sqrt{5}}

(b) We are to find the arc length between

(0, 3, 0)

and

(1, 3, - 1 / 3) .

\frac{d s}{d t} = | r^{'} (t) | = \sqrt{4 t^{2} + t^{4}},

that arc length is the integral of

\sqrt{4 t^{2} + t^{4}}

with

t

running between the value of

t

for which

r (t) = (0, 3, 0)

and the value of

t

for which

r (t) = (1, 3, - 1 / 3) .

To find those two values of

t,

we observe that

the first component of $r (t),$ namely $t^{2},$ matches the first component of $(0, 3, 0),$ namely $0,$ only when $t = 0 .$ So we would guess that the $t$ corresponding to $(0, 3, 0)$ is $t = 0 .$ As $r (0) = (0^{2}, 3, \frac{1}{3} 0^{3}) = (0, 3, 0),$ this is indeed the case.
The first component of $r (t),$ namely $t^{2},$ matches the first component of $(1, 3, - 1 / 3),$ namely $1,$ only when $t = \pm 1 .$ For the third component of $r (t),$ namely $\frac{1}{3} t^{3},$ to match the third component of $(1, 3, - 1 / 3),$ namely $- 1 / 3,$ we need $t < 0 .$ So we would guess that the $t$ corresponding to $(1, 3, - 1 / 3)$ is $t = - 1 .$ To check that this is indeed the case, we compute $r (- 1) = ((- 1)^{2}, 3, \frac{1}{3} (- 1)^{3}) = (1, 3, - 1 / 3) .$

t

runs from

- 1

0

and

\begin{aligned} arc length & = \int_{- 1}^{0} \sqrt{4 t^{2} + t^{4}} d t \end{aligned}

The integrand is even, so

\begin{aligned} arc length & = \int_{0}^{1} \sqrt{4 t^{2} + t^{4}} d t = \int_{0}^{1} t \sqrt{4 + t^{2}} d t = [\frac{1}{3} {(4 + t^{2})}^{3 / 2}]_{0}^{1} \\ = \frac{1}{3} [5^{3 / 2} - 8] \end{aligned}

1.1.14.

Solution.

By Lemma 1.1.4 the arclength of

r (t)

from

t = 0

t = 1

\int_{0}^{1} | \frac{d r}{d t} (t) | d t .

We’ll calculate this in a few pieces to make the steps clearer.

\begin{aligned} r (t) & = (t, \sqrt{\frac{3}{2}} t^{2}, t^{3}) \\ \frac{d r}{d t} (t) & = (1, \sqrt{6} t, 3 t^{2}) \\ | \frac{d r}{d t} (t) | & = \sqrt{1^{2} + (\sqrt{6} t)^{2} + (3 t^{2})^{2}} = \sqrt{1 + 6 t^{2} + 9 t^{4}} \\ = \sqrt{(3 t^{2} + 1)^{2}} = 3 t^{2} + 1 \\ \int_{0}^{1} | \frac{d r}{d t} (t) | d t & = \int_{0}^{1} (3 t^{2} + 1) d t = 2 \end{aligned}

1.1.15.

Solution.

Since

\begin{aligned} x^{'} (t) & = a [\cos^{2} t - \sin^{2} t] = a \cos 2 t \\ y^{'} (t) & = 2 a \sin t \cos t = a \sin 2 t \\ z^{'} (t) & = b \end{aligned}

we have

\frac{d s}{d t} (t) = \sqrt{x^{'} (t)^{2} + y^{'} (t)^{2} + z^{'} (t)^{2}} = \sqrt{a^{2} + b^{2}}

As the speed

\frac{d s}{d t} (t)

is constant, the length is just

\frac{d s}{d t} T = \sqrt{a^{2} + b^{2}} T .

1.1.16.

Solution.

Since

r (t)

is the position of the particle, its acceleration is

r^{″} (t) .

\begin{aligned} r (t) & = (t + \sin t, \cos t) \\ r^{'} (t) & = (1 + \cos t, - \sin t) \\ r^{″} (t) & = (- \sin t, - \cos t) \\ | r^{″} (t) | & = \sqrt{\sin^{2} t + \cos^{2} t} = 1 \end{aligned}

The magnitude of acceleration is constant, but its direction is changing, since

r^{″} (t)

is a vector with changing direction.

1.1.17. (✳).

Solution.

(a) The speed is

\begin{aligned} \frac{d s}{d t} (t) = | r^{'} (t) | & = | (2 \cos t - 2 t \sin t, 2 \sin t + 2 t \cos t, t^{2}) | \\ = \sqrt{(2 \cos t - 2 t \sin t)^{2} + (2 \sin t + 2 t \cos t)^{2} + t^{4}} \\ = \sqrt{4 + 4 t^{2} + t^{4}} \\ = 2 + t^{2} \end{aligned}

so the length of the curve is

\begin{aligned} length & = \int_{0}^{2} \frac{d s}{d t} d t = \int_{0}^{2} (2 + t^{2}) d t = {[2 t + \frac{t^{3}}{3}]}_{0}^{2} = \frac{20}{3} \end{aligned}

(b) A tangent vector to the curve at

r (π) = (- 2 π, 0, \frac{π^{3}}{3})

r^{'} (π) = (2 \cos π - 2 π \sin π, 2 \sin π + 2 π \cos π, π^{2}) = (- 2, - 2 π, π^{2})

So parametric equations for the tangent line at

r (π)

are

\begin{aligned} x (t) & = - 2 π - 2 t \\ y (t) & = - 2 π t \\ z (t) & = \frac{π^{3}}{3} + π^{2} t \end{aligned}

1.1.18. (✳).

Solution.

(a) As

r (t) = (3 \cos t, 3 \sin t, 4 t),

the velocity of the particle is

r^{'} (t) = (- 3 \sin t, 3 \cos t, 4)

(b) As

\frac{d s}{d t},

the rate of change of arc length per unit time, is

\frac{d s}{d t} (t) = | r^{'} (t) | = | (- 3 \sin t, 3 \cos t, 4) | = 5

the arclength of its path between

t = 1

and

t = 2

\int_{1}^{2} d t \frac{d s}{d t} (t) = \int_{1}^{2} d t 5 = 5

1.1.19.

Solution.

(a) We can parametrize the circle

x^{2} + y^{2} = 9

x (θ) = 3 \cos θ,

y (θ) = 3 \sin θ

with

θ

running from

0

2 π .

z = 2 x + 3 y,

the ellipse can be parametrized by

\begin{aligned} x (θ) = 3 \cos θ, y (θ) = 3 \sin θ, z (θ) = 2 x (θ) + 3 y (θ) = 6 \cos θ + 9 \sin θ \\ 0 \leq θ \leq 2 π \end{aligned}

(b) As

\begin{aligned} \frac{d s}{d θ} & = \sqrt{x^{'} (θ)^{2} + y^{'} (θ)^{2} + z^{'} (θ)^{2}} \\ = \sqrt{9 \sin^{2} θ + 9 \cos^{2} θ + 36 \sin^{2} θ + 81 \cos^{2} θ - 108 \sin θ \cos θ} \\ = \sqrt{45 + 45 \cos^{2} θ - 108 \sin θ \cos θ} \end{aligned}

the circumference is

s = \int_{0}^{2 π} \sqrt{45 + 45 \cos^{2} θ - 108 \sin θ \cos θ} d θ

1.1.20. (✳).

Solution.

(a) As

\begin{aligned} r^{'} (t) & = - \sin t \cos^{2} t \hat{ı ı} + \sin^{2} t \cos t \hat{ı ı} + 3 \sin^{2} t \cos t \hat{k} \\ = \sin t \cos t (- \cos t \hat{ı ı} + \sin t \hat{ȷ ȷ} + 3 \sin t \hat{k}) \\ \frac{d s}{d t} (t) & = | \sin t \cos t | \sqrt{\cos^{2} t + \sin^{2} t + 9 \sin^{2} t} = | \sin t \cos t | \sqrt{1 + 9 \sin^{2} t} \end{aligned}

the arclength from

t = 0

t = \frac{π}{2}

\begin{aligned} \int_{0}^{π / 2} \frac{d s}{d t} (t) d t = \int_{0}^{π / 2} \sin t \cos t \sqrt{1 + 9 \sin^{2} t} d t \\ = \frac{1}{18} \int_{1}^{10} \sqrt{u} d u with u = 1 + 9 \sin^{2} t, d u = 18 \sin t \cos t d t \\ = \frac{1}{18} [\frac{2}{3} u^{3 / 2}]_{1}^{10} \\ = \frac{1}{27} (10 \sqrt{10} - 1) \end{aligned}

(b) The arclength from

t = 0

t = π

\begin{aligned} \int_{0}^{π} \frac{d s}{d t} (t) d t & = \int_{0}^{π} | \sin t \cos t | \sqrt{1 + 9 \sin^{2} t} d t \\ Don't forget the absolute value signs! \\ = 2 \int_{0}^{π / 2} | \sin t \cos t | \sqrt{1 + 9 \sin^{2} t} d t \\ = 2 \int_{0}^{π / 2} \sin t \cos t \sqrt{1 + 9 \sin^{2} t} d t \end{aligned}

since the integrand is invariant under

t \to π - t .

So the arc length from

t = 0

t = π

is just twice the arc length from part (a), namely

\frac{2}{27} (10 \sqrt{10} - 1) .

1.1.21. (✳).

Solution.

Since

\begin{aligned} r (t) & = \frac{t^{3}}{3} \hat{ı ı} + \frac{t^{2}}{2} \hat{ȷ ȷ} + \frac{t}{2} \hat{k} \\ r^{'} (t) & = t^{2} \hat{ı ı} + t \hat{ȷ ȷ} + \frac{1}{2} \hat{k} \\ \frac{d s}{d t} (t) = | r^{'} (t) | & = \sqrt{t^{4} + t^{2} + \frac{1}{4}} = \sqrt{(t^{2} + \frac{1}{2})^{2}} = t^{2} + \frac{1}{2} \end{aligned}

the length of the curve is

\begin{matrix} s (t) = \int_{0}^{t} \frac{d s}{d t} (u) d u = \int_{0}^{t} (u^{2} + \frac{1}{2}) d u = \frac{t^{3}}{3} + \frac{t}{2} \end{matrix}

1.1.22. (✳).

Solution.

Since

\begin{aligned} r (t) & = t^{m} \hat{ı ı} + t^{m} \hat{ȷ ȷ} + t^{3 m / 2} \hat{k} \\ r^{'} (t) & = m t^{m - 1} \hat{ı ı} + m t^{m - 1} \hat{ȷ ȷ} + \frac{3 m}{2} t^{3 m / 2 - 1} \hat{k} \\ \frac{d s}{d t} = | r^{'} (t) | & = \sqrt{2 m^{2} t^{2 m - 2} + \frac{9 m^{2}}{4} t^{3 m - 2}} = m t^{m - 1} \sqrt{2 + \frac{9}{4} t^{m}} \end{aligned}

the arc length is

\begin{aligned} \int_{a}^{b} \frac{d s}{d t} (t) d t & = \int_{a}^{b} m t^{m - 1} \sqrt{2 + \frac{9}{4} t^{m}} d t \\ = \frac{4}{9} \int_{2 + \frac{9}{4} a^{m}}^{2 + \frac{9}{4} b^{m}} \sqrt{u} d u with u = 2 + \frac{9}{4} t^{m}, d u = \frac{9 m}{4} t^{m - 1} \\ = \frac{4}{9} [\frac{2}{3} u^{3 / 2}]_{2 + \frac{9}{4} a^{m}}^{2 + \frac{9}{4} b^{m}} \\ = \frac{8}{27} [(2 + \frac{9}{4} b^{m})^{3 / 2} - (2 + \frac{9}{4} a^{m})^{3 / 2}] \end{aligned}

1.1.23.

Solution.

(a) Since

y = \sqrt{x}

and

z = \frac{2}{3} x y = \frac{2}{3} x^{3 / 2},

\begin{aligned} r (x) & = x \hat{ı ı} + \sqrt{x} \hat{ȷ ȷ} + \frac{2}{3} x^{3 / 2} \hat{k} \end{aligned}

For the remaining parts of this problem we will also need

\begin{aligned} r^{'} (x) & = \hat{ı ı} + \frac{1}{2 \sqrt{x}} \hat{ȷ ȷ} + \sqrt{x} \hat{k} \\ r^{″} (x) & = - \frac{1}{4 x^{3 / 2}} \hat{ȷ ȷ} + \frac{1}{2 \sqrt{x}} \hat{k} \\ \frac{d s}{d x} = | r^{'} (x) | & = \sqrt{1 + \frac{1}{4 x} + x} = \sqrt{{(\frac{1}{2 \sqrt{x}} + \sqrt{x})}^{2}} = \frac{1}{2 \sqrt{x}} + \sqrt{x} \\ \frac{d s}{d x} (1) & = \frac{3}{2} \end{aligned}

(b)

\begin{aligned} \int_{C} d s & = \int_{0}^{9} \frac{d s}{d x} d x = \int_{0}^{9} (\frac{1}{2 \sqrt{x}} + \sqrt{x}) d x = {[\sqrt{x} + \frac{2}{3} x^{3 / 2}]}_{0}^{9} \\ = 3 + 18 = 21 \end{aligned}

$r (x)$ the position of the particle when its first coordinate is $x,$
$R (t)$ the position of the particle at time $t,$
$x (t)$ the $x$ --coordinate of the particle at time $t,$ and
$s (x)$ the arc length of the curve from the origin to $r (x) .$

We are told that

| R^{'} (t) | = 9

for all

t .

\begin{aligned} R (t) = r (x (t)) ⟹ R^{'} (t) = r^{'} (x (t)) \frac{d x}{d t} (t) \\ ⟹ 9 = | R^{'} (t) | = \frac{d s}{d x} (x (t)) \frac{d x}{d t} (t) = (\frac{1}{2 \sqrt{x (t)}} + \sqrt{x (t)}) \frac{d x}{d t} (t) \end{aligned}

In particular, if the particle is at

(1, 1, \frac{2}{3})

at time

0,

then

x (0) = 1

and

9 = (\frac{1}{2 \sqrt{1}} + \sqrt{1}) \frac{d x}{d t} (0) ⟹ \frac{d x}{d t} (0) = 6

so that

R^{'} (0) = r^{'} (1) \frac{d x}{d t} (0) = (\hat{ı ı} + \frac{1}{2} \hat{ȷ ȷ} + \hat{k}) 6 = 6 \hat{ı ı} + 3 \hat{ȷ ȷ} + 6 \hat{k}

(d) By the product and chain rules,

R^{'} (t) = r^{'} (x (t)) \frac{d x}{d t} (t) ⟹ R^{″} (t) = r^{″} (x (t)) {(\frac{d x}{d t} (t))}^{2} + r^{'} (x (t)) \frac{d^{2} x}{d t^{2}} (t)

We saw in part (c) that

9 = | R^{'} (t) | = (\frac{1}{2 \sqrt{x (t)}} + \sqrt{x (t)}) \frac{d x}{d t} (t)

so that

\frac{d x}{d t} (t) = 9 {(\frac{1}{2 \sqrt{x (t)}} + \sqrt{x (t)})}^{- 1}

Differentiating that gives

\frac{d^{2} x}{d t^{2}} (t) = - 9 {(\frac{1}{2 \sqrt{x (t)}} + \sqrt{x (t)})}^{- 2} (- \frac{1}{4 x (t)^{3 / 2}} + \frac{1}{2 \sqrt{x (t)}}) \frac{d x}{d t} (t)

In particular, when

t = 0,

x (0) = 1

and

\frac{d x}{d t} (0) = 6

\frac{d^{2} x}{d t^{2}} (0) = - 9 {(\frac{3}{2})}^{- 2} (\frac{1}{4}) 6 = - 6

\begin{aligned} R^{″} (0) & = r^{″} (1) (6)^{2} + r^{'} (1) (- 6) = 36 (- \frac{1}{4} \hat{ȷ ȷ} + \frac{1}{2} \hat{k}) - 6 (\hat{ı ı} + \frac{1}{2} \hat{ȷ ȷ} + \hat{k}) \\ = - 6 \hat{ı ı} - 12 \hat{ȷ ȷ} + 12 \hat{k} \end{aligned}

1.1.24.

Solution.

Given the position of the particle, we can find its velocity:

v (t) = r^{'} (t) = (\cos t, - \sin t, 1)

Applying the given formula,

L (t) = r \times v = (\sin t, \cos t, t) \times (\cos t, - \sin t, 1) .

[Solution 1:] We can first compute the cross product, then differentiate:
$\begin{aligned} L (t) & = (\cos t + t \sin t) \hat{ı ı} + (t \cos t - \sin t) \hat{ȷ ȷ} - \hat{k} \\ L^{'} (t) & = t \cos t \hat{ı ı} - t \sin t \hat{ȷ ȷ} \\ | L^{'} (t) | & = \sqrt{t^{2} (\sin^{2} t + \cos^{2} t)} = \sqrt{t^{2}} = | t | \end{aligned}$
[Solution 2:] Using the product rule:
$\begin{aligned} L^{'} (t) & = r^{'} (t) \times v (t) + r (t) \times v^{'} (t) \\ = \underset{0}{\underset{⏟}{r^{'} (t) \times r^{'} (t)}} + r (t) \times v^{'} (t) \\ = (\sin t, \cos t, t) \times (- \sin t, - \cos t, 0) \\ = t \cos t \hat{ı ı} - t \sin t \hat{ȷ ȷ} \\ | L^{'} (t) | & = \sqrt{t^{2} \cos^{2} t + t^{2} \sin t^{2}} = | t | \end{aligned}$

1.1.25. (✳).

Solution.

(a) Since

z = 6 u,

y = \frac{z^{2}}{12} = 3 u^{2}

and

x = \frac{y z}{18} = u^{3},

\begin{aligned} r (u) & = u^{3} \hat{ı ı} + 3 u^{2} \hat{ȷ ȷ} + 6 u \hat{k} \end{aligned}

(b)

\begin{aligned} r^{'} (u) & = 3 u^{2} \hat{ı ı} + 6 u \hat{ȷ ȷ} + 6 \hat{k} \\ r^{″} (u) & = 6 u \hat{ı ı} + 6 \hat{ȷ ȷ} \\ \frac{d s}{d u} (u) = | r^{'} (u) | & = \sqrt{9 u^{4} + 36 u^{2} + 36} = 3 (u^{2} + 2) \end{aligned}

\int_{C} d s = \int_{0}^{1} \frac{d s}{d u} d u = \int_{0}^{1} 3 (u^{2} + 2) d u = [u^{3} + 6 u]_{0}^{1} = 7

R (t)

the position of the particle at time

t .

Then

R (t) = r (u (t)) ⟹ R^{'} (t) = r^{'} (u (t)) \frac{d u}{d t}

In particular, if the particle is at

(1, 3, 6)

at time

t_{1},

then

u (t_{1}) = 1

and

6 \hat{ı ı} + 12 \hat{ȷ ȷ} + 12 \hat{k} = R^{'} (t_{1}) = r^{'} (1) \frac{d u}{d t} (t_{1}) = (3 \hat{ı ı} + 6 \hat{ȷ ȷ} + 6 \hat{k}) \frac{d u}{d t} (t_{1})

which implies that

\frac{d u}{d t} (t_{1}) = 2 .

(d) By the product and chain rules,

R^{'} (t) = r^{'} (u (t)) \frac{d u}{d t} ⟹ R^{″} (t) = r^{″} (u (t)) (\frac{d u}{d t})^{2} + r^{'} (u (t)) \frac{d^{2} u}{d t^{2}}

In particular,

\begin{aligned} 27 \hat{ı ı} + 30 \hat{ȷ ȷ} + 6 \hat{k} & = R^{″} (t_{1}) = r^{″} (1) (\frac{d u}{d t} (t_{1}))^{2} + r^{'} (1) \frac{d^{2} u}{d t^{2}} (t_{1}) \\ = (6 \hat{ı ı} + 6 \hat{ȷ ȷ}) 2^{2} + (3 \hat{ı ı} + 6 \hat{ȷ ȷ} + 6 \hat{k}) \frac{d^{2} u}{d t^{2}} (t_{1}) \end{aligned}

Simplifying

3 \hat{ı ı} + 6 \hat{ȷ ȷ} + 6 \hat{k} = (3 \hat{ı ı} + 6 \hat{ȷ ȷ} + 6 \hat{k}) \frac{d^{2} u}{d t^{2}} (t_{1}) ⟹ \frac{d^{2} u}{d t^{2}} (t_{1}) = 1

1.1.26. (✳).

Solution.

(a) According to Newton,

m r^{″} (t) = F (t) so that r^{″} (t) = - 3 t \hat{ı ı} + \sin t \hat{ȷ ȷ} + 2 e^{2 t} \hat{k}

Integrating once gives

\begin{matrix} r^{'} (t) = - 3 \frac{t^{2}}{2} \hat{ı ı} - \cos t \hat{ȷ ȷ} + e^{2 t} \hat{k} + c \end{matrix}

for some constant vector

c .

We are told that

r^{'} (0) = v_{0} = \frac{π^{2}}{2} \hat{ı ı} .

This forces

c = \frac{π^{2}}{2} \hat{ı ı} + \hat{ȷ ȷ} - \hat{k}

so that

\begin{matrix} r^{'} (t) = (\frac{π^{2}}{2} - \frac{3 t^{2}}{2}) \hat{ı ı} + (1 - \cos t) \hat{ȷ ȷ} + (e^{2 t} - 1) \hat{k} \end{matrix}

Integrating a second time gives

\begin{matrix} r (t) = (\frac{π^{2} t}{2} - \frac{t^{3}}{2}) \hat{ı ı} + (t - \sin t) \hat{ȷ ȷ} + (\frac{1}{2} e^{2 t} - t) \hat{k} + c \end{matrix}

for some (other) constant vector

c .

We are told that

r (0) = r_{0} = \frac{1}{2} \hat{k} .

This forces

c = 0

so that

\begin{matrix} r (t) = (\frac{π^{2} t}{2} - \frac{t^{3}}{2}) \hat{ı ı} + (t - \sin t) \hat{ȷ ȷ} + (\frac{1}{2} e^{2 t} - t) \hat{k} \end{matrix}

(b) The particle is in the plane

x = 0

when

\begin{matrix} 0 = (\frac{π^{2} t}{2} - \frac{t^{3}}{2}) = \frac{t}{2} (π^{2} - t^{2}) ⟺ t = 0, \pm π \end{matrix}

So the desired time is

t = π .

t = π,

the velocity is

\begin{aligned} r^{'} (π) & = (\frac{π^{2}}{2} - \frac{3 π^{2}}{2}) \hat{ı ı} + (1 - \cos π) \hat{ȷ ȷ} + (e^{2 π} - 1) \hat{k} \\ = - π^{2} \hat{ı ı} + 2 \hat{ȷ ȷ} + (e^{2 π} - 1) \hat{k} \end{aligned}

1.1.27. (✳).

Solution.

(a) Parametrize

C

x .

Since

y = x^{2}

and

z = \frac{2}{3} x^{3},

\begin{aligned} r (x) & = x \hat{ı ı} + x^{2} \hat{ȷ ȷ} + \frac{2}{3} x^{3} \hat{k} \\ r^{'} (x) & = \hat{ı ı} + 2 x \hat{ȷ ȷ} + 2 x^{2} \hat{k} \\ r^{″} (x) & = 2 \hat{ȷ ȷ} + 4 x \hat{k} \\ \frac{d s}{d x} & = | r^{'} (x) | = \sqrt{1 + 4 x^{2} + 4 x^{4}} = 1 + 2 x^{2} \end{aligned}

and

\int_{C} d s = \int_{0}^{3} \frac{d s}{d x} d x = \int_{0}^{3} (1 + 2 x^{2}) d x = {[x + \frac{2}{3} x^{3}]}_{0}^{3} = 21

(b) The particle travelled a distance of 21 units in

\frac{7}{2}

time units. This corresponds to a speed of

\frac{21}{7 / 2} = 6 .

R (t)

the position of the particle at time

t .

Then

R (t) = r (x (t)) ⟹ R^{'} (t) = r^{'} (x (t)) \frac{d x}{d t}

By parts (a) and (b) and the chain rule

6 = \frac{d s}{d t} = \frac{d s}{d x} \frac{d x}{d t} = (1 + 2 x^{2}) \frac{d x}{d t} ⟹ \frac{d x}{d t} = \frac{6}{1 + 2 x^{2}}

In particular, the particle is at

(1, 1, \frac{2}{3})

x = 1 .

At this time

\frac{d x}{d t} = \frac{6}{1 + 2 \times 1} = 2

and

R^{'} = r^{'} (1) \frac{d x}{d t} = (\hat{ı ı} + 2 \hat{ȷ ȷ} + 2 \hat{k}) 2 = 2 \hat{ı ı} + 4 \hat{ȷ ȷ} + 4 \hat{k}

(d) By the product and chain rules,

R^{'} (t) = r^{'} (x (t)) \frac{d x}{d t} ⟹ R^{″} (t) = r^{″} (x (t)) {(\frac{d x}{d t})}^{2} + r^{'} (x (t)) \frac{d^{2} x}{d t^{2}}

Applying

\frac{d}{d t}

6 = (1 + 2 x (t)^{2}) \frac{d x}{d t} (t)

gives

0 = 4 x {(\frac{d x}{d t})}^{2} + (1 + 2 x^{2}) \frac{d^{2} x}{d t^{2}}

In particular, when

x = 1

and

\frac{d x}{d t} = 2,

0 = 4 \times 1 (2)^{2} + (3) \frac{d^{2} x}{d t^{2}}

gives

\frac{d^{2} x}{d t^{2}} = - \frac{16}{3}

and

R^{″} = (2 \hat{ȷ ȷ} + 4 \hat{k}) (2)^{2} - (\hat{ı ı} + 2 \hat{ȷ ȷ} + 2 \hat{k}) \frac{16}{3} = - \frac{8}{3} (2 \hat{ı ı} + \hat{ȷ ȷ} - 2 \hat{k})

1.1.28.

Solution.

The question is already set up as an

x y

-plane, with the camera at the origin, so the vector in the direction the camera is pointing is

(x (t), y (t)) .

Let

θ

be the angle the camera makes with the positive

x

-axis (due east). The camera, the object, and the due-east direction (positive

x

-axis) make a right triangle.

\begin{aligned} \tan θ & = \frac{y}{x} \end{aligned}

Differentiating implicitly with respect to $t :$

\begin{aligned} \sec^{2} θ \frac{d θ}{d t} & = \frac{x y^{'} - y x^{'}}{x^{2}} \\ \frac{d θ}{d t} & = \cos^{2} θ (\frac{x y^{'} - y x^{'}}{x^{2}}) = {(\frac{x}{\sqrt{x^{2} + y^{2}}})}^{2} (\frac{x y^{'} - y x^{'}}{x^{2}}) \\ = \frac{x y^{'} - y x^{'}}{x^{2} + y^{2}} \end{aligned}

1.1.29.

Solution.

Using the Theorem of Pappus, we can calculate the surface area and volume of a pipe with the same length and radius as this pipe. So, we need to find the length of the pipe,

L .

\begin{aligned} \frac{d r}{d t} & = (\sqrt{2 t}, t, 1) \\ | \frac{d r}{d t} | & = \sqrt{2 t + t^{2} + 1} = | t + 1 | \\ L & = \int_{0}^{10} (t + 1) d t = 60 \end{aligned}

A pipe with radius 3 and length 60 has surface area

60 (2 π \cdot 3) = 360 π

and volume

60 (π \cdot 3^{2}) = 540 π .

1.1.30.

Solution.

In general a helix can be parametrized by

r (θ) = a \cos θ \hat{ı ı} + a \sin θ \hat{ȷ ȷ} + b θ \hat{k}

Our first task is to determine

a

and

b .

The radius of the helix is

3

cm, so

a = 3

cm. After 10 turns (i.e.

θ = 20 π

) the height,

b θ,

is 1 cm. So

b (20 π) = 1

and

b = \frac{1}{20 π}

cm/rad. Thus

r (θ) = 3 \cos θ \hat{ı ı} + 3 \sin θ \hat{ȷ ȷ} + \frac{1}{20 π} θ \hat{k} .

With each full turn of the helix (i.e. each increase of

θ

2 π

) the height of the helix increases by

2 π b = \frac{1}{10} cm .

So if we can determine the length of wire in one full turn of the helix, we can easily determine how many turns the helix goes through in total, and from that we can determine the total height of the helix.

r^{'} (θ) = - 3 \sin θ \hat{ı ı} + 3 \cos θ \hat{ȷ ȷ} + \frac{1}{20 π} \hat{k}

we have

\frac{d s}{d θ} = | r^{'} (θ) | = \sqrt{9 + \frac{1}{400 π^{2}}} .

So the length of one full turn of the helix is

\begin{aligned} \int_{0}^{2 π} \sqrt{9 + \frac{1}{400 π^{2}}} d θ & = 2 π \sqrt{9 + \frac{1}{400 π^{2}}} \end{aligned}

and 1000cm of wire generates

\frac{1000}{2 π \sqrt{9 + \frac{1}{400 π^{2}}}} = \frac{500}{π \sqrt{9 + \frac{1}{400 π^{2}}}}

turns. Each turn adds

\frac{1}{10} cm

to the height, so the total height is

\frac{500}{π \sqrt{9 + \frac{1}{400 π^{2}}}} \cdot \frac{1}{10} = \frac{50}{π \sqrt{9 + \frac{1}{400 π^{2}}}} \approx 5.3 cm

Remark. We can check that this answer is reasonable by taking advantage of the fact that each coil adds only a very small height (relative to the radius). So we expect the length of one coil to be about the same as the circumference of a circle of the same radius, namely

6 π .

If we were making actual circles of the wire, there would be

\frac{1000}{6 π}

of them. Stacking up at 10 per centimetre, this would make a pile of height

\frac{1000}{6 π \cdot 10}

cm. Since this number is also approximately

5.3

cm, we feel our result is reasonable.

1.1.31.

Solution.

Define

u (t) = e^{α t} \frac{d r}{d t} (t) .

Then

\begin{aligned} \frac{d u}{d t} (t) & = α e^{α t} \frac{d r}{d t} (t) + e^{α t} \frac{d^{2} r}{d t^{2}} (t) \\ = α e^{α t} \frac{d r}{d t} (t) - g e^{α t} \hat{k} - α e^{α t} \frac{d r}{d t} (t) \\ = - g e^{α t} \hat{k} \end{aligned}

Integrating both sides of this equation from

t = 0

t = T

gives

u (T) - u (0) = - g \frac{e^{α T} - 1}{α} \hat{k}

so that

u (T) = u (0) - g \frac{e^{α T} - 1}{α} \hat{k} = \frac{d r}{d t} (0) - g \frac{e^{α T} - 1}{α} \hat{k} = v_{0} - g \frac{e^{α T} - 1}{α} \hat{k}

Substituting in

u (T) = e^{α t} \frac{d r}{d t} (T)

and multiplying through by

e^{- α T}

gives

\frac{d r}{d t} (T) = e^{- α T} v_{0} - g \frac{1 - e^{- α T}}{α} \hat{k}

Integrating both sides of this equation from

T = 0

T = t

gives

\begin{matrix} r (t) - r (0) = \frac{e^{- α t} - 1}{- α} v_{0} - g \frac{t}{α} \hat{k} + g \frac{e^{- α t} - 1}{- α^{2}} \hat{k} \end{matrix}

so that

\begin{matrix} r (t) = r_{0} - \frac{e^{- α t} - 1}{α} v_{0} + g \frac{1 - α t - e^{- α t}}{α^{2}} \hat{k} \end{matrix}

1.2 Reparametrization

Exercises

1.2.1.

Solution.

By Lemma 1.1.4.c,

| r^{'} (s) | = 1

under arclength parametrization. So

\int_{1}^{t} | r^{'} (s) | d s = \int_{1}^{t} d s = t - 1 .

1.2.2.

Solution.

The arclength from

P

P

will be 0, so

P

is the point where

s = 0 .

That is,

r (0),

(\sin (1 / 2), \cos (1 / 2), \sqrt{3} / 2) .

1.2.3.

Solution 1.

We consider the situation geometrically. If we plot

R

in space (of the relevant dimension), regardless of its parametrization, the derivative at a point will give a vector tangent to

R,

in the direction the curve moves when the parameter is increasing. Since

a (t_{0})

and

b (s_{0})

describe the same spot on the curve,

a^{'} (t_{0})

and

b^{'} (s_{0})

will be parallel

Since we specified the derivatives are nonzero, there’s no messiness about vectors being parallel to a zero vector.

— they’re both tangent to the same piece of curve. Furthermore, as

t

increases, so does

s,

so the direction of increasing

t

is the same as the direction of increasing

s .

Therefore, A. holds.

Now we consider the magnitudes of the vectors, to rule out E. Recall

| a^{'} (t) |

is the speed at which the curve changes relative to

t;

this could be any (nonnegative) number. By the same token,

| b^{'} (s) | = 1 .

So,

b^{'} (s_{0})

is a unit vector, while

a^{'} (t_{0})

may or may not be. Then the two vectors are not necessarily equal (although they could be).

So, the best answer is A.

Solution 2.

The chain rule gives us a relationship between

b^{'} (s)

and

a^{'} (t) .

\begin{aligned} \frac{d b}{d s} & = \frac{d}{d s} [a (t (s))] = \frac{d a}{d t} \frac{d t}{d s} \end{aligned}

So, the vectors

\frac{d b}{d s}

and

\frac{d a}{d t}

differ only by the scalar function

\frac{d t}{d s} .

So, at any point along the curve, these vectors are parallel.

Furthermore, we know that

t

and

s

are positively correlated: as

t

increases, so does

s,

because we’re covering more arclength. So,

\frac{d t}{d s}

is nonnegative. Furthermore, since the derivatives are nonzero,

\frac{d t}{d s}

is nonzero. So,

b^{'} (s_{0})

and

a^{'} (t_{0})

are positive scalar multiples of each other. That is, they are parallel, and pointing in the same direction. However, unless

\frac{d t}{d s} = 1

(that is,

t (s) = s + C

for some constant

C

), the vectors do not have the same magnitude, and hence are not equal.

So, A is the best solution.

1.2.4. (✳).

Solution.

(a) The velocity vector is

\begin{aligned} r^{'} (t) & = (6 \sin^{2} (t) \cos t, - 6 \sin t \cos^{2} (t), 3 \cos^{2} t - 3 \sin^{2} t) \\ = 3 (\sin t \sin (2 t), - \cos t \sin (2 t), \cos (2 t)) \end{aligned}

In particular, since

\sin (π / 3) = \sin (2 π / 3) = \frac{\sqrt{3}}{2}

and

\cos (π / 3) = - \cos (2 π / 3) = \frac{1}{2},

\begin{aligned} r^{'} (π / 3) & = 3 (\frac{3}{4}, - \frac{\sqrt{3}}{4}, - \frac{1}{2}) \end{aligned}

and the specified unit tangent vector is

\begin{aligned} \hat{T} & = \frac{(\frac{3}{4}, - \frac{\sqrt{3}}{4}, - \frac{1}{2})}{| (\frac{3}{4}, - \frac{\sqrt{3}}{4}, - \frac{1}{2}) |} = (\frac{3}{4}, - \frac{\sqrt{3}}{4}, - \frac{1}{2}) \end{aligned}

(b) The speed is

\begin{aligned} \frac{d s}{d t} & = | r^{'} (t) | = 3 \sqrt{\sin^{2} t \sin^{2} (2 t) + \cos^{2} t \sin^{2} (2 t) + \cos^{2} (2 t)} \\ = 3 \sqrt{\sin^{2} (2 t) + \cos^{2} (2 t)} \\ = 3 \end{aligned}

s = 3 t

and the reparametrized form is

\begin{matrix} R (s) = (2 \sin^{3} (\frac{s}{3}), 2 \cos^{3} (\frac{s}{3}), 3 \sin (\frac{s}{3}) \cos (\frac{s}{3})) \end{matrix}

1.2.5. (✳).

Solution.

(a) We have

| r (t) | = e^{t} \leq 1

for

t \leq 0 .

So the part of the spiral contained in the unit circle is the part of the spiral with

- \infty < t \leq 0 .

\begin{matrix} r^{'} (t) = e^{t} (\cos t, \sin t) + e^{t} (- \sin t, \cos t) = e^{t} (\cos t - \sin t, \sin t + \cos t) \end{matrix}

the speed

\begin{matrix} \frac{d s}{d t} = | r^{'} (t) | = e^{t} \sqrt{(\cos t - \sin t)^{2} + (\sin t + \cos t)^{2}} = \sqrt{2} e^{t} \end{matrix}

and the arclength from

t = - \infty

r (t)

s (t) = \int_{- \infty}^{t} \frac{d s}{d t} (\tilde{t}) d \tilde{t} = \int_{- \infty}^{t} \sqrt{2} e^{\tilde{t}} d \tilde{t} = \sqrt{2} e^{t}

In particular the length of the part of the spiral contained in the unit circle is

s (0) = \sqrt{2} .

(b) The inverse function of

s (t) = \sqrt{2} e^{t}

t (s) = \ln (\frac{s}{\sqrt{2}})

with

s > 0 .

(As

t \to - \infty,

the arc length

s \to 0

and as

t \to + \infty,

the arc length

s \to + \infty .

) So the reparametrization is

\begin{aligned} R (s) = e^{t} (\cos t, \sin t) |_{t = \ln (\frac{s}{\sqrt{2}})} = \frac{s}{\sqrt{2}} (\cos (\ln (\frac{s}{\sqrt{2}})), \sin (\ln (\frac{s}{\sqrt{2}}))) \\ with s > 0 \end{aligned}

1.2.6.

Solution.

Using

\arctan t = z,

and so

t = \tan z :

\begin{aligned} r (t) & = (\frac{1}{\sqrt{1 + t^{2}}}, \frac{\arctan t}{\sqrt{1 + t^{- 2}}}, \arctan t) \\ = (\frac{1}{\sqrt{1 + \tan^{2} z}}, \frac{z}{\sqrt{1 + \cot^{2} z}}, z) \\ = (\frac{1}{| \sec z |}, \frac{z}{| \csc z |}, z) \\ = (| \cos z |, z | \sin z |, z) \end{aligned}

Since $0 \leq t,$ and $\arctan t < π / 2$ we have $0 \leq z < π / 2,$ so $\cos z$ and $\sin z$ are both nonnegative.

\begin{aligned} = (\cos z, z \sin z, z) \end{aligned}

If we didn’t have the restricted domain, this would make a spiral going up:

z

is both the height of the spiral and a radian measure. The

\hat{ı ı}

-component of the spiral stays between

- 1

and

1,

while the

\hat{ȷ ȷ}

-component increases. So, our spiral gets increasingly “wide,” while staying the same “thickness.”

Due to the restricted domain, our actual curve is only one-quarter of a “turn” of this spiral, indicated in red above.

The parameter

z

is a measure of height, and it is also a radian measure as the spiral turns.

1.2.7.

Solution.

\begin{aligned} r (t) & = (\frac{1}{2} t^{2}, \frac{1}{3} t^{3}) \\ r^{'} (t) & = (t, t^{2}) \\ | r^{'} (t) | & = \sqrt{t^{2} + t^{4}} = | t | \sqrt{1 + t^{2}} \\ s (t) & = \int_{- 1}^{t} | x | \sqrt{1 + x^{2}} d x \\ = {\begin{cases} \int_{- 1}^{t} - x \sqrt{1 + x^{2}} d x & when t \leq 0 \\ \int_{- 1}^{0} - x \sqrt{1 + x^{2}} d x + \int_{0}^{t} x \sqrt{1 + x^{2}} d x & when t > 0 \end{cases} \end{aligned}

Let $u = 1 + x^{2}, \frac{1}{2} d u = x d x$

\begin{aligned} = {\begin{cases} - \int_{2}^{1 + t^{2}} \frac{1}{2} \sqrt{u} d u & when t \leq 0 \\ - \int_{2}^{1} \frac{1}{2} \sqrt{u} d u + \int_{1}^{1 + t^{2}} \frac{1}{2} \sqrt{u} d u & when t > 0 \end{cases} \\ = {\begin{cases} - \frac{1}{3} u^{3 / 2} |_{2}^{1 + t^{2}} & when t \leq 0 \\ - \frac{1}{3} u^{3 / 2} |_{2}^{1} + \frac{1}{3} u^{3 / 2} |_{1}^{1 + t^{2}} & when t > 0 \end{cases} \\ = {\begin{cases} \frac{2^{3 / 2}}{3} - \frac{1}{3} (1 + t^{2})^{3 / 2} & when t \leq 0 \\ - \frac{2}{3} + \frac{2^{3 / 2}}{3} + \frac{1}{3} (1 + t^{2})^{3 / 2} & when t > 0 \end{cases} \end{aligned}

Solving for $t$ in terms of $s :$

\begin{aligned} 1 + t^{2} & = {\begin{cases} (2 \sqrt{2} - 3 s)^{2 / 3} & when t \leq 0 \\ (3 s + 2 - 2 \sqrt{2})^{2 / 3} & when t > 0 \end{cases} \\ t^{2} & = {\begin{cases} (2 \sqrt{2} - 3 s)^{2 / 3} - 1 & when t \leq 0 \\ (3 s + 2 - 2 \sqrt{2})^{2 / 3} - 1 & when t > 0 \end{cases} \end{aligned}

Remembering that $\sqrt{t^{2}} = | t | :$

\begin{aligned} t & = {\begin{cases} - \sqrt{(2 \sqrt{2} - 3 s)^{2 / 3} - 1} & when t \leq 0 \\ \sqrt{(3 s + 2 - 2 \sqrt{2})^{2 / 3} - 1} & when t > 0 \end{cases} \end{aligned}

Noting that

t = 0

when

s = \frac{1}{3} (2 \sqrt{2} - 1),

we find our reparametrization of

(\frac{1}{2} t^{2}, \frac{1}{3} t^{3}) .

When

s \leq \frac{1}{3} (2 \sqrt{2} - 1),

R (s) = (\frac{1}{2} [(2 \sqrt{2} - 3 s)^{2 / 3} - 1], - \frac{1}{3} {[(2 \sqrt{2} - 3 s)^{2 / 3} - 1]}^{3 / 2})

and when

s > \frac{1}{3} (2 \sqrt{2} - 1),

R (s) = (\frac{1}{2} [(3 s + 2 - 2 \sqrt{2})^{2 / 3} - 1], \frac{1}{3} {[(3 s + 2 - 2 \sqrt{2})^{2 / 3} - 1]}^{3 / 2})

Remark: after a computation with this much detail, it’s nice to find a few points to check, to verify that our answer is reasonable. For instance, when

s = 0,

t

should be

- 1,

and vice-versa. Also, we found that

t = 0

corresponds to

s = \frac{1}{3} (2 \sqrt{2} - 1) .

So, we should be able to verify that

r (0) = R (\frac{1}{3} (2 \sqrt{2} - 1))

and

r (- 1) = R (0) .

1.3 Curvature

Exercises

1.3.1.

Solution.

The curve is a circle of radius 3, centred at the origin. So, the “circle of best fit” is just the curve itself.

\hat{T}

is the unit vector tangent to the circle in direction of increasing

t,

and

\hat{N}

is the unit vector pointing towards the origin.

The radius of the (osculating) circle is 3, so

ρ = 3

and

κ = \frac{1}{ρ} = \frac{1}{3} .

1.3.2.

Solution.

The arclength of

r (t)

traced out by an interval of

t

of length

θ

3 θ .

That is,

s = 3 t .

Our reparametrization of the circle in terms of arclength is

R (s) = (3 \sin (s / 3), 3 \cos (s / 3)) .

We can calculate the vectors tangent to the circle, then normalize them (i.e. make them length one) to find

\hat{T} .

\begin{aligned} v (t) & = r^{'} (t) = (3 \cos t, - 3 \sin t) \\ \hat{T} (t) & = \frac{r^{'} (t)}{| r^{'} (t) |} = \frac{(3 \cos t, - 3 \sin t)}{3} = (\cos t, - \sin t) \\ \hat{T} (s) & = R^{'} (s) = (\cos (s / 3), - \sin (s / 3)) \end{aligned}

Note

R^{'} (s),

because it’s parametrized in terms of arclength, has derivative vectors of length one. So, we don’t need to normalize them (although if we did, it wouldn’t change anything).

Note also that we can check out answers using Question 1.3.1. In that question, we found

\hat{T}

was

\hat{ı ı}

when

t = s = 0;

this fits with the vectors we just found.

As in Question 1.3.1,

κ = \frac{1}{3} .

So, using Theorem 1.3.3 Part (b):

\begin{aligned} \frac{d \hat{T}}{d s} (s) & = κ (s) \hat{N} (s) \\ (- \frac{1}{3} \sin (s / 3), - \frac{1}{3} \cos (s / 3)) & = \frac{1}{3} \hat{N} (s) \\ (- \sin (s / 3), - \cos (s / 3)) & = \hat{N} (s) \end{aligned}

Remember

s = 3 t .

Using Theorem 1.3.3 Part (c):

\begin{aligned} \frac{d \hat{T}}{d t} & = κ \frac{d s}{d t} \hat{N} (t) \\ (- \sin t, - \cos t) & = \frac{1}{3} (3) \hat{N} (t) \\ (- \sin t, - \cos t) & = \hat{N} (t) \end{aligned}

1.3.3.

Solution.

t

increases, the arms of the spiral “flatten out,” looking like a circle of bigger and bigger radius. So, we would expect the curvature to decrease:

lim_{t \to \infty} κ (t) = 0 .

1.3.4.

Solution.

\frac{d s}{d t} = | v (t) | = | r^{'} (t) | = | (e^{t}, 3, \cos t) | = \sqrt{e^{2 t} + 9 + \cos^{2} t}

1.3.5.

Solution.

\begin{aligned} \hat{T} (t) & = \frac{v (t)}{| v (t) |} = \frac{r^{'} (t)}{| r^{'} (t) |} \\ = \frac{(e^{t} (\cos t - \sin t), e^{t} (\cos t + \sin t))}{\sqrt{e^{2 t} (\cos t - \sin t)^{2} + e^{2 t} (\cos t + \sin t)^{2}}} \\ = \frac{1}{\sqrt{2}} (\cos t - \sin t, \cos t + \sin t) \\ \frac{d \hat{T}}{d t} & = \frac{1}{\sqrt{2}} (- \sin t - \cos t, - \sin t + \cos t) \end{aligned}

Since $R (s)$ is parametrized with respect to arclength, $| R^{'} (s) | = 1 .$

\begin{aligned} \hat{T} (s) & = R^{'} (s) \end{aligned}

Making ample use of the chain rule, and setting $U (s) = (\ln (s / \sqrt{2})),$ we have $U^{'} (s) = \frac{1}{s} :$

\begin{aligned} \hat{T} (s) & = \frac{1}{\sqrt{2}} (\cos U (s) - \sin U (s), \cos U (s) + \sin U (s)) \\ \frac{d \hat{T}}{d s} & = \frac{1}{\sqrt{2} s} (- \sin U (s) - \cos U (s), - \sin U (s) + \cos U (s)) \end{aligned}

1.3.6.

Solution.

The circle of radius

r

centred at

(0, r)

x^{2} + (y - r)^{2} = r^{2} .

The bottom half of this circle is

y = g (x) = r - \sqrt{r^{2} - x^{2}}

\begin{aligned} g^{'} (x) & = \frac{x}{\sqrt{r^{2} - x^{2}}} & g^{'} (0) & = 0 \\ g^{″} (x) & = \frac{1}{\sqrt{r^{2} - x^{2}}} + \frac{x^{2}}{[{r^{2} - x^{2}]}^{3 / 2}} & g^{″} (0) & = \frac{1}{r} \end{aligned}

f (x)

and

g (x)

have the same second order Taylor approximation at

x = 0,

f^{″} (0) = g^{″} (0) = \frac{1}{r} .

We may parametrize the curve by

r (x) = x \hat{ı ı} + f (x) \hat{ȷ ȷ} .

\begin{aligned} r^{'} (x) & = \hat{ı ı} + f^{'} (x) \hat{ȷ ȷ} & r^{'} (0) & = \hat{ı ı} + f^{'} (0) \hat{ȷ ȷ} = \hat{ı ı} \\ r^{″} (x) & = f^{″} (x) \hat{ȷ ȷ} & r^{″} (0) & = f^{″} (0) \hat{ȷ ȷ} \end{aligned}

and

\begin{aligned} κ (0) & = \frac{| r^{'} (0) \times r^{″} (0) |}{| r^{'} (0) |^{3}} = \frac{| f^{″} (0) \hat{ı ı} \times \hat{ȷ ȷ} |}{| \hat{ı ı} |^{3}} = f^{″} (0) \end{aligned}

κ (0) = f^{″} (0) = \frac{1}{r}

and

r

is indeed the radius of curvature of

y = f (x)

x = 0 .

1.3.7.

Solution.

$v (t) = r^{'} (t) = (e^{t}, 2 t + 1)$
$a (t) = r^{″} (t) = (e^{t}, 2)$
$\frac{d s}{d t} = | v (t) | = \sqrt{e^{2 t} + (2 t + 1)^{2}}$
$\begin{aligned} \hat{T} (t) & = \frac{v (t)}{| v (t) |} = \frac{(e^{t}, 2 t + 1)}{\sqrt{e^{2 t} + (2 t + 1)^{2}}} \\ = (\frac{e^{t}}{\sqrt{e^{2 t} + (2 t + 1)^{2}}}, \frac{2 t + 1}{\sqrt{e^{2 t} + (2 t + 1)^{2}}}) \end{aligned}$
$κ (t) = \frac{| v (t) \times a (t) |}{{(\frac{d s}{d t})}^{3}} = \frac{| (e^{t}, 2 t + 1) \times (e^{t}, 2) |}{{\sqrt{e^{2 t} + (2 t + 1)^{2}}}^{3}} = \frac{e^{t} | 1 - 2 t |}{(e^{2 t} + (2 t + 1)^{2})^{3 / 2}}$

1.3.8.

Solution 1.

Note that

(\cos t + \sin t)^{2} + (\sin t - \cos t)^{2} = 2

for all

t .

So, the points

(x, y)

of our curve lie on

x^{2} + y^{2} = 2,

which is a circle of radius

\sqrt{2} .

Indeed

\begin{aligned} x (t) & = \cos t + \sin t = \sqrt{2} [\cos t \cos \frac{π}{4} + \sin t \sin \frac{π}{4}] \\ = \sqrt{2} \cos (t - \frac{π}{4}) \\ y (t) & = \sin t - \cos t = \sqrt{2} [\sin t \cos \frac{π}{4} - \cos t \sin \frac{π}{4}] \\ = \sqrt{2} \sin (t - \frac{π}{4}) \end{aligned}

So,

r (t)

circumnavigates a circle of radius

\sqrt{2}

and consequently has curvature

κ = \frac{1}{\sqrt{2}} .

Solution 2.

We use the formula

κ = \frac{| v (t) \times a (t) |}{| {(\frac{d s}{d t})}^{3} |},

remembering that

v (t) = r^{'} (t),

a (t) = r^{″} (t),

and

\frac{d s}{d t} = | r^{'} (t) | .

\begin{aligned} v (t) & = r^{'} (t) = (- \sin t + \cos t, \cos t + \sin t) \\ a (t) & = r^{″} (t) = (- \cos t - \sin t, - \sin t + \cos t) \\ v (t) \times a (t) & = [(- \sin t + \cos t)^{2} + (\cos t + \sin t)^{2}] \hat{k} = 2 \hat{k} \\ \frac{d s}{d t} & = | \frac{d v}{d t} | = \sqrt{(- \sin t + \cos t)^{2} + (\cos t + \sin t)^{2}} = \sqrt{2} \\ κ & = | \frac{v (t) \times a (t)}{{(\frac{d s}{d t})}^{3}} | = | \frac{2 \hat{k}}{{\sqrt{2}}^{3}} | = \frac{1}{\sqrt{2}} \end{aligned}

1.3.9.

Solution.

For the given ellipse

\begin{aligned} r (t) & = a \cos t \hat{ı ı} + b \sin t \hat{ȷ ȷ} \\ v (t) & = - a \sin t \hat{ı ı} + b \cos t \hat{ȷ ȷ} \\ | v (t) | & = \sqrt{a^{2} \sin^{2} t + b^{2} \cos^{2} t} \\ a (t) & = - a \cos t \hat{ı ı} - b \sin t \hat{ȷ ȷ} \\ v (t) \times a (t) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - a \sin t & b \cos t & 0 \\ - a \cos t & - b \sin t & 0 \end{array}] = a b \hat{k} \\ κ (t) & = \frac{| v (t) \times a (t) |}{| v (t) |^{3}} = \frac{a b}{{[a^{2} \sin^{2} t + b^{2} \cos^{2} t]}^{3 / 2}} \end{aligned}

So the maximum (minimum) curvature is achieved when the denominator is a minimum (maximum) which is the case when

\sin t = 0

(

\cos t = 0

). So

κ_{m a x} = \frac{a}{b^{2}}

and

κ_{m i n} = \frac{b}{a^{2}} .

1.3.10. (✳).

Solution.

Parametrize the curve by

r (t) = t \hat{ı ı} + e^{t} \hat{ȷ ȷ} .

Then

\begin{aligned} v (t) & = \hat{ı ı} + e^{t} \hat{ȷ ȷ} & v (0) & = \hat{ı ı} + \hat{ȷ ȷ} \\ \frac{d s}{d t} & = | v (t) | = \sqrt{1 + e^{2 t}} & \frac{d s}{d t} (0) & = \sqrt{2} \\ \hat{T} (t) & = \frac{v (t)}{| v (t) |} = \frac{\hat{ı ı} + e^{t} \hat{ȷ ȷ}}{\sqrt{1 + e^{2 t}}} & \hat{T} (0) & = \frac{v (0)}{| v (0) |} = \frac{\hat{ı ı} + \hat{ȷ ȷ}}{\sqrt{2}} \\ a (t) & = e^{t} \hat{ȷ ȷ} & a (0) & = \hat{ȷ ȷ} \end{aligned}

(a) We’re given

y

in terms of

x,

so let’s use Part (e) of Theorem 1.3.3:

\begin{aligned} κ & = \frac{| \frac{d^{2} y}{d x^{2}} |}{{[1 + (\frac{d y}{d x})^{2}]}^{3 / 2}} = \frac{e^{x}}{{[1 + (e^{x})^{2}]}^{3 / 2}} \\ κ (0) & = \frac{1}{[1 + 1]^{3 / 2}} = 2^{- 3 / 2} \end{aligned}

(b)

The radius of the circle we want is $ρ = \frac{1}{κ} = 2^{3 / 2} .$ If its centre is at $(a, b),$ then the circle will have equation $(x - a)^{2} + (y - b)^{2} = 2^{3} .$ So, we will find its centre.
The unit vector $\hat{N}$ points from our point $(0, 1)$ towards the centre of the circle. Since the radius of the circle is $2^{3 / 2},$ the centre of the circle will be at $(0, 1) + 2^{3 / 2} \hat{N} .$ So, we’ll find $\hat{N} .$
Since $\hat{N}$ is a unit vector perpendicular to $\hat{T} = \frac{\hat{ı ı} + \hat{ȷ ȷ}}{\sqrt{2}},$ we know $\hat{N}$ will be either $\frac{\hat{ı ı} - \hat{ȷ ȷ}}{\sqrt{2}}$ or $\frac{- \hat{ı ı} + \hat{ȷ ȷ}}{\sqrt{2}} .$
Using Part (d) of the proof of Theorem 1.3.3:
$\begin{aligned} v (t) \times a (t) & = κ {(\frac{d s}{d t})}^{3} \hat{T} \times \hat{N} \\ (\hat{ı ı} + \hat{ȷ ȷ}) \times (\hat{ȷ ȷ}) & = 2^{- 3 / 2} {(\sqrt{2})}^{3} \frac{\hat{ı ı} + \hat{ȷ ȷ}}{\sqrt{2}} \times \hat{N} \\ \hat{k} & = \frac{1}{\sqrt{2}} (\hat{ı ı} + \hat{ȷ ȷ}) \times \hat{N} \\ \hat{N} & = \frac{- \hat{ı ı} + \hat{ȷ ȷ}}{\sqrt{2}} \end{aligned}$
So, the centre of our circle is at point $(0, 1) + ρ \hat{N} = (0, 1) + 2^{3 / 2} \frac{- \hat{ı ı} + \hat{ȷ ȷ}}{2^{1 / 2}} = (- 2, 3) .$ Then the equation of the circle is $(x + 2)^{2} + (y - 3)^{2} = 8 .$

1.3.11. (✳).

Solution.

(a) Think of

r (t) = (t, 1) - (\sin t, \cos t)

The

(t, 1)

part gives the position of the centre of the wheel at time

t .

The other part gives the position of the thumbtack with respect to the centre of the wheel. In particular,

at time $t = 0,$ $r (0) = (0, 0) .$ The thumbtack is on the ground (i.e. at $y = 0$ ).
At time $t = π,$ $r (π) = (π, 2) .$ The thumbtack is at its highest point (i.e. at $y = 2$ ) and is above the centre of the wheel at $x = π .$
At time $t = 2 π,$ $r (2 π) = (2 π, 0) .$ The thumbtack is back on the ground (i.e. at $y = 0$ ) and is below the centre of the wheel at $x = 2 π .$
At time $t = 3 π,$ $r (3 π) = (3 π, 2) .$ The thumbtack is again at its highest point (i.e. at $y = 2$ ) and is above the centre of the wheel at $x = 3 π .$
At time $t = 4 π,$ $r (4 π) = (4 π, 0) .$ The thumbtack is back on the ground (i.e. at $y = 0$ ) and is below the centre of the wheel at $x = 4 π .$

Here is a sketch of the curve.

(b) Since

\begin{aligned} r (t) & = (t - \sin t, 1 - \cos t) \\ v (t) = r^{'} (t) & = (1 - \cos t, \sin t) \\ \frac{d s}{d t} (t) = | v (t) | & = \sqrt{2 - 2 \cos t} \\ a (t) = v^{'} (t) & = (\sin t, \cos t) \\ v (t) \times a (t) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 1 - \cos t & \sin t & 0 \\ \sin t & \cos t & 0 \end{array}] = (\cos t - 1) \hat{k} \end{aligned}

the curvature

\begin{matrix} κ (t) = \frac{| v (t) \times a (t) |}{| v (t) |^{3}} = \frac{| \cos t - 1 |}{(2 - 2 \cos t)^{3 / 2}} = \frac{1}{2^{3 / 2} \sqrt{1 - \cos t}} \end{matrix}

t = π

ρ (π) = \frac{1}{κ (π)} = \frac{1}{1 / 2^{3 / 2} \sqrt{2}} = 4

(d) At time

π,

the tack is at

r (π) = (π, 2),

which is at the top of its trajectory. Looking at the sketch in part (a), we see that, at that time

\hat{N} (π) = - \hat{ȷ ȷ} .

So the osculating circle at time

t = π

has center

r (π) + ρ (π) \hat{N} (π) = (π, 2) + 4 (0, - 1) = (π, - 2)

and radius

ρ (π) = 4 .

So the equation of the osculating circle at time

π

(x - π)^{2} + (y + 2)^{2} = 16

1.3.12.

Solution.

The velocity vector is

v (θ) = x^{'} (θ) \hat{ı ı} + y^{'} (θ) \hat{ȷ ȷ} = \cos (\frac{1}{2} π θ^{2}) \hat{ı ı} + \sin (\frac{1}{2} π θ^{2}) \hat{ȷ ȷ}

Consequently the speed

\frac{d s}{d θ} (θ) = | v (θ) | = 1 ⟹ s (θ) = θ + s (0)

Since

s (θ)

is zero when

θ = 0,

we have

s (θ) = θ

and hence

\hat{T} (s) = v (s) = \cos (\frac{1}{2} π s^{2}) \hat{ı ı} + \sin (\frac{1}{2} π s^{2}) \hat{ȷ ȷ}

so that

κ (s) = | \frac{d \hat{T}}{d s} (s) | = | - π s \sin (\frac{1}{2} π s^{2}) \hat{ı ı} + π s \cos (\frac{1}{2} π s^{2}) \hat{ȷ ȷ} | = π s

1.3.13. (✳).

Solution.

The curve is

y = y (x) = \frac{x^{3}}{3} .

Since

y^{'} (x) = x^{2}

and

y^{″} (x) = 2 x,

the curvature is

\begin{aligned} κ (x) & = \frac{| \frac{d^{2} y}{d x^{2}} (x) |}{[1 + (\frac{d y}{d x} (x))^{2}]^{3 / 2}} = \frac{| 2 x |}{[1 + x^{4}]^{3 / 2}} \end{aligned}

We’d like to find the critical points of

κ (x),

but differentiating it looks messy. Since

κ (x)

has only nonnegative values, its maxima correspond the maxima of the function

κ^{2} (x) .

So, we find the critical points of

κ^{2} (x)

instead, to save ourselves some computational toil.

\begin{aligned} 0 & = \frac{d}{d x} κ (x)^{2} = \frac{d}{d x} \frac{4 x^{2}}{{(1 + x^{4})}^{3}} = \frac{8 x}{{(1 + x^{4})}^{3}} - 3 \frac{16 x^{5}}{{(1 + x^{4})}^{4}} \\ = \frac{8 x (1 + x^{4}) - 3 \times 16 x^{5}}{{(1 + x^{4})}^{4}} = \frac{8 x (1 - 5 x^{4})}{{(1 + x^{4})}^{4}} \end{aligned}

Note that

κ (0) = 0

and

κ (x) \to 0

x \to \pm \infty .

So the maximum occurs when

x = \pm 1 / \sqrt[4]{5} .

1.4 Curves in Three Dimensions

Exercises

1.4.1.

Solution.

\hat{T}

is tangent to the curve, while

\hat{N}

is perpendicular to it.

Using the right-hand rule and

\hat{B} = \hat{T} \times \hat{N},

\hat{B}

points out of the page (towards the reader).

To see this, point the fingers of your right hand in the direction of

\hat{T},

and curl them inwards until they are in the direction of

\hat{N} .

To do this, your thumb must be pointing towards you, not away from you. Your thumb shows the direction of

\hat{T} \times \hat{N} .

1.4.2.

Solution.

In this equation,

s

stands for arclength.

When we take a very small interval from

t

t + h,

the change in arclength

s (t + h) - s (t)

is approximately

| r (t + h) - r (t) |,

because our curve is approximated by a straight line. So,

\frac{s (t + h) - s (t)}{h} \approx \frac{| r (t + h) - r (t) |}{h},

leading to

\frac{d s}{d t} = | \frac{d r}{d t} | = | v (t) | .

The magnitude of velocity is speed; in this text we generally call this

v .

That is,

v = | v (t) | .

This leads to the potentially confusing (but standard) convention that

s

stands for arclength, while

v

stands for speed.

1.4.3.

Solution 1.

Curves

a

and

b

are the same curve, just parametrized differently (replace

t

with

- t

to convince yourself if the picture isn’t enough). So, they ought to have the same torsion.

As in Example 1.4.4, we imagine that the curve is the thread on a bolt. Take a look at your right hand. If your thumb is pointing up (corresponding to the

+ z

direction), and you’re looking at the tip of your thumb, your fingers curl anticlockwise. Imagine a screw has threads matching the curves

a

and

b,

and we turn it anticlockwise. The screw would move down — not in the same direction as our thumb. So these curves are not right-handed helices, so they have negative torsion.

The curve

c

sits entirely in a plane (the plane

x = 0

) so its torsion is zero everywhere.

Solution 2.

Here is the conventional computation for both

a (t)

and

b (t) .

(The upper sign is for

a

and the lower sign is for

b .

)

\begin{aligned} r (t) & = (\cos t, \mp 2 \sin t, \pm t / 2) \\ v (t) & = (- \sin t, \mp 2 \cos t, \pm 1 / 2) \\ a (t) & = (- \cos t, \pm 2 \sin t, 0) \\ v (t) \times a (t) & = (- \sin t, \mp \cos t / 2, \mp 2) \\ \frac{d a}{d t} (t) & = (\sin t, \pm 2 \cos t, 0) \\ v (t) \times a (t) \cdot \frac{d a}{d t} (t) & = - 1 \\ τ (t) = \frac{v (t) \times a (t) \cdot \frac{d a}{d t} (t)}{| v (t) \times a (t) |^{2}} & = - \frac{1}{\sin^{2} t + \frac{1}{4} \cos^{2} t + 4} < 0 \end{aligned}

1.4.4.

Solution.

(a) If

κ (s) \equiv 0,

then

\frac{d \hat{T}}{d s} = κ (s) \hat{N} (s) \equiv 0

so that

\hat{T}

is a constant. As a result

\frac{d r}{d s} (s) = \hat{T}

and

r (s) = s \hat{T} + r (0)

so that the curve is the straight line with direction vector

\hat{T}

that passes through

r (0) .

(b) If

τ (s) \equiv 0,

then

\frac{d \hat{B}}{d s} = - τ (s) \hat{N} (s) \equiv 0

so that

\hat{B}

is a constant. As

\hat{T} (s) ⊥ \hat{B},

\frac{d}{d s} (r (s) - r (0)) \cdot \hat{B} = \hat{T} (s) \cdot \hat{B} = 0

and

(r (s) - r (0)) \cdot \hat{B}

must be a constant. The constant must be zero (set

s = 0

), so

(r (s) - r (0)) \cdot \hat{B} = 0

and

r (s)

always lies in the plane through

r (0)

with normal vector

\hat{B} .

s

r_{c} (s) = r (s) + \frac{1}{κ (s)} \hat{N} (s)

Since

κ (s) = κ_{0}

is a constant and

τ (s) \equiv 0,

\frac{d}{d s} r_{c} (s) = \hat{T} (s) + \frac{1}{κ_{0}} [τ (s) \hat{B} - κ (s) \hat{T}] = \hat{T} (s) + \frac{1}{κ_{0}} [0 \hat{B} - κ_{0} \hat{T}] = 0

Thus

r_{c} (s) = r_{c}

is a constant and

| r (s) - r_{c} | = \frac{1}{κ_{0}}

lies on the sphere of radius

\frac{1}{κ_{0}}

centred on

r_{c} .

Since

τ (s) \equiv 0,

the curve also lies on a plane, so it is a circle.

1.4.5. (✳).

Solution.

(a), (b):

\hat{T}

points in the direction of the curve;

\hat{N}

is perpendicular to it, in the same plane, pointing towards the centre of curvature. Using the right-hand rule in the picture, we see

\hat{B}

is pointing to the left.

x = y

1.4.6. (✳).

Solution.

(a) As

\begin{aligned} r^{'} (t) & = (e^{t} + e^{- t}) \hat{ı ı} + (e^{t} - e^{- t}) \hat{ȷ ȷ} + 2 \hat{k} \\ \frac{d s}{d t} (t) & = | r^{'} (t) | = \sqrt{4 + 2 e^{2 t} + 2 e^{- 2 t}} = \sqrt{2} (e^{t} + e^{- t}) \\ r^{″} (t) & = (e^{t} - e^{- t}) \hat{ı ı} + (e^{t} + e^{- t}) \hat{ȷ ȷ} \\ r^{'} (t) \times r^{″} (t) & = - 2 (e^{t} + e^{- t}) \hat{ı ı} + 2 (e^{t} - e^{- t}) \hat{ȷ ȷ} + 4 \hat{k} \end{aligned}

the curvature

\begin{aligned} κ (t) & = \frac{| v (t) \times a (t) |}{(\frac{d s}{d t})^{3}} = \frac{2 \sqrt{4 + 2 e^{2 t} + 2 e^{- 2 t}}}{[4 + 2 e^{2 t} + 2 e^{- 2 t}]^{3 / 2}} = \frac{1}{2 + e^{2 t} + e^{- 2 t}} \end{aligned}

(b) The length of

C

between

r (0)

and

r (1)

\begin{aligned} \int_{0}^{1} \frac{d s}{d t} (t) d t & = \sqrt{2} \int_{0}^{1} (e^{t} + e^{- t}) d t = \sqrt{2} [e^{t} - e^{- t}]_{0}^{1} = \sqrt{2} [e - \frac{1}{e}] \end{aligned}

1.4.7.

Solution.

The point

(2, 4, 8)

occurs when

t = 2 .

\begin{aligned} v (t) & = (1, 2 t, 3 t^{2}) & v (2) & = (1, 4, 12) \\ a (t) & = (0, 2, 6 t) & a (2) & = (0, 2, 12) \\ \frac{d a}{d t} (t) & = (0, 0, 6) & \frac{d a}{d t} (2) & = (0, 0, 6) \\ v (2) \times a (2) & = (24, - 12, 2) \\ | v (2) \times a (2) | & = 2 \sqrt{181} \end{aligned}

Now, we use a formula for torsion:

\begin{aligned} τ (t) & = \frac{(v (t) \times a (t)) \cdot \frac{d a}{d t} (t)}{| v (t) \times a (t) |^{2}} \\ τ (2) & = \frac{(24, - 12, 2) \cdot (0, 0, 6)}{(2 \sqrt{181})^{2}} = \frac{3}{181} \end{aligned}

1.4.8.

Solution.

For the specified curve

\begin{aligned} r (t) & = t \hat{ı ı} + \frac{t^{2}}{2} \hat{ȷ ȷ} + \frac{t^{3}}{3} \hat{k} \\ v (t) = r^{'} (t) & = \hat{ı ı} + t \hat{ȷ ȷ} + t^{2} \hat{k} \\ a (t) = r^{″} (t) & = \hat{ȷ ȷ} + 2 t \hat{k} \\ v (t) \times a (t) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 1 & t & t^{2} \\ 0 & 1 & 2 t \end{array}] \\ = t^{2} \hat{ı ı} - 2 t \hat{ȷ ȷ} + \hat{k} \\ a^{'} (t) & = 2 \hat{k} \end{aligned}

From this, we read off

\begin{aligned} \hat{T} (t) & = \frac{v (t)}{| v (t) |} = \frac{\hat{ı ı} + t \hat{ȷ ȷ} + t^{2} \hat{k}}{\sqrt{1 + t^{2} + t^{4}}} \\ κ (t) & = \frac{| v (t) \times a (t) |}{| v (t) |^{3}} = \frac{\sqrt{1 + 4 t^{2} + t^{4}}}{[1 + t^{2} + t^{4}]^{3 / 2}} \\ \hat{B} (t) & = \frac{v (t) \times a (t)}{| v (t) \times a (t) |} = \frac{t^{2} \hat{ı ı} - 2 t \hat{ȷ ȷ} + \hat{k}}{\sqrt{1 + 4 t^{2} + t^{4}}} \\ \hat{N} (t) & = \hat{B} (t) \times \hat{T} (t) \\ = \frac{1}{\sqrt{1 + t^{2} + t^{4}} \sqrt{1 + 4 t^{2} + t^{4}}} det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ t^{2} & - 2 t & 1 \\ 1 & t & t^{2} \end{array}] \\ = \frac{- (t + 2 t^{3}) \hat{ı ı} + (1 - t^{4}) \hat{ȷ ȷ} + (2 t + t^{3}) \hat{k}}{\sqrt{1 + t^{2} + t^{4}} \sqrt{1 + 4 t^{2} + t^{4}}} \\ τ (t) & = \frac{(v (t) \times a (t)) \cdot a^{'} (t)}{| v (t) \times a (t) |^{2}} = \frac{2}{1 + 4 t^{2} + t^{4}} \end{aligned}

1.4.9.

Solution.

First, some preliminaries:

\begin{aligned} r (t) & = (t^{3}, t, e^{c t}) & r (5) & = (5^{3}, 5, e^{5 c}) \\ v (t) & = (3 t^{2}, 1, c e^{c t}) & v (5) & = (3 \cdot 5^{2}, 1, c e^{5 c}) \\ a (t) & = (6 t, 0, c^{2} e^{c t}) & a (5) & = (6 \cdot 5, 0, c^{2} e^{5 c}) \\ \frac{d a}{d t} (t) & = (6, 0, c^{3} e^{c t}) & \frac{d a}{d t} (5) & = (6, 0, c^{3} e^{5 c}) \\ v (5) \times a (5) & = (c^{2} e^{5 c}, 15 c e^{5 c} (2 - 5 c), - 30) \end{aligned}

Second, we figure out what value of

c

makes

τ (5) = 0 .

\begin{aligned} 0 & = τ (5) = \frac{(v (5) \times a (5)) \cdot \frac{d a}{d t} (5)}{| v (5) \times a (5) |^{2}} \\ 0 & = (v (5) \times a (5)) \cdot \frac{d a}{d t} (5) \\ = (c^{2} e^{5 c}, 15 c e^{5 c} (2 - 5 c), - 30) \cdot (6, 0, c^{3} e^{5 c}) \\ = 6 c^{2} e^{5 c} (1 - 5 c) \\ c & = 0 or c = \frac{1}{5} \end{aligned}

c = 0,

then

r (t) = (t^{3}, t, 1),

and so the entire curve is contained inside the plane

z = 1 .

(Its torsion is zero everywhere — not just at

t = 5 .

)

Consider the case

c = \frac{1}{5} .

When

t = 5,

our curve (and its osculating circle) passes through the point

r (5) = (5^{3}, 5, e) .

The normal vector to the plane of the osculating curve is the binormal vector

\hat{B} (5) = \frac{v (5) \times a (5)}{| v (5) \times a (5) |} .

Since we don’t need the normal vector to the plane to be a unit vector, we can take as the normal vector to the plane simply

v (5) \times a (5),

(e / 25, 3 e, - 30) .

Then, an equation of the plane containing the osculating circle is

(e / 25) x + (3 e) y - 30 z = - 10 e .

An equivalent equation for this plane is

(1 / 25) x + 3 y - (30 / e) z = - 10 .

1.4.10. (✳).

Solution.

(a) Since

r^{'} (t) = (2 t, 1, 3 t^{2}),

we have

r^{'} (1) = (2, 1, 3) .

So the normal plane must pass through

r (1) = (1, 1, 1)

and be perpendicular to

(2, 1, 3) .

The equation of the normal plane is then

2 (x - 1) + (y - 1) + 3 (z - 1) = 0 or 2 x + y + 3 z = 6

(b) As

\begin{aligned} v (t) = r^{'} (t) & = (2 t, 1, 3 t^{2}) & \frac{d s}{d t} & = \sqrt{1 + 4 t^{2} + 9 t^{4}} \\ a (t) = v^{'} (t) & = (2, 0, 6 t) & v (t) \times a (t) & = (6 t, - 6 t^{2}, - 2) \end{aligned}

the curvature

\begin{aligned} κ (t) & = \frac{| v (t) \times a (t) |}{(\frac{d s}{d t})^{3}} = \frac{2 \sqrt{1 + 9 t^{2} + 9 t^{4}}}{[1 + 4 t^{2} + 9 t^{4}]^{3 / 2}} \end{aligned}

1.4.11. (✳).

Solution.

First some preliminaries.

\begin{aligned} v (t) & = r^{'} (t) = - \sin t \hat{ı ı} + \cos t \hat{ȷ ȷ} + \hat{k} \\ a (t) & = r^{″} (t) = - \cos t \hat{ı ı} - \sin t \hat{ȷ ȷ} \end{aligned}

(a), (b) From

v (t)

we read off

\begin{matrix} \frac{d s}{d t} = | v (t) | = \sqrt{2} \end{matrix}

From

a (t) = \frac{d^{2} s}{d t^{2}} (t) \hat{T} (t) + κ (t) (\frac{d s}{d t} (t))^{2} \hat{N} (t),

and the fact that

\frac{d^{2} s}{d t^{2}} = 0,

we read off that

\begin{matrix} κ (t) = (\frac{d s}{d t} (t))^{- 2} | a | = \frac{1}{2} \hat{N} (t) = \frac{a}{| a |} = - \cos t \hat{ı ı} - \sin t \hat{ȷ ȷ} \end{matrix}

So the radius of curvature is

\frac{1}{κ} = 2

and the centre of curvature is

\begin{aligned} {[r (t) + \frac{1}{κ (t)} \hat{N} (t)]}_{t = \frac{π}{6}} & = {[(\cos t \hat{ı ı} + \sin t \hat{ȷ ȷ} + t \hat{k}) + 2 (- \cos t \hat{ı ı} - \sin t \hat{ȷ ȷ})]}_{t = \frac{π}{6}} \\ = {[- \cos t \hat{ı ı} - \sin t \hat{ȷ ȷ} + t \hat{k}]}_{t = \frac{π}{6}} \\ = - \frac{\sqrt{3}}{2} \hat{ı ı} - \frac{1}{2} \hat{ȷ ȷ} + \frac{π}{6} \hat{k} \end{aligned}

\begin{aligned} v (t) \times a (t) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - \sin t & \cos t & 1 \\ - \cos t & - \sin t & 0 \end{array}] = \sin t \hat{ı ı} - \cos t \hat{ȷ ȷ} + \hat{k} \\ | v (t) \times a (t) |^{2} & = 2 \end{aligned}

we read off

\begin{aligned} \hat{B} (t) & = \frac{v (t) \times a (t)}{| v (t) \times a (t) |} = \frac{1}{\sqrt{2}} \sin t \hat{ı ı} - \frac{1}{\sqrt{2}} \cos t \hat{ȷ ȷ} + \frac{1}{\sqrt{2}} \hat{k} \end{aligned}

so that

\begin{matrix} \hat{B} (\frac{π}{6}) = \frac{1}{2 \sqrt{2}} \hat{ı ı} - \frac{\sqrt{3}}{2 \sqrt{2}} \hat{ȷ ȷ} + \frac{1}{\sqrt{2}} \hat{k} \end{matrix}

1.4.12. (✳).

Solution.

(a) The velocity vector is

\begin{matrix} r^{'} (t) = (- \sin (t), \cos (t), 2 t) \end{matrix}

So a tangent vector at

t = π

T = (0, - 1, 2 π)

and a parametric form for the tangent line is

\begin{matrix} R (t) = r (π) + t T = (- 1, 0, π^{2}) + t (0, - 1, 2 π) \end{matrix}

(b) The speed is

\begin{matrix} \frac{d s}{d t} = | r^{'} (t) | = \sqrt{1 + 4 t^{2}} \end{matrix}

By Theorem 1.3.3, the tangential component of acceleration is

\begin{matrix} a_{T} (t) = \frac{d^{2} s}{d t^{2}} = \frac{d}{d t} \sqrt{1 + 4 t^{2}} = \frac{4 t}{\sqrt{1 + 4 t^{2}}} \end{matrix}

1.4.13. (✳).

Solution.

(a) The velocity vector of the particle at time

t

\begin{aligned} r^{'} (t) & = (\cos t - \cos t + t \sin t) \hat{ı ı} + (- \sin t + \sin t + t \cos t) \hat{ȷ ȷ} + 2 t \hat{k} \\ = t \sin t \hat{ı ı} + t \cos t \hat{ȷ ȷ} + 2 t \hat{k} \end{aligned}

so its speed at time

1 \leq t < \infty

\begin{aligned} \frac{d s}{d t} = | r^{'} (t) | & = \sqrt{t^{2} \sin^{2} t + t^{2} \cos^{2} t + 4 t^{2}} = \sqrt{5} t \end{aligned}

(b) The unit tangent at time

t

\hat{T} (T) = \frac{r^{'} (t)}{| r^{'} (t) |} = \frac{1}{\sqrt{5}} (\sin t \hat{ı ı} + \cos t \hat{ȷ ȷ} + 2 \hat{k})

So the tangential component of acceleration at time

t

a_{T} (t) = \frac{d^{2} s}{d t^{2}} (t) \hat{T} (t) = \sin t \hat{ı ı} + \cos t \hat{ȷ ȷ} + 2 \hat{k}

r^{″} (t) = \frac{d}{d t} r^{'} (t) = (\sin t + t \cos t) \hat{ı ı} + (\cos t - t \sin t) \hat{ȷ ȷ} + 2 \hat{k}

So the normal component of acceleration at time

t

a_{N} (t) = a (t) - a_{T} (t) = t \cos t \hat{ı ı} - t \sin t \hat{ȷ ȷ}

(d) Another formula for the normal component of acceleration is

κ (t) (\frac{d s}{d t} (t))^{2} \hat{N} (t) .

So the magnitude of the normal component of acceleration is

κ (t) (\frac{d s}{d t} (t))^{2}

and, by part (c),

\begin{matrix} κ (t) {(\frac{d s}{d t} (t))}^{2} = | t \cos t \hat{ı ı} - t \sin t \hat{ȷ ȷ} | = t \end{matrix}

Consequently, by part (a),

κ (t) = \frac{t}{{(\frac{d s}{d t} (t))}^{2}} = \frac{1}{5 t}

1.4.14. (✳).

Solution.

(a) If the point

(x, y, z)

is on the curve, it obeys both

z = x^{2} + y^{2}

and

z = 8 - 2 x

and hence is also obeys

x^{2} + y^{2} = 8 - 2 x or (x + 1)^{2} + y^{2} = 9

So the curve

C

is also the intersection of

(x + 1)^{2} + y^{2} = 9 and z = 8 - 2 x

(x + 1)^{2} + y^{2} = 9

is the circle of radius

3

centred on

(- 1, 0)

and can be parametrized by

x (θ) = - 1 + 3 \cos θ,

y (θ) = 3 \sin θ,

0 \leq θ \leq 2 π .

C

can be parametrized by

\begin{aligned} x (θ) & = - 1 + 3 \cos θ \\ y (θ) & = 3 \sin θ \\ z (θ) & = 8 - 2 x (θ) = 10 - 6 \cos θ \\ or r (θ) & = [- 1 + 3 \cos θ] \hat{ı ı} + 3 \sin θ \hat{ȷ ȷ} + [10 - 6 \cos θ] \hat{k} \end{aligned}

with

0 \leq θ < 2 π .

Remark: if we tried to parametrize the equation as

(x, y, z) = (x, \sqrt{8 - 2 x - x^{2}}, 8 - 2 x),

then we would miss the negative

y

-values.

(b) Note that

r (θ)

(2, 0, 4)

when

θ = 0 .

\begin{aligned} v (θ) & = r^{'} (θ) = - 3 \sin θ \hat{ı ı} + 3 \cos θ \hat{ȷ ȷ} + 6 \sin θ \hat{k} & v (0) & = 3 \hat{ȷ ȷ} \\ a (θ) & = v^{'} (θ) = - 3 \cos θ \hat{ı ı} - 3 \sin θ \hat{ȷ ȷ} + 6 \cos θ \hat{k} & a (0) & = - 3 \hat{ı ı} + 6 \hat{k} \end{aligned}

the unit tangent vector at

(2, 0, 4)

\begin{aligned} \hat{T} (0) & = \frac{v (0)}{| v (0) |} = \hat{ȷ ȷ} \end{aligned}

and, since

v (0) \times a (0) = 9 \hat{k} + 18 \hat{ı ı},

the unit binormal vector and curvature at

(2, 0, 4)

are

\begin{aligned} \hat{B} (0) & = \frac{v (0) \times a (0)}{| v (0) \times a (0) |} = \frac{2 \hat{ı ı} + \hat{k}}{\sqrt{5}} κ (0) = \frac{| v (0) \times a (0) |}{| v (0) |^{3}} = \frac{9 \sqrt{5}}{3^{3}} = \frac{\sqrt{5}}{3} \end{aligned}

and the unit normal vector

\hat{N}

(2, 0, 4)

\begin{matrix} \hat{N} (0) = \hat{B} (0) \times \hat{T} (0) = \frac{1}{\sqrt{5}} (2 \hat{ı ı} + \hat{k}) \times \hat{ȷ ȷ} = \frac{1}{\sqrt{5}} (2 \hat{k} - \hat{ı ı}) \end{matrix}

1.4.15. (✳).

Solution.

We have

\begin{aligned} v (t) & = r^{'} (t) = t^{2} \hat{ı ı} + \sqrt{2} t \hat{ȷ ȷ} + \hat{k} & | v (t) | & = \sqrt{t^{4} + 2 t^{2} + 1} = t^{2} + 1 \\ a (t) & = v^{'} (t) = 2 t \hat{ı ı} + \sqrt{2} \hat{ȷ ȷ} \end{aligned}

(a) The unit tangent vector is

\begin{aligned} \hat{T} (t) & = \frac{v (t)}{| v (t) |} = \frac{t^{2} \hat{ı ı} + \sqrt{2} t \hat{ȷ ȷ} + \hat{k}}{t^{2} + 1} \end{aligned}

(b) Since

\begin{aligned} v (t) \times a (t) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ t^{2} & \sqrt{2} t & 1 \\ 2 t & \sqrt{2} & 0 \end{array}] = - \sqrt{2} \hat{ı ı} + 2 t \hat{ȷ ȷ} - \sqrt{2} t^{2} \hat{k} \end{aligned}

The curvature is

\begin{matrix} κ (t) = \frac{| v (t) \times a (t) |}{| v (t) |^{3}} = \frac{\sqrt{2 + 4 t^{2} + 2 t^{4}}}{{(t^{2} + 1)}^{3}} = \frac{\sqrt{2}}{{(t^{2} + 1)}^{2}} \end{matrix}

r (2)

(\frac{8}{3}, 2 \sqrt{2}, 2) .

Solution 1: Since
$\begin{aligned} {\hat{T}}^{'} (t) & = \frac{2 t \hat{ı ı} + \sqrt{2} \hat{ȷ ȷ}}{t^{2} + 1} - 2 t \frac{t^{2} \hat{ı ı} + \sqrt{2} t \hat{ȷ ȷ} + \hat{k}}{{(t^{2} + 1)}^{2}} \\ {\hat{T}}^{'} (2) & = \frac{4 \hat{ı ı} + \sqrt{2} \hat{ȷ ȷ}}{5} - 4 \frac{4 \hat{ı ı} + 2 \sqrt{2} \hat{ȷ ȷ} + \hat{k}}{25} \\ = \frac{4 \hat{ı ı} - 3 \sqrt{2} \hat{ȷ ȷ} - 4 \hat{k}}{25} \\ | T^{'} (2) | & = \frac{5 \sqrt{2}}{25} \end{aligned}$
the principal normal vector $\hat{N}$ at $(\frac{8}{3}, 2 \sqrt{2}, 2)$ is
$\begin{aligned} \hat{N} (2) & = \frac{{\hat{T}}^{'} (2)}{| {\hat{T}}^{'} (2) |} = \frac{4 \hat{ı ı} - 3 \sqrt{2} \hat{ȷ ȷ} - 4 \hat{k}}{5 \sqrt{2}} \end{aligned}$
Solution 2: Perhaps we’d rather not differentiate $\hat{T} (t) .$
$\begin{aligned} \hat{B} & = \frac{v (t) \times a (t)}{| v (t) \times a (t) |} and \hat{N} = \hat{B} \times \hat{T} \end{aligned}$
Using our previous work:
$\begin{aligned} \hat{B} (2) & = \frac{v (2) \times a (2)}{| v (2) \times a (2) |} = \frac{- \sqrt{2} \hat{ı ı} + 4 \hat{ȷ ȷ} - 4 \sqrt{2} \hat{k}}{\sqrt{2 + 16 + 32}} \\ = \frac{1}{5} (- \hat{ı ı} + 2 \sqrt{2} \hat{ȷ ȷ} - 4 \hat{k}) \\ \hat{T} (2) & = \frac{1}{5} (4 \hat{ı ı} + 2 \sqrt{2} \hat{ȷ ȷ} + \hat{k}) \\ \hat{N} (2) & = \hat{B} (2) \times \hat{T} (2) = \frac{1}{5} (- \hat{ı ı} + 2 \sqrt{2} \hat{ȷ ȷ} - 4 \hat{k}) \times \frac{1}{5} (4 \hat{ı ı} + 2 \sqrt{2} \hat{ȷ ȷ} + \hat{k}) \\ = (\frac{2 \sqrt{2}}{5}) \hat{ı ı} + (- \frac{3}{5}) \hat{ȷ ȷ} + (- \frac{2 \sqrt{2}}{5}) \hat{k} \end{aligned}$

1.4.16. (✳).

Solution.

(a) The curve

x^{2} + y^{2} = 1

is a circle of radius

1 .

So we can parametrize it by

x (θ) = \cos θ,

y (θ) = \sin θ,

0 \leq θ < 2 π .

The

z

-coordinate of any point on the intersection is determined by

z = 1 - x - y .

So we can use

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + (1 - \cos θ - \sin θ) \hat{k} 0 \leq θ < 2 π

(b) As

\begin{aligned} v (θ) = r^{'} (θ) & = (- \sin θ, \cos θ, \sin θ - \cos θ) \\ a (θ) = v^{'} (θ) & = (- \cos θ, - \sin θ, \cos θ + \sin θ) \end{aligned}

we have

\begin{aligned} \frac{d s}{d θ} & = | r^{'} (θ) | = \sqrt{\sin^{2} θ + \cos^{2} θ + (\sin θ - \cos θ)^{2}} \\ = \sqrt{2 - 2 \sin θ \cos θ} \\ = \sqrt{2 - \sin (2 θ)} \\ v (θ) \times a (θ) & = (1, 1, 1) \end{aligned}

and the curvature

\begin{aligned} κ (θ) & = \frac{| v (θ) \times a (θ) |}{(\frac{d s}{d θ})^{3}} = \frac{\sqrt{3}}{[2 - \sin (2 θ)]^{3 / 2}} \end{aligned}

a maximum (minimum) when $2 - \sin (2 θ)$ is a minimum (maximum),
which is the case when $\sin (2 θ) = 1$ ( $\sin (2 θ) = - 1$ ),
which in turn is the case when $θ = \frac{π}{4}, \frac{5 π}{4}$ ( $θ = \frac{3 π}{4}, \frac{7 π}{4}$ ).

\begin{aligned} maximum curvature & = \frac{\sqrt{3}}{[2 - 1]^{3 / 2}} = \sqrt{3} & at \frac{\hat{ı ı}}{\sqrt{2}} + \frac{\hat{ȷ ȷ}}{\sqrt{2}} + (1 - \sqrt{2}) \hat{k} \\ and - \frac{\hat{ı ı}}{\sqrt{2}} - \frac{\hat{ȷ ȷ}}{\sqrt{2}} + (1 + \sqrt{2}) \hat{k} \\ minimum curvature & = \frac{\sqrt{3}}{[2 - (- 1)]^{3 / 2}} = \frac{1}{3} & at - \frac{\hat{ı ı}}{\sqrt{2}} + \frac{\hat{ȷ ȷ}}{\sqrt{2}} + \hat{k} \\ and \frac{\hat{ı ı}}{\sqrt{2}} - \frac{\hat{ȷ ȷ}}{\sqrt{2}} + \hat{k} \end{aligned}

1.4.17. (✳).

Solution.

For

r (t)

to be well-defined, we need

t > 0

(because of the

\ln t .

)

\begin{aligned} v (t) = r^{'} (t) & = 2 t \hat{ı ı} + 2 \hat{ȷ ȷ} + \frac{1}{t} \hat{k} & \frac{d s}{d t} & = \sqrt{4 t^{2} + 4 + \frac{1}{t^{2}}} = 2 t + \frac{1}{t} \end{aligned}

The unit tangent vector is

\begin{matrix} \hat{T} (t) = \frac{r^{'} (t)}{| r^{'} (t) |} = \frac{2 t \hat{ı ı} + 2 \hat{ȷ ȷ} + \frac{1}{t} \hat{k}}{2 t + \frac{1}{t}} = \frac{2 t^{2} \hat{ı ı} + 2 t \hat{ȷ ȷ} + \hat{k}}{2 t^{2} + 1} \end{matrix}

so, from §1.5,

\begin{aligned} \frac{d s}{d t} (t) κ (t) \hat{N} (t) & = {\hat{T}}^{'} (t) = \frac{4 t \hat{ı ı} + 2 \hat{ȷ ȷ}}{2 t^{2} + 1} - 4 t \frac{2 t^{2} \hat{ı ı} + 2 t \hat{ȷ ȷ} + \hat{k}}{{(2 t^{2} + 1)}^{2}} \\ = \frac{4 t \hat{ı ı} + (- 4 t^{2} + 2) \hat{ȷ ȷ} - 4 t \hat{k}}{{(2 t^{2} + 1)}^{2}} \\ = 2 \frac{2 t \hat{ı ı} - (2 t^{2} - 1) \hat{ȷ ȷ} - 2 t \hat{k}}{{(2 t^{2} + 1)}^{2}} \end{aligned}

Since the length of

2 t \hat{ı ı} - (2 t^{2} - 1) \hat{ȷ ȷ} - 2 t \hat{k}

\begin{aligned} \sqrt{4 t^{2} + (2 t^{2} - 1)^{2} + 4 t^{2}} & = \sqrt{8 t^{2} + 4 t^{4} - 4 t^{2} + 1} = \sqrt{4 t^{4} + 4 t^{2} + 1} \\ = \sqrt{(2 t^{2} + 1)^{2}} = 2 t^{2} + 1 \end{aligned}

we have

\begin{matrix} \hat{N} (t) = \frac{2 t \hat{ı ı} - (2 t^{2} - 1) \hat{ȷ ȷ} - 2 t \hat{k}}{2 t^{2} + 1} \end{matrix}

and

\begin{matrix} κ (t) = \frac{| {\hat{T}}^{'} (t) |}{\frac{d s}{d t} (t)} = \frac{\frac{2}{2 t^{2} + 1}}{2 t + \frac{1}{t}} = \frac{2 t}{{(2 t^{2} + 1)}^{2}} \end{matrix}

1.4.18. (✳).

Solution.

(a) Since

\begin{aligned} r (t) & = \hat{ı ı} + \frac{t^{2}}{2} \hat{ȷ ȷ} + \frac{t^{3}}{3} \hat{k} \\ r^{'} (t) & = t \hat{ȷ ȷ} + t^{2} \hat{k} \\ \frac{d s}{d t} (t) = | r^{'} (t) | & = \sqrt{t^{2} + t^{4}} = t \sqrt{1 + t^{2}} \end{aligned}

the length of the curve is

\begin{matrix} \int_{0}^{1} \frac{d s}{d t} (t) d t = \int_{0}^{1} t \sqrt{1 + t^{2}} d t = \frac{1}{3} {[1 + t^{2}]}^{3 / 2} |_{0}^{1} = \frac{1}{3} [2^{3 / 2} - 1] \end{matrix}

(b) For the specified curve

\begin{aligned} r (t) & = \cos (t) \hat{ı ı} + \sin (t) \hat{ȷ ȷ} + t \hat{k} \\ r^{'} (t) & = - \sin (t) \hat{ı ı} + \cos (t) \hat{ȷ ȷ} + 1 \hat{k} \\ \hat{T} (t) & = \frac{- \sin (t) \hat{ı ı} + \cos (t) \hat{ȷ ȷ} + 1 \hat{k}}{\sqrt{2}} \\ {\hat{T}}^{'} (t) & = \frac{- \cos (t) \hat{ı ı} - \sin (t) \hat{ȷ ȷ}}{\sqrt{2}} \\ {\hat{T}}^{'} (\frac{π}{4}) & = \frac{- \hat{ı ı} - \hat{ȷ ȷ}}{2} \\ \hat{N} (\frac{π}{4}) & = \frac{{\hat{T}}^{'} (\frac{π}{4})}{| {\hat{T}}^{'} (\frac{π}{4}) |} = \frac{- \hat{ı ı} - \hat{ȷ ȷ}}{\sqrt{2}} \end{aligned}

\begin{matrix} {\hat{T}}^{'} (t) = κ (t) \frac{d s}{d t} (t) \hat{N} (t) \end{matrix}

we have, by part (d),

\begin{aligned} κ (\frac{π}{4}) & = \frac{| {\hat{T}}^{'} (\frac{π}{4}) |}{| r^{'} (\frac{π}{4}) |} = \frac{1 / \sqrt{2}}{\sqrt{2}} = \frac{1}{2} \end{aligned}

1.4.19. (✳).

Solution.

(a), (b), (c) We have

\begin{aligned} r (t) & = (t + 2, 1 - t, t^{2} / 2) \\ v (t) = r^{'} (t) & = (1, - 1, t) \\ \frac{d s}{d t} (t) = | v (t) | & = \sqrt{2 + t^{2}} \\ a (t) = v^{'} (t) & = (0, 0, 1) \end{aligned}

(d) By §1.5, the curvature

\begin{aligned} κ (t) & = \frac{| v (t) \times a (t) |}{(\frac{d s}{d t} (t))^{3}} = \frac{| (- 1, - 1, 0) |}{{[2 + t^{2}]}^{3 / 2}} = \frac{\sqrt{2}}{{[2 + t^{2}]}^{3 / 2}} \end{aligned}

(e) Since

\frac{d s}{d t} (t) = \sqrt{2 + t^{2}},

we have

\frac{d^{2} s}{d t^{2}} (t) = \frac{t}{\sqrt{2 + t^{2}}}

and

\begin{aligned} (0, 0, 1) = a (t) & = \frac{d^{2} s}{d t^{2}} \hat{T} (t) + κ (t) (\frac{d s}{d t})^{2} \hat{N} (t) \\ = \frac{t}{\sqrt{2 + t^{2}}} \frac{(1, - 1, t)}{\sqrt{2 + t^{2}}} + \frac{\sqrt{2}}{{[2 + t^{2}]}^{3 / 2}} (\sqrt{2 + t^{2}})^{2} \hat{N} (t) \end{aligned}

\begin{matrix} \frac{\sqrt{2}}{\sqrt{2 + t^{2}}} \hat{N} (t) = (0, 0, 1) - \frac{(t, - t, t^{2})}{2 + t^{2}} = \frac{(- t, t, 2)}{2 + t^{2}} \end{matrix}

which implies

\begin{matrix} \hat{N} (t) = \frac{(- t, t, 2)}{\sqrt{2 (2 + t^{2})}} \end{matrix}

(f) At

t = 0

\begin{aligned} r (0) & = (2, 1, 0) \\ \hat{T} (0) & = \frac{(1, - 1, 0)}{\sqrt{2}} \\ \hat{N} (0) & = (0, 0, 1) \\ \hat{B} (0) & = \hat{T} (0) \times \hat{N} (0) = \frac{1}{\sqrt{2}} (1, - 1, 0) \times (0, 0, 1) = \frac{1}{\sqrt{2}} (- 1, - 1, 0) \end{aligned}

The osculating plane is the plane through

r (0)

which is perpendicular to

\hat{B} (0),

which is

\frac{1}{\sqrt{2}} (- 1, - 1, 0) \cdot {(x, y, z) - (2, 1, 0)} = 0 or x + y = 3

(g) The osculating circle has centre

r (0) + \frac{1}{κ (0)} \hat{N} (0) = (2, 1, 0) + \frac{1}{1 / 2} (0, 0, 1) = (2, 1, 2)

1.4.20. (✳).

Solution.

First some preliminary computations.

\begin{aligned} r (t) & = \frac{t^{3}}{3} \hat{ı ı} + \frac{t^{2}}{\sqrt{2}} \hat{ȷ ȷ} + t \hat{k} \\ r^{'} (t) & = t^{2} \hat{ı ı} + \sqrt{2} t \hat{ȷ ȷ} + \hat{k} \\ | r^{'} (t) | & = \sqrt{t^{4} + 2 t^{2} + 1} = t^{2} + 1 \\ r^{″} (t) & = 2 t \hat{ı ı} + \sqrt{2} \hat{ȷ ȷ} \\ r^{'} (t) \times r^{″} (t) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ t^{2} & \sqrt{2} t & 1 \\ 2 t & \sqrt{2} & 0 \end{array}] = - \sqrt{2} \hat{ı ı} + 2 t \hat{ȷ ȷ} - \sqrt{2} t^{2} \hat{k} \end{aligned}

(a) The unit tangent vector is

\begin{matrix} \hat{T} (t) = \frac{r^{'} (t)}{| r^{'} (t) |} = \frac{t^{2} \hat{ı ı} + \sqrt{2} t \hat{ȷ ȷ} + \hat{k}}{t^{2} + 1} \end{matrix}

(b) The curvature is (see §1.5)

\begin{aligned} κ (t) & = \frac{| r^{'} (t) \times r^{″} (t) |}{| r^{'} (t) |^{3}} = \frac{| - \sqrt{2} \hat{ı ı} + 2 t \hat{ȷ ȷ} - \sqrt{2} t^{2} \hat{k} |}{{(t^{2} + 1)}^{3}} = \frac{\sqrt{2 + 4 t^{2} + 2 t^{4}}}{{(t^{2} + 1)}^{3}} \\ = \frac{\sqrt{2}}{{(t^{2} + 1)}^{2}} \end{aligned}

t = 0

κ (0) = \sqrt{2}

For ease of computation, we’ll find

\hat{B}

first, then use it to find

\hat{N} .

(e) At

t = 0,

the binormal vector is (see §1.5)

\begin{matrix} \hat{B} (0) = \frac{r^{'} (0) \times r^{″} (0)}{| r^{'} (0) \times r^{″} (0) |} = \frac{- \sqrt{2} \hat{ı ı}}{\sqrt{2}} = - \hat{ı ı} \end{matrix}

(d) At

t = 0,

the principal normal vector is (see §1.5)

\begin{matrix} \hat{N} (0) = \hat{B} (0) \times \hat{T} (0) = - \hat{ı ı} \times \hat{k} = \hat{ȷ ȷ} \end{matrix}

1.4.21. (✳).

Solution.

The curve has

\begin{aligned} r (t) & = (t^{2}, t, t^{3}) \\ v (t) = r^{'} (t) & = (2 t, 1, 3 t^{2}) \\ a (t) = r^{″} (t) & = (2, 0, 6 t) \end{aligned}

(a) In particular, a (non unit) tangent vector at

r (- 1) = (1, - 1, - 1)

r^{'} (- 1) = (- 2, 1, 3) .

So the tangent line to the curve at

(1, - 1, - 1)

\begin{matrix} (x, y, z) - (1, - 1, - 1) = t (- 2, 1, 3) \end{matrix}

\begin{aligned} x & = 1 - 2 t \\ y & = - 1 + t \\ z & = - 1 + 3 t \end{aligned}

(b) At

r (1) = (1, 1, 1),

\begin{aligned} v (1) = r^{'} (1) & = (2, 1, 3) \\ a (1) = r^{″} (1) & = (2, 0, 6) \\ v (1) \times a (1) & = (6, - 6, - 2) \end{aligned}

So the unit binormal vector is

\hat{B} (1) = \frac{v (1) \times a (1)}{| v (1) \times a (1) |} = \frac{(3, - 3, - 1)}{| (3, - 3, - 1) |} = \frac{1}{\sqrt{19}} (3, - 3, - 1)

An equation for the osculating plane is

(3, - 3, - 1) \cdot (x - 1, y - 1, z - 1) = 0 or 3 x - 3 y - z = - 1

1.4.22. (✳).

Solution.

(a) For this curve

\begin{aligned} r^{'} (t) & = t \sin t \hat{ı ı} + t \cos t \hat{ȷ ȷ} + 2 t \hat{k} \\ \frac{d s}{d t} (t) = | r^{'} (t) | & = \sqrt{5} t \end{aligned}

so the length of the curve from

t = 0

t = π

\begin{matrix} \int_{0}^{π} \frac{d s}{d t} (t) d t = \sqrt{5} \int_{0}^{π} t d t = \frac{\sqrt{5} π^{2}}{2} \end{matrix}

(b) The unit tangent vector is

\hat{T} (t) = \frac{r^{'} (t)}{| r^{'} (t) |} = \frac{1}{\sqrt{5}} (\sin t \hat{ı ı} + \cos t \hat{ȷ ȷ} + 2 \hat{k})

so that

\begin{matrix} κ (t) \frac{d s}{d t} (t) \hat{N} (t) = \frac{d \hat{T}}{d t} (t) = \frac{1}{\sqrt{5}} (\cos t \hat{ı ı} - \sin t \hat{ȷ ȷ}) \end{matrix}

which implies that

\begin{matrix} κ (t) \overset{\frac{d s}{d t} (t)}{\overset{⏞}{\sqrt{5} t}} = \frac{1}{\sqrt{5}} | (\cos t \hat{ı ı} - \sin t \hat{ȷ ȷ}) | = \frac{1}{\sqrt{5}} ⟹ κ (t) = \frac{1}{5 t} \end{matrix}

1.4.23. (✳).

Solution.

(a) For the specified curve

\begin{aligned} r (t) & = (\frac{4 \sqrt{2}}{3} t^{3 / 2}, \frac{4 \sqrt{2}}{3} t^{3 / 2}, t (2 - t)) \\ v (t) & = (2 \sqrt{2} t^{1 / 2}, 2 \sqrt{2} t^{1 / 2}, 2 - 2 t) \\ | v | & = \sqrt{8 t + 8 t + 4 - 8 t + 4 t^{2}} = \sqrt{4 (1 + 2 t + t^{2})} = 2 (1 + t) \end{aligned}

The rocket is at

z = 0

when

t = 0

and when

t = 2 .

So the distance travelled is

\begin{matrix} \int_{0}^{2} | v (t) | d t = \int_{0}^{2} 2 (1 + t) d t = 2 {[t + \frac{t^{2}}{2}]}_{0}^{2} = 8 \end{matrix}

(b) The rocket is at its maximum height when

\frac{d z}{d t} = 2 - 2 t = 0 .

That is, when

t = 1 .

Its velocity then is

(2 \sqrt{2}, 2 \sqrt{2}, 0) .

A unit vector in this direction is

\hat{T} (1) = \frac{1}{\sqrt{2}} (1, 1, 0) .

That is the unit tangent vector.

At general

t,

the unit tangent is

\hat{T} (t) = \frac{v (t)}{| v (t) |} = \frac{(\sqrt{2} t^{1 / 2}, \sqrt{2} t^{1 / 2}, 1 - t)}{1 + t}

\begin{aligned} {\hat{T}}^{'} (t) & = \frac{(\sqrt{2} t^{- 1 / 2} / 2, \sqrt{2} t^{- 1 / 2} / 2, - 1)}{1 + t} - \frac{(\sqrt{2} t^{1 / 2}, \sqrt{2} t^{1 / 2}, 1 - t)}{(1 + t)^{2}} \\ {\hat{T}}^{'} (1) & = \frac{(\sqrt{2} / 2, \sqrt{2} / 2, - 1)}{2} - \frac{(\sqrt{2}, \sqrt{2}, 0)}{4} \\ = (0, 0, - \frac{1}{2}) \end{aligned}

So the principal unit normal vector is

\hat{N} (1) = \frac{{\hat{T}}^{'} (1)}{| {\hat{T}}^{'} (1) |} = (0, 0, - 1)

\begin{matrix} \frac{d \hat{T}}{d t} (1) = (0, 0, - \frac{1}{2}) \frac{d s}{d t} (1) = | v (1) | = 4 \end{matrix}

the curvature

\begin{aligned} κ (1) & = \frac{| {\hat{T}}^{'} (1) |}{| v (1) |} = \frac{1}{8} \end{aligned}

1.4.24. (✳).

Solution.

(a) For the specified curve

\begin{aligned} r (t) & = (\cos^{3} t, \sin^{3} t, 2 \sin^{2} t) \\ v (t) & = (- 3 \cos^{2} t \sin t, 3 \sin^{2} t \cos t, 4 \sin t \cos t) \\ = \sin t \cos t (- 3 \cos t, 3 \sin t, 4) \\ | v (t) | & = \sin t \cos t \sqrt{9 \cos^{2} t + 9 \sin^{2} t + 16} = 5 \sin t \cos t \end{aligned}

So the distance travelled is

\int_{0}^{π / 2} | v (t) | d t = \int_{0}^{π / 2} 5 \sin t \cos t d t = \frac{5}{2} \sin^{2} t |_{0}^{π / 2} = \frac{5}{2}

(b) Since

\begin{aligned} v (t) & = \sin t \cos t (- 3 \cos t, 3 \sin t, 4) & | v (t) | & = 5 \sin t \cos t \end{aligned}

we have

\begin{aligned} \hat{T} (t) & = \frac{v (t)}{| v (t) |} = \frac{1}{5} (- 3 \cos t, 3 \sin t, 4) & \hat{T} (\frac{π}{6}) & = \frac{1}{5} (- \frac{3}{2}, \frac{3 \sqrt{3}}{2}, 4) \\ {\hat{T}}^{'} (t) & = \frac{1}{5} (3 \sin t, 3 \cos t, 0) & {\hat{T}}^{'} (\frac{π}{6}) & = \frac{1}{5} (\frac{3 \sqrt{3}}{2}, \frac{3}{2}, 0) \\ = \frac{3}{10} (\sqrt{3}, 1, 0) \\ \hat{N} (\frac{π}{6}) & = \frac{{\hat{T}}^{'} (\frac{π}{6})}{| {\hat{T}}^{'} (\frac{π}{6}) |} = \frac{1}{2} (\sqrt{3}, 1, 0) & \hat{B} (\frac{π}{6}) & = \hat{T} (\frac{π}{6}) \times \hat{N} (\frac{π}{6}) \\ = \frac{1}{10} (- 4, 4 \sqrt{3}, - 6) \\ = \frac{1}{5} (- 2, 2 \sqrt{3}, - 3) \end{aligned}

1.4.25.

Solution.

(a) The curve

x^{2} + y^{2} = 1

is a circle of radius

1 .

So we can parametrize it by

x (θ) = \cos θ,

y (θ) = \sin θ,

0 \leq θ < 2 π .

The

z

-coordinate of any point on the intersection is determined by

z = x^{2} - y^{2} .

So we can use the parametrization

\begin{aligned} r (θ) & = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + [\cos^{2} θ - \sin^{2} θ] \hat{k} \\ = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + \cos (2 θ) \hat{k} 0 \leq θ < 2 π \end{aligned}

(b) Note that

r (θ) = (1 / \sqrt{2}, 1 / \sqrt{2}, 0)

when

θ = \frac{π}{4} .

For general

θ,

the velocity and acceleration are

\begin{aligned} v (θ) = r^{'} (θ) & = - \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ} - 2 \sin (2 θ) \hat{k} \\ a (θ) = v^{'} (θ) & = - \cos θ \hat{ı ı} - \sin θ \hat{ȷ ȷ} - 4 \cos (2 θ) \hat{k} \end{aligned}

In particular,

\begin{aligned} v (π / 4) & = - \frac{1}{\sqrt{2}} \hat{ı ı} + \frac{1}{\sqrt{2}} \hat{ȷ ȷ} - 2 \hat{k} \\ a (π / 4) & = - \frac{1}{\sqrt{2}} \hat{ı ı} - \frac{1}{\sqrt{2}} \hat{ȷ ȷ} \\ \frac{d s}{d θ} (π / 4) & = | v (π / 4) | = \sqrt{5} \\ v (π / 4) \times a (π / 4) & = - \sqrt{2} \hat{ı ı} + \sqrt{2} \hat{ȷ ȷ} + \hat{k} \\ | v (π / 4) \times a (π / 4) | & = \sqrt{5} \end{aligned}

So the curvature

κ (π / 4) = \frac{| v (π / 4) \times a (π / 4) |}{| v (π / 4) |^{3}} = \frac{1}{5}

C

(1 / \sqrt{2}, 1 / \sqrt{2}, 0)

\begin{matrix} \hat{B} (π / 4) = \frac{v (π / 4) \times a (π / 4)}{| v (π / 4) \times a (π / 4) |} = \frac{- \sqrt{2} \hat{ı ı} + \sqrt{2} \hat{ȷ ȷ} + \hat{k}}{\sqrt{5}} \end{matrix}

So the osculating plane to

C

(1 / \sqrt{2}, 1 / \sqrt{2}, 0)

\begin{aligned} (- \sqrt{2}, \sqrt{2}, 1) \cdot (x - 1 / \sqrt{2}, y - 1 / \sqrt{2}, z - 0) = 0 o r \\ z = \sqrt{2} x - \sqrt{2} y \end{aligned}

(d) From the computations in parts (b) and (c), we have

\begin{aligned} \hat{T} (π / 4) & = \frac{v (π / 4)}{| v (π / 4) |} = \frac{- 1 / \sqrt{2} \hat{ı ı} + 1 / \sqrt{2} \hat{ȷ ȷ} - 2 \hat{k}}{\sqrt{5}} \\ \hat{B} (π / 4) & = \frac{v (π / 4) \times a (π / 4)}{| v (π / 4) \times a (π / 4) |} = \frac{- \sqrt{2} \hat{ı ı} + \sqrt{2} \hat{ȷ ȷ} + \hat{k}}{\sqrt{5}} \\ \hat{N} (π / 4) & = \hat{B} (π / 4) \times \hat{T} (π / 4) = \frac{- \hat{ı ı} - \hat{ȷ ȷ}}{\sqrt{2}} \end{aligned}

So the osculating circle has radius

1 / κ (π / 4) = 5

and centre

\begin{aligned} r_{c} (π / 4) & = r (π / 4) + \frac{\hat{N} (π / 4)}{κ (π / 4)} = (1 / \sqrt{2}, 1 / \sqrt{2}, 0) - 5 (1 / \sqrt{2}, 1 / \sqrt{2}, 0) \\ = (- 2 \sqrt{2}, - 2 \sqrt{2}, 0) \end{aligned}

1.4.26. (✳).

Solution.

We’ll solve this problem twice, using two different strategies. (The second strategy will be much more efficient than the first one.) Both strategies use that

F = m a .

Since we are told that

m = 2,

we just have to find the acceleration

a

(1, 1, 1) .

Strategy 1: In the first strategy, we’ll find the position

r (t),

as a function of time

t

and then differentiate twice to get the acceleration

a (t) .

First we’ll find any old parametrization. We are told that, on the path, $z = x$ and $z = y^{2} .$ So let’s use $y$ as the parameter. Then $x = z = y^{2} .$ So the parametrization is $R (y) = y^{2} \hat{ı ı} + y \hat{ȷ ȷ} + y^{2} \hat{k} .$ (We’ll save the notation “ $r (t)$ ” for the parametrization with respect to time.)
Next we’ll reparametrize to get the time $t$ as the parameter. Since
$\begin{aligned} \frac{d R}{d y} & = 2 y \hat{ı ı} + \hat{ȷ ȷ} + 2 y \hat{k} \\ ⟹ \frac{d s}{d y} & = | 2 y \hat{ı ı} + \hat{ȷ ȷ} + 2 y \hat{k} | = \sqrt{1 + 8 y^{2}} \end{aligned}$
We are told that the speed $\frac{d s}{d t} = 3$ for all $t .$ So, choosing our zero point for time to coincide with our zero point for $s,$ we have $s = 3 t,$ or $t = s / 3$ so that
$\frac{d t}{d y} = \frac{1}{3} \sqrt{1 + 8 y^{2}}$
We could now integrate to get $t$ as a function of $y .$ But that looks quite messy. Fortunately we only need the acceleration at one point, namely $(1, 1, 1) .$ We’ll now see that that saves quite a bit of work. Pretend that we have integrated to get $t$ as a function of $y$ and call the answer $t (y) .$ Call the inverse function, which gives $y$ as a function of $t,$ $y (t) .$
We now have $r (t) = R (y (t)) .$ So, by the chain rule,
$\begin{aligned} r^{'} (t) & = R^{'} (y (t)) y^{'} (t) \\ r^{″} (t) & = R^{'} (y (t)) y^{″} (t) + R^{″} (y (t)) y^{'} (t)^{2} \end{aligned}$
We’re only interest in the time, call it $t_{0},$ at which $y (t_{0}) = 1 .$ The acceleration at time $t_{0}$ is
$\begin{aligned} r^{″} (t_{0}) & = R^{'} (y (t_{0})) y^{″} (t_{0}) + R^{″} (y (t_{0})) y^{'} (t_{0})^{2} \\ = R^{'} (1) y^{″} (t_{0}) + R^{″} (1) y^{'} (t_{0})^{2} \\ = [2 \hat{ı ı} + \hat{ȷ ȷ} + 2 \hat{k}] y^{″} (t_{0}) + [2 \hat{ı ı} + 2 \hat{k}] y^{'} (t_{0})^{2} \end{aligned}$
so we just have to find $y^{'} (t_{0})$ and $y^{″} (t_{0}) .$
We know that $\frac{d t}{d y} = \frac{1}{3} \sqrt{1 + 8 y^{2}} .$ So by the inverse function theorem
$\begin{aligned} \frac{d y}{d t} (t) & = \frac{3}{\sqrt{1 + 8 y (t)^{2}}} \\ \frac{d^{2} y}{d t^{2}} (t) & = - \frac{1}{2} \frac{3 (16 y (t) y^{'} (t))}{{[1 + 8 y (t)^{2}]}^{3 / 2}} \end{aligned}$
In particular
$\begin{aligned} y^{'} (t_{0}) & = \frac{3}{\sqrt{1 + 8 y (t_{0})^{2}}} = \frac{3}{\sqrt{1 + 8}} = 1 \\ y^{″} (t_{0}) & = - \frac{24 y (t_{0}) y^{'} (t_{0})}{{[1 + 8 y (t_{0})^{2}]}^{3 / 2}} = - \frac{24 \times 1 \times 1}{{(1 + 8)}^{3 / 2}} = - \frac{8}{9} \end{aligned}$
Finally, the force is
$\begin{aligned} 2 r^{″} (t_{0}) & = 2 [2 \hat{ı ı} + \hat{ȷ ȷ} + 2 \hat{k}] y^{″} (t_{0}) + 2 [2 \hat{ı ı} + 2 \hat{k}] y^{'} (t_{0})^{2} \\ = - \frac{16}{9} [2 \hat{ı ı} + \hat{ȷ ȷ} + 2 \hat{k}] + 2 [2 \hat{ı ı} + 2 \hat{k}] \\ = \frac{4}{9} \hat{ı ı} - \frac{16}{9} \hat{ȷ ȷ} + \frac{4}{9} \hat{k} \end{aligned}$

Strategy 2: The second strategy will be based on (see §1.5)

a = \frac{d^{2} s}{d t^{2}} \hat{T} + κ (\frac{d s}{d t})^{2} \hat{N}

In this problem, we are told that

\frac{d s}{d t} = 3

for all

t,

so that

\frac{d^{2} s}{d t^{2}} = 0

and

a = 9 κ \hat{N}

So we just have to find the curvature,

κ,

and unit normal,

\hat{N},

(1, 1, 1) .

We have already found one parametrization of the path in strategy 1, namely

R (y) = y^{2} \hat{ı ı} + y \hat{ȷ ȷ} + y^{2} \hat{k}

Note that

R (1) = (1, 1, 1) .

Since

\begin{aligned} R^{'} (y) & = 2 y \hat{ı ı} + \hat{ȷ ȷ} + 2 y \hat{k} \\ \hat{T} (y) & = \frac{R^{'} (y)}{| R^{'} (y) |} = \frac{2 y \hat{ı ı} + \hat{ȷ ȷ} + 2 y \hat{k}}{\sqrt{1 + 8 y^{2}}} \\ {\hat{T}}^{'} (y) & = \frac{2 \hat{ı ı} + 2 \hat{k}}{\sqrt{1 + 8 y^{2}}} - \frac{16 y}{2} \frac{2 y \hat{ı ı} + \hat{ȷ ȷ} + 2 y \hat{k}}{[1 + 8 y^{2}]^{3 / 2}} \\ {\hat{T}}^{'} (1) & = \frac{2 \hat{ı ı} + 2 \hat{k}}{3} - 8 \frac{2 \hat{ı ı} + \hat{ȷ ȷ} + 2 \hat{k}}{27} = \frac{2 \hat{ı ı} - 8 \hat{ȷ ȷ} + 2 \hat{k}}{27} \end{aligned}

we have (again see §1.5)

\begin{aligned} κ (1) & = \frac{| {\hat{T}}^{'} (1) |}{| R^{'} (1) |} \\ \hat{N} (1) & = \frac{{\hat{T}}^{'} (1)}{| {\hat{T}}^{'} (1) |} \\ F & = m a = 2 \times 9 κ (1) \hat{N} (1) = 18 \frac{{\hat{T}}^{'} (1)}{| R^{'} (1) |} = 18 \frac{2 \hat{ı ı} - 8 \hat{ȷ ȷ} + 2 \hat{k}}{27 \sqrt{1 + 8 \times 1^{2}}} \\ = \frac{4}{9} (\hat{ı ı} - 4 \hat{ȷ ȷ} + \hat{k}) \end{aligned}

1.4.27. (✳).

Solution.

(a) As

\begin{matrix} r (t) = 2 t \hat{ı ı} + t^{2} \hat{ȷ ȷ} + \sqrt{3} t^{2} \hat{k} \\ r^{'} (t) = 2 \hat{ı ı} + 2 t \hat{ȷ ȷ} + 2 \sqrt{3} t \hat{k} \end{matrix}

the unit tangent vector is

\begin{aligned} \hat{T} (t) & = \frac{\hat{ı ı} + t \hat{ȷ ȷ} + \sqrt{3} t \hat{k}}{| \hat{ı ı} + t \hat{ȷ ȷ} + \sqrt{3} t \hat{k} |} = \frac{\hat{ı ı} + t \hat{ȷ ȷ} + \sqrt{3} t \hat{k}}{\sqrt{1 + 4 t^{2}}} \end{aligned}

(b) Since

\begin{aligned} \frac{d \hat{T}}{d t} (t) & = \frac{\hat{ȷ ȷ} + \sqrt{3} \hat{k}}{\sqrt{1 + 4 t^{2}}} - 4 t \frac{\hat{ı ı} + t \hat{ȷ ȷ} + \sqrt{3} t \hat{k}}{{(1 + 4 t^{2})}^{3 / 2}} = \frac{- 4 t \hat{ı ı} + \hat{ȷ ȷ} + \sqrt{3} \hat{k}}{{(1 + 4 t^{2})}^{3 / 2}} \end{aligned}

the unit normal is

\begin{aligned} \hat{N} (t) & = \frac{- 4 t \hat{ı ı} + \hat{ȷ ȷ} + \sqrt{3} \hat{k}}{| - 4 t \hat{ı ı} + \hat{ȷ ȷ} + \sqrt{3} \hat{k} |} = \frac{- 4 t \hat{ı ı} + \hat{ȷ ȷ} + \sqrt{3} \hat{k}}{2 \sqrt{1 + 4 t^{2}}} \end{aligned}

\begin{aligned} \hat{B} (t) & = \hat{T} (t) \times \hat{N} (t) \\ = \frac{1}{2 (1 + 4 t^{2})} det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 1 & t & \sqrt{3} t \\ - 4 t & 1 & \sqrt{3} \end{array}] \\ = \frac{- \sqrt{3} (1 + 4 t^{2}) \hat{ȷ ȷ} + (1 + 4 t^{2}) \hat{k}}{2 (1 + 4 t^{2})} \\ = - \frac{\sqrt{3}}{2} \hat{ȷ ȷ} + \frac{1}{2} \hat{k} \end{aligned}

which is (3).

(d) The plane contains the point

r (0) = 0

and is perpendicular to the vector

- \frac{\sqrt{3}}{2} \hat{ȷ ȷ} + \frac{1}{2} \hat{k}

and so is

- \sqrt{3} y + z = 0

(e) The curvature is

\begin{aligned} κ (t) & = | \frac{d \hat{T}}{d t} (t) | / | \frac{d s}{d t} | = \frac{| - 4 t \hat{ı ı} + \hat{ȷ ȷ} + \sqrt{3} \hat{k} |}{{(1 + 4 t^{2})}^{3 / 2}} \frac{1}{| 2 \hat{ı ı} + 2 t \hat{ȷ ȷ} + 2 \sqrt{3} t \hat{k} |} \\ = \frac{\sqrt{4 + 16 t^{2}}}{{(1 + 4 t^{2})}^{3 / 2}} \frac{1}{2 \sqrt{1 + 4 t^{2}}} = \frac{1}{{(1 + 4 t^{2})}^{3 / 2}} \end{aligned}

(f), (g) The denominator

{(1 + 4 t^{2})}^{3 / 2}

κ (t)

is a minimum at

t = 0

and grows without bound as

| t |

increases. So the denominator never achieves a maximum. Consquently, the curvature

κ (t)

achieves its maximum value when

t = 0

and so at

r (0) = (0, 0, 0) .

The curvature never achieves a minimum.

(h) Since

\sqrt{3} v + w = 4 \hat{k}

and

v - \sqrt{3} w = 4 \hat{ȷ ȷ},

\hat{ı ı} = \frac{u}{2} \hat{ȷ ȷ} = \frac{v - \sqrt{3} w}{4} \hat{k} = \frac{\sqrt{3} v + w}{4}

Since

u = 2 \hat{ı ı}

and

v = \hat{ȷ ȷ} + \sqrt{3} \hat{k},

\begin{aligned} r (t) = t u + t^{2} v = a (t) u + b (t) v + c (t) w \\ with a (t) = t, b (t) = t^{2}, c (t) = 0 \end{aligned}

The curve

(a (t), b (t)) = (t, t^{2})

is the curve

y = x^{2} .

It is “curviest” at the origin, which is consistent with part (f). It becomes flatter and flatter as

| t |

increases, but never achieves “perfect flatness”, which is consistent with (g).

1.4.28. (✳).

Solution.

The three unit vectors

\hat{T},

\hat{N}

and

\hat{B}

are mutually perpendicular and form a right handed triple.

\hat{N} = \hat{B} \times \hat{T} \hat{N} \times \hat{T} = - \hat{B} \hat{B} \times \hat{N} = - \hat{T}

and

\begin{aligned} \frac{d \hat{N}}{d s} & = \frac{d \hat{B}}{d s} \times \hat{T} + \hat{B} \times \frac{d \hat{T}}{d s} = - τ \hat{N} \times \hat{T} + \hat{B} \times (κ \hat{N}) = τ \hat{B} - κ \hat{T} \end{aligned}

1.4.29. (✳).

Solution.

(a) Parametrizing the curve by

θ

gives

\begin{aligned} r (θ) & = (\sin (2 θ), 1 - \cos (2 θ), 2 \cos θ), \\ v = r^{'} (θ) & = (2 \cos (2 θ), 2 \sin (2 θ), - 2 \sin θ), \\ a = r^{″} (θ) & = (- 4 \sin (2 θ), 4 \cos (2 θ), - 2 \cos θ) . \end{aligned}

At the point

P,

we have

θ = π / 4,

giving instantaneous values

r = (1, 1, \sqrt{2}), v = (0, 2, - \sqrt{2}), v = | v | = \sqrt{6}, a = (- 4, 0, - \sqrt{2}) .

Hence

\hat{T} = \frac{v}{| v |} = \frac{1}{\sqrt{6}} (0, 2, - \sqrt{2}) .

Now

\hat{B} = \frac{v \times a}{| v \times a |} = \frac{1}{\sqrt{26}} (- \sqrt{2}, 2 \sqrt{2}, 4) = \frac{1}{\sqrt{13}} (- 1, 2, 2 \sqrt{2}),

since

v \times a = | \begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 0 & 2 & - \sqrt{2} \\ - 4 & 0 & - \sqrt{2} \end{matrix} | = (- 2 \sqrt{2}, 4 \sqrt{2}, 8), | v \times a | = \sqrt{104} = 2 \sqrt{26} .

This leads to

\begin{aligned} \hat{N} & = \hat{B} \times \hat{T} = \frac{1}{\sqrt{78}} | \begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - 1 & 2 & 2 \sqrt{2} \\ 0 & 2 & - \sqrt{2} \end{array} | = - \frac{1}{\sqrt{78}} (6 \sqrt{2}, \sqrt{2}, 2) \\ = - \frac{1}{\sqrt{39}} (6, 1, \sqrt{2}) . \end{aligned}

Finally,

κ = \frac{| v \times a |}{v^{3}} = \frac{2 \sqrt{26}}{{(\sqrt{6})}^{3}} = \frac{2 \sqrt{2} \sqrt{13}}{6 \sqrt{2} \sqrt{3}} = \frac{\sqrt{13}}{3 \sqrt{3}} = \frac{\sqrt{39}}{9} .

(b) Now parametrize the curve by time,

t,

and write

v = r^{'} (t),

v = | r^{'} (t) |

and

a = r^{″} (t) .

Note that in part (a) we used

v,

v

and

a

with different meanings. We use the dot product to extract the tangential and normal components of

a = \frac{d v}{d t} \hat{T} + v^{2} κ \hat{N} :

\begin{aligned} a \cdot \hat{T} & = (\frac{d v}{d t} \hat{T} + v^{2} κ \hat{N}) \cdot \hat{T} \\ = \frac{d v}{d t} \hat{T} \cdot \hat{T} + (v^{2} κ) \hat{N} \cdot \hat{T} \end{aligned}

Since $\hat{T}$ is a unit vector, $\hat{T} \cdot \hat{T} = ‖ \hat{T} ‖^{2} = 1;$ since $\hat{T}$ and $\hat{N}$ are perpendicular, $\hat{T} \cdot \hat{N} = 0 .$

\begin{aligned} = \frac{d v}{d t} \end{aligned}

This gives us a nice way to compute $\frac{d v}{d t},$ the rate of change of speed.

\begin{aligned} \frac{d v}{d t} & = a \cdot \hat{T} = (- 2, 3, - 2 \sqrt{2}) \cdot \frac{1}{\sqrt{6}} (0, 2, - \sqrt{2}) \\ = \frac{1}{\sqrt{6}} [0 + 6 + 4] = \frac{10}{\sqrt{6}} = \frac{5}{3} \sqrt{6} . \end{aligned}

Similarly,

a \cdot \hat{N} = v^{2} κ,

v^{2} = \frac{1}{κ} a \cdot \hat{N} = \frac{9}{\sqrt{39}} \frac{- 1}{\sqrt{39}} (- 2, 3, - 2 \sqrt{2}) \cdot (6, 1, \sqrt{2}) = \frac{9 \times 13}{39} = 3.

Hence

| v | = \sqrt{3};

since

v = | v |,

v = \sqrt{3} .

Then

v = | v | \hat{T} = v \hat{T} = \frac{\sqrt{3}}{\sqrt{6}} (0, 2, - \sqrt{2}) = (0, \sqrt{2}, - 1) .

1.4.30. (✳).

Solution.

(a) The position, velocity and acceleration are

\begin{aligned} r (t) & = (\cos t, \sin t, c \sin t) \\ v (t) = r^{'} (t) & = (- \sin t, \cos t, c \cos t) \\ a (t) = r^{″} (t) & = (- \cos t, - \sin t, - c \sin t) \end{aligned}

(b) The speed is

\begin{matrix} v (t) = | v (t) | = \sqrt{1 + c^{2} \cos^{2} t} \end{matrix}

\begin{matrix} \frac{d^{2} s}{d t^{2}} = \frac{d}{d t} \sqrt{1 + c^{2} \cos^{2} t} = \frac{- c^{2} \sin t \cos t}{\sqrt{1 + c^{2} \cos^{2} t}} \end{matrix}

(d)

y (t) = \sin t

and

z (t) = c \sin t

obey

z (t) = c y (t)

for all

t .

So the curve lies on the plane

z = c y .

1.4.31. (✳).

Solution.

(a) For the specified curve

r (π) = (- 4, 0, \frac{1}{4})

and

\begin{aligned} r (θ) & = (4 \cos θ, 2 \sin θ, \frac{1}{4} \cos (2 θ)) \\ v (θ) = r^{'} (θ) & = (- 4 \sin θ, 2 \cos θ, - \frac{1}{2} \sin (2 θ)) \\ a (θ) = r^{″} (θ) & = (- 4 \cos θ, - 2 \sin θ, - \cos (2 θ)) \\ v (π) & = (0, - 2, 0) \\ a (π) & = (4, 0, - 1) \\ v (π) \times a (π) & = (2, 0, 8) \end{aligned}

So the curvature at

θ = π

\begin{aligned} κ (π) & = \frac{| v (π) \times a (π) |}{| v (π) |^{3}} = \frac{| (2, 0, 8) |}{| (0, - 2, 0) |^{3}} = \frac{\sqrt{17}}{4} \end{aligned}

(b) The radius is

\frac{1}{κ (π)} = \frac{4}{\sqrt{17}}

R (t) = r (t^{2}) .

Then

\begin{aligned} R^{'} (t) & = 2 t r^{'} (t^{2}) \\ R^{″} (t) & = 2 r^{'} (t^{2}) + 4 t^{2} r^{″} (t^{2}) \end{aligned}

In particular,

\begin{aligned} R (\sqrt{π}) & = (- 4, 0, \frac{1}{4}) \\ R^{'} (\sqrt{π}) & = 2 \sqrt{π} v (π) = (0, - 4 \sqrt{π}, 0) \\ speed = | R^{'} (\sqrt{π}) | & = 4 \sqrt{π} \\ acceleration = R^{″} (\sqrt{π}) & = 2 v (π) + 4 π a (π) = (16 π, - 4, - 4 π) \end{aligned}

The normal component of the acceleration has magnitude

\begin{matrix} κ (\frac{d s}{d t})^{2} = \frac{\sqrt{17}}{4} (4 \sqrt{π})^{2} = 4 \sqrt{17} π \end{matrix}

1.6 Integrating Along a Curve

Exercises

1.6.1.

Solution.

We want to add up all the tiny pieces of arclength

d s

along a curve

C .

So, the integral would simply be

\int_{C} d s .

To see this another way, if we define

r = (x (t), y (t), z (t))

for

a \leq t \leq b

to be the equation of

C,

we could calculate the arclength as:

\begin{aligned} \int_{a}^{b} | r^{'} (t) | d t & = \int_{a}^{b} \sqrt{x^{'} (t)^{2} + y^{'} (t)^{2} + z^{'} (t)^{2}} d t \end{aligned}

This fits the form of Definition 1.6.1 with

f (x, y, z) = 1,

so we write it as a line integral as

\int_{C} 1 d s,

which is equivalent to

\int_{C} d s .

1.6.2.

Solution.

(a) The curve is

r (θ) = x (θ) \hat{ı ı} + y (θ) \hat{ȷ ȷ}

with

x (θ) = r (θ) \cos θ,

y (θ) = r (θ) \sin θ

and

θ_{1} \leq θ \leq θ_{2} .

On this curve

\begin{aligned} v (θ) = \frac{d r}{d θ} (θ) & = x^{'} (θ) \hat{ı ı} + y^{'} (θ) \hat{ȷ ȷ} \\ = [r^{'} (θ) \cos θ - r (θ) \sin θ] \hat{ı ı} + [r^{'} (θ) \sin θ + r (θ) \cos θ] \hat{ȷ ȷ} \\ ⟹ \frac{d s}{d θ} (θ) & = \sqrt{[r^{'} (θ) \cos θ - r (θ) \sin θ]^{2} + [r^{'} (θ) \sin θ + r (θ) \cos θ]^{2}} \\ = \sqrt{r^{'} (θ)^{2} + r (θ)^{2}} \end{aligned}

Hence

\begin{aligned} \int_{C} f (x, y) d s & = \int_{θ_{1}}^{θ_{2}} f (x (θ), y (θ)) \frac{d s}{d θ} d θ \\ = \int_{θ_{1}}^{θ_{2}} f (r (θ) \cos θ, r (θ) \sin θ) \sqrt{r (θ)^{2} + {(\frac{d r}{d θ} (θ))}^{2}} d θ \end{aligned}

(b) In this case

f (x, y) = 1,

r (θ) = 1 + \cos θ,

θ_{1} = 0

and

θ_{2} = 2 π,

\begin{aligned} \int_{C} d s & = \int_{0}^{2 π} \sqrt{[1 + \cos θ]^{2} + [- \sin θ]^{2}} d θ = \int_{0}^{2 π} \sqrt{2 (1 + \cos θ)} d θ \\ = \int_{0}^{2 π} \sqrt{4 \cos^{2} \frac{θ}{2}} d θ = 2 \int_{0}^{2 π} | \cos \frac{θ}{2} | d θ = 4 \int_{0}^{π} \cos \frac{θ}{2} d θ \\ = 8 \sin \frac{θ}{2} |_{0}^{π} = 8 \end{aligned}

1.6.3.

Solution.

Following Definition 1.6.1:

\begin{aligned} \int_{C} (\frac{x y}{z}) d s & = \int_{1}^{2} (\frac{\frac{2}{3} t^{3} \cdot \sqrt{3} t^{2}}{3 t}) \sqrt{(2 t^{2})^{2} + (2 \sqrt{3} t)^{2} + (3)^{2}} d t \\ = \int_{1}^{2} (\frac{2}{3 \sqrt{3}} t^{4}) (2 t^{2} + 3) d t \\ = \frac{4}{21 \sqrt{3}} (2^{7} - 1) + \frac{2}{5 \sqrt{3}} (2^{5} - 1) \end{aligned}

1.6.4.

Solution.

We parametrize the unit circle as

(\cos t, \sin t),

0 \leq t \leq 2 π .

A tiny slice of the hoop with length

d s

has mass

(x^{2} kg / m) (d s m) = x^{2} d s kg .

So, the entire hoop has mass:

\begin{aligned} \int_{C} x^{2} d s & = \int_{0}^{2 π} \cos^{2} t \sqrt{(- \sin t)^{2} + (\cos t)^{2}} d t = \int_{0}^{2 π} \cos^{2} t d t \\ = \int_{0}^{2 π} \frac{1 + \cos (2 t)}{2} d t = {[\frac{t}{2} + \frac{\sin (2 t)}{4}]}_{0}^{2 π} = π kg \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} t d t,

see Example 2.4.4.

1.6.5.

Solution.

To parametrize

C,

we note the vector between the two points is

(2 - 1, 4 - 2, 5 - 3) = (1, 2, 2) .

So, the line is

(1, 2, 3) + t (1, 2, 2)

for

0 \leq t \leq 1 .

That is,

x (t) = 1 + t,

y (t) = 2 + 2 t,

and

z (t) = 3 + 2 t .

\begin{aligned} \int_{C} (x y + z) d s & = \int_{0}^{1} ((1 + t) (2 + 2 t) + (3 + 2 t)) \sqrt{1^{1} + 2^{2} + 2^{2}} d t \\ = \int_{0}^{1} 3 (5 + 6 t + 2 t^{2}) d t = 26 \end{aligned}

1.6.6.

Solution.

(a) In this case

r (t) = t \hat{ı ı} + t^{2} \hat{ȷ ȷ},

so that

v (t) = \frac{d r}{d t} (t) = \hat{ı ı} + 2 t \hat{ȷ ȷ}

and

\frac{d s}{d t} = \sqrt{1 + 4 t^{2}} .

Hence

\begin{aligned} \int_{C} f (x, y, z) d s & = \int_{0}^{1} x (t) \cos z (t) \frac{d s}{d t} d t = \int_{0}^{1} t (\cos 0) \sqrt{1 + 4 t^{2}} d t \\ = \frac{1}{8} \frac{{(1 + 4 t^{2})}^{3 / 2}}{3 / 2} |_{0}^{1} = \frac{5^{3 / 2} - 1}{12} \end{aligned}

(b) In this case

r (t) = (t, \frac{2}{3} t^{3 / 2}, t),

so that

v (t) = \frac{d r}{d t} (t) = (1, t^{1 / 2}, 1)

and

\frac{d s}{d t} = \sqrt{2 + t} .

Hence

\begin{aligned} \int_{C} f (x, y, z) d s & = \int_{1}^{2} \frac{x (t) + y (t)}{y (t) + z (t)} \frac{d s}{d t} d t = \int_{1}^{2} \frac{t + \frac{2}{3} t^{3 / 2}}{\frac{2}{3} t^{3 / 2} + t} \sqrt{2 + t} d t \\ = \frac{(2 + t)^{3 / 2}}{3 / 2} |_{1}^{2} = \frac{8 - 3^{3 / 2}}{3 / 2} \end{aligned}

1.6.7.

Solution.

In the figure below, we construct a triangle with

θ = arcsec t;

the hypotenuse has length

t,

while the side adjacent to

θ

has length 1. By the Pythagorean Theorem, the remaining side has length

\sqrt{t^{2} - 1},

\sin θ = \sin (arcsec t) = \frac{\sqrt{t^{2} - 1}}{t} .

Remember

\frac{d}{d t} {\ln t} = \frac{1}{t}

and

\frac{d}{d t} {arcsec t} = \frac{1}{| t | \sqrt{t^{2} - 1}} .

In our range,

1 \leq t \leq \sqrt{2},

we have

| t | = t .

\begin{aligned} \int_{C} \sin x d s & = \int_{1}^{\sqrt{2}} \sin (arcsec t) \sqrt{{(\frac{1}{t \sqrt{t^{2} - 1}})}^{2} + {(\frac{1}{t})}^{2}} d t \\ = \int_{1}^{\sqrt{2}} \frac{\sqrt{t^{2} - 1}}{t} \sqrt{\frac{1}{t^{2} (t^{2} - 1)} + \frac{1}{t^{2}}} d t \\ = \int_{1}^{\sqrt{2}} \frac{1}{t} d t = \frac{1}{2} \ln 2 \end{aligned}

1.6.8. (✳).

Solution.

You may have noticed that the orientation of (i.e. the direction of motion along) the curve

C

was not specified in the question. That’s because the value of the integral is independent of the orientation of

C :

At each point of $C,$ $F (x, y) = x y^{2} \hat{ı ı} + y e^{x} \hat{ȷ ȷ},$ independent of the orientation.
At each point of $C,$ $\hat{n}$ is the outward pointing normal, independent of the orientation.
$d s > 0,$ independent of the orientation.

Here is a sketch of the rectangle

R .

We have arbitrarily chosen to traverse

C

in the counterclockwise direction. The boundary of

R

consists of four line segments.

$L_{1}$ from $(0, - 1)$ to $(3, - 1),$ with $\hat{n} = - \hat{ȷ ȷ}$
$L_{2}$ from $(3, - 1)$ to $(3, 1),$ with $\hat{n} = \hat{ı ı}$
$L_{3}$ from $(3, 1)$ to $(0, 1),$ with $\hat{n} = \hat{ȷ ȷ}$
$L_{4}$ from $(0, 1)$ to $(0, - 1),$ with $\hat{n} = - \hat{ı ı}$

\begin{aligned} \int_{C} F \cdot \hat{n} d s & = \int_{L_{1}} F \cdot (- \hat{ȷ ȷ}) d s + \int_{L_{2}} F \cdot \hat{ı ı} d s + \int_{L_{3}} F \cdot (\hat{ȷ ȷ}) d s + \int_{L_{4}} F \cdot (- \hat{ı ı}) d s \\ = \int_{0}^{3} - \overset{y}{\overset{⏞}{(- 1)}} e^{x} d x + \int_{- 1}^{1} \overset{x}{\overset{⏞}{(3)}} y^{2} d y + \int_{3}^{0} \overset{y}{\overset{⏞}{(1)}} e^{x} \overset{d s}{\overset{⏞}{(- d x)}} \\ + \int_{1}^{- 1} \overset{x}{\overset{⏞}{(0)}} y^{2} \overset{d s}{\overset{⏞}{(- d y)}} \\ = [e^{3} - 1] + [1^{3} - (- 1)^{3}] + [e^{3} - 1] + 0 \\ = 2 e^{3} \end{aligned}

The trickiest part of this computation is getting

d s

correct on

L_{3}

and

L_{4}

(remembering that

d s

is the arc length traveled and so is positive, while

d x < 0

L_{3}

and

d y < 0

L_{4}

). To make a more detailed computation of

\int_{L_{3}} F \cdot (\hat{ȷ ȷ}) d s,

parametrize

L_{3}

r (t) = (3, 1) + t {(0, 1) - (3, 1)} = (3 - 3 t, 1) 0 \leq t \leq 1

so that

r (0) = (3, 1)

is the initial point of

L_{3}

and

r (1) = (0, 1)

is the final point of

L_{3} .

Then

r^{'} (t) = (- 3, 0) \frac{d s}{d t} (t) = | r^{'} (t) | = 3

and

\begin{aligned} \int_{L_{3}} F \cdot \hat{ȷ ȷ} d s & = \int_{0}^{1} F (r (t)) \cdot \hat{ȷ ȷ} \frac{d s}{d t} (t) d t = \int_{0}^{1} \overset{y (t) e^{x (t)}}{\overset{⏞}{e^{3 - 3 t}}} \overset{\frac{d s}{d t} (t)}{\overset{⏞}{3}} d t \\ = - e^{3 - 3 t} |_{0}^{1} = e^{3} - 1 \end{aligned}

As a check, we’ll now verify that we get the same answer for

\int_{L_{3}} F \cdot (\hat{ȷ ȷ}) d s,

when we parametrize

L_{3}

r (t) = (0, 1) + t {(3, 1) - (0, 1)} = (3 t, 1) 0 \leq t \leq 1

so that, this time,

r (0) = (0, 1)

is the initial point of

L_{3}

and

r (1) = (3, 1)

is the final point of

L_{3} .

This time

r^{'} (t) = (3, 0) \frac{d s}{d t} (t) = | r^{'} (t) | = 3

and

\int_{L_{3}} F \cdot \hat{ȷ ȷ} d s = \int_{0}^{1} F (r (t)) \cdot \hat{ȷ ȷ} \frac{d s}{d t} (t) d t = \int_{0}^{1} \overset{y (t) e^{x (t)}}{\overset{⏞}{e^{3 t}}} \overset{\frac{d s}{d t} (t)}{\overset{⏞}{3}} d t = e^{3 t} |_{0}^{1} = e^{3} - 1

So both orientations of

L_{3}

give the same value of

\int_{L_{3}} F \cdot (\hat{ȷ ȷ}) d s .

1.6.9. (✳).

Solution.

(a) Since

r (t) = t \cos t \hat{ı ı} + t \sin t \hat{ȷ ȷ} + t^{2} \hat{k}

\begin{aligned} r^{'} (t) & = (\cos t - t \sin t) \hat{ı ı} + (\sin t + t \cos t) \hat{ȷ ȷ} + 2 t \hat{k} \\ \frac{d s}{d t} & = | r^{'} (t) | = \sqrt{(\cos t - t \sin t)^{2} + (\sin t + t \cos t)^{2} + (2 t)^{2}} \\ = \sqrt{1 + 5 t^{2}} \\ r^{'} (π) & = - \hat{ı ı} - π \hat{ȷ ȷ} + 2 π \hat{k} \\ \hat{T} (π) & = \frac{r^{'} (t)}{| r^{'} (t) |} = \frac{1}{\sqrt{1 + 5 π^{2}}} (- \hat{ı ı} - π \hat{ȷ ȷ} + 2 π \hat{k}) \end{aligned}

(b)

\begin{aligned} \int_{C} \sqrt{x^{2} + y^{2}} d s & = \int_{0}^{π} \sqrt{x^{2} (t) + y^{2} (t)} \frac{d s}{d t} d t = \int_{0}^{π} t \sqrt{1 + 5 t^{2}} d t \\ = {[\frac{1}{15} (1 + 5 t^{2})^{3 / 2}]}_{0}^{π} = \frac{1}{15} [(1 + 5 π^{2})^{3 / 2} - 1] \end{aligned}

t,

the coordinates

x (t) = t \cos t,

y (t) = t \sin t,

z (t) = t^{2}

obey

x (t)^{2} + y (t)^{2} = t^{2} = z (t)

and so the curve lies on

z = x^{2} + y^{2} .

(d) First concentrate on

(x (t), y (t)) .

t

runs from

0

π,

the curve

(r \cos t, r \sin t)

sweeps out half of a circle of radius

r .

Our

(x (t), y (t))

does something similar, but the radius

r = t

increases from

0

π .

Thus our

(x (t), y (t))

sweeps out the beginning of a spiral. At the same time

z (t)

increases from

0

π^{2} .

So the curve

C

looks like

1.6.10.

Solution.

We use the centre of mass formulae

\bar{x} = \frac{\int_{C} x ρ d s}{\int_{C} ρ d s},

etc. To make the working clearer, we’ll break these calculations into several steps.

\begin{aligned} x (t) & = t + \frac{1}{2} t^{2} & x^{'} (t) & = 1 + t \\ y (t) & = t - \frac{1}{2} t^{2} & y^{'} (t) & = 1 - t \\ z (t) & = \frac{4}{3} t^{3 / 2} & z^{'} (t) & = 2 \sqrt{t} \end{aligned}

\begin{aligned} \sqrt{x^{'} (t)^{2} + y^{'} (t)^{2} + z^{'} (t)^{2}} & = \sqrt{1 + 2 t + t^{2} + 1 - 2 t + t^{2} + 4 t} \\ = \sqrt{2 (t^{2} + 2 t + 1)} = \sqrt{2} (t + 1) \\ ρ (x (t), y (t), z (t)) & = \frac{x (t) + y (t)}{2} = \frac{(t + t^{2} / 2) + (t - t^{2} / 2)}{2} = t \end{aligned}

\begin{aligned} \int_{C} ρ d s & = \int_{0}^{4} t \sqrt{2} (t + 1) d t = \frac{2^{3} \cdot 11 \sqrt{2}}{3} \\ \int_{C} x ρ d s & = \int_{0}^{4} (t + \frac{1}{2} t^{2}) t \sqrt{2} (t + 1) d t = \sqrt{2} \int_{0}^{2^{2}} (\frac{t^{4}}{2} + \frac{3}{2} t^{3} + t^{2}) d t \\ = \sqrt{2} (\frac{2^{9}}{5} + 3 (2^{5}) + \frac{2^{6}}{3}) = \frac{2^{5} \cdot 103 \sqrt{2}}{15} \\ \int_{C} y ρ d s & = \int_{0}^{4} (t - \frac{1}{2} t^{2}) t \sqrt{2} (t + 1) d t = \sqrt{2} \int_{0}^{2^{2}} (- \frac{t^{4}}{2} + \frac{t^{3}}{2} + t^{2}) d t \\ = \sqrt{2} (- \frac{2^{9}}{5} + 2^{5} + \frac{2^{6}}{3}) = - \frac{2^{5} \cdot 23 \sqrt{2}}{15} \\ \int_{C} z ρ d s & = \int_{0}^{4} (\frac{4}{3} t^{3 / 2}) t \sqrt{2} (t + 1) d t = \frac{4 \sqrt{2}}{3} \int_{0}^{2^{2}} (t^{7 / 2} + t^{5 / 2}) d t \\ = \frac{4 \sqrt{2}}{3} (\frac{2^{10}}{9} + \frac{2^{8}}{7}) = \frac{2^{10} \cdot 37 \sqrt{2}}{7 \cdot 3^{3}} \\ \overset{―}{x} & = \frac{\int x ρ d s}{\int ρ d s} = \frac{\frac{2^{5} \cdot 103 \sqrt{2}}{15}}{\frac{2^{3} \cdot 11 \sqrt{2}}{3}} = \frac{412}{55} \approx 7.5 \\ \overset{―}{y} & = \frac{\int y ρ d s}{\int ρ d s} = \frac{- \frac{2^{5} \cdot 23 \sqrt{2}}{15}}{\frac{2^{3} \cdot 11 \sqrt{2}}{3}} = - \frac{92}{55} \approx - 1.7 \\ \overset{―}{z} & = \frac{\int z ρ d s}{\int ρ d s} = \frac{\frac{2^{10} \cdot 37 \sqrt{2}}{7 \cdot 3^{3}}}{\frac{2^{3} \cdot 11 \sqrt{2}}{3}} = \frac{4736}{693} \approx 6.8 \end{aligned}

After these long calculations, it’s nice to do a sanity check. Using

0 \leq t \leq 4,

we see our wire takes up space in the following intervals:

0 \leq x \leq 12,

- 4 \leq y \leq 1 / 2,

and

0 \leq z \leq 32 / 3 .

The coordinates of our centre of mass all fall in these intervals, which doesn’t guarantee our answer is correct, but it is a nice sign. If, say

\overset{―}{x}

had been negative, or

\overset{―}{z}

were greater than 11, we would have known there was something wrong.

1.7 Sliding on a Curve
1.7.4 Exercises

1.7.4.1.

Solution.

We don’t have enough information to gauge the size of the vectors, but we can figure out their direction. Gravity pulls straight down, so the vector

- m g \hat{ȷ ȷ}

points straight down. The normal force will be normal to the curve.

1.7.4.2.

Solution.

This equation stems from

F = m a .

In that equation,

a

is acceleration — the second derivative of position with respect to time. So,

v

is the derivative of position with respect to time.

We previously used

v

as the derivative of position with respect to the parameter we use to define our position — which was often called

t,

but was not the necessarily time. So this is a good point to keep straight.

1.7.4.3.

Solution 1.

For large, negative values of

x,

the wire is closer and closer to a vertical line. If the bead were sliding down a vertical wire, it could do so without even touching the wire, so the force exerted on the bead would be zero. As

x

approaches 0 from the left, the wire approximates a horizontal line. If the bead were sitting on a horizontal line, the wire would be pushing up to counter gravity. So, we imagine the magnitude of the force exerted by the wire might increase as

x

increases. That is,

\frac{d W}{d x} > 0 .

Solution 2.

The net force exerted on the bead is

\begin{aligned} F = m a & = W \hat{N} - m g \hat{ȷ ȷ} \end{aligned}

We dot both sides with $\hat{N} .$

\begin{aligned} W \hat{N} \cdot \hat{N} - m g \hat{ȷ ȷ} \cdot \hat{N} & = m a \cdot \hat{N} \end{aligned}

Using the equation $a (t) = \frac{d^{2} s}{d t^{2}} \hat{T} + κ {(\frac{d s}{d t})}^{2} \hat{N},$

\begin{aligned} W - m g \hat{ȷ ȷ} \cdot \hat{N} & = m κ {(\frac{d s}{d t})}^{2} \\ W & = m g \hat{n} \cdot \hat{N} + m κ {(\frac{d s}{d t})}^{2} \\ = m g \cos θ + m κ {(\frac{d s}{d t})}^{2} \end{aligned}

where

θ

is the angle between

\hat{ȷ ȷ}

and

\hat{N} .

x

moves from a highly negative number to zero,

θ

moves from nearly

π / 2

to nearly

0 .

Therefore

\cos θ

increases from nearly zero to nearly one. Then

m g \cos θ

is increasing.

Furthermore, as

x

increases, we see from the picture that the curvature

κ

increases, and speed

\frac{d s}{d t}

increases as well (kinetic energy is increasing as potential energy decreases).

So,

\frac{d W}{d x} > 0 .

1.7.4.4.

Solution.

Equation 1.7.1 defines

E = \frac{1}{2} m | v |^{2} + m g y .

The skater reaches their highest point when

| v | = 0,

so when

y = \frac{E}{m g} .

This is the same equation as a sufficiently large circular culvert: it’s the height where all the kinetic energy has been converted into potential energy. That’s why we never even used the equation

y = x^{2}!

1.7.4.5.

Solution.

The skateboarder starts going back down at

y_{S} = \frac{E}{m g},

so we solve

3 m = \frac{E}{100 kg \cdot 9.8 \frac{m}{s^{2}}}

to find

E = 2940 \frac{kg \cdot m^{2}}{s^{2}} = 2940 J

Remark: we needed the diameter to be greater than 3m for the skateboarder to not be going all the way around the culvert, but choosing

r = 5

leads to an answer no different from, say,

r = 50 .

1.7.4.6.

Solution.

From the text, the skateboarder will make it all the way around when

\frac{5}{2} (5) \leq \frac{E}{m g} .

Energy

E

is given by

E = \frac{1}{2} m | v |^{2} + m g y,

the sum of the kinetic and potential energy of the system. At

y = 0,

all the energy is kinetic, so

E = \frac{1}{2} m | v |^{2},

where

| v |

is the skater’s velocity at the bottom of the culvert.

So, we solve:

\begin{aligned} \frac{25}{2} & \leq \frac{E}{m g} = \frac{\frac{1}{2} m | v |^{2}}{m \cdot 9.8} \\ | v | & \geq 5 \sqrt{9.8} \end{aligned}

So, a speed of

5 \sqrt{9.8}

m/s or higher is needed. (That’s about 56 kph.)

1.7.4.7.

Solution.

Equation 1.7.2 tells us the normal force exerted by the track is

W \hat{N},

where

W = m κ | v |^{2} + m g \hat{k} \cdot \hat{N} .

(Note in our problem, the vertical direction is

\hat{k},

not

\hat{ȷ ȷ}

as in §1.7.1.) So, we ought to find

κ

and

N .

\begin{aligned} r (θ) & = (3 \cos θ, 5 \sin θ, 4 + 4 \cos θ) \\ v (θ) & = (- 3 \sin θ, 5 \cos θ, - 4 \sin θ) \\ | v (θ) | = \frac{d s}{d θ} & = \sqrt{9 \sin^{2} θ + 25 \cos^{2} θ + 16 \sin^{2} θ} = 5 \\ a (θ) & = (- 3 \cos θ, - 5 \sin θ, - 4 \cos θ) \\ v \times a & = 5 (- 4, 0, 3) \\ κ (θ) & = \frac{| v \times a |}{{(\frac{d s}{d θ})}^{3}} = \frac{25}{5^{3}} = \frac{1}{5} \end{aligned}

Since $\frac{d^{2} s}{d θ^{2}} = 0,$ we use the following theorem to find $\hat{N} :$

\begin{aligned} a (θ) & = \frac{d^{2} s}{d θ^{2}} \hat{T} + κ {(\frac{d s}{d θ})}^{2} \hat{N} \\ (- 3 \cos θ, - 5 \sin θ, - 4 \cos θ) & = 0 + \frac{25}{5} \hat{N} \\ \hat{N} (θ) & = (- \frac{3}{5} \cos θ, - \sin θ, - \frac{4}{5} \cos θ) \end{aligned}

Using the given quantity $| v (t) | = 5$ at the specified point,

\begin{aligned} W |_{θ = π / 4} & = m κ | v |^{2} + m g \hat{k} \cdot \hat{N} \\ = (1) \frac{1}{5} 5^{2} + 1 (9.8) (- \frac{4}{5} \cos (π / 4)) = 5 - \frac{39.2}{5 \sqrt{2}} \\ W \hat{N} |_{θ = π / 4} & = (5 - \frac{39.2}{5 \sqrt{2}}) (- \frac{3}{5} \cos (π / 4), - \sin (π / 4), - \frac{4}{5} \cos (π / 4)) \\ = (5 - \frac{39.2}{5 \sqrt{2}}) (- \frac{3}{5 \sqrt{2}}, - \frac{1}{\sqrt{2}}, - \frac{4}{5 \sqrt{2}}) \\ = (- \frac{3}{\sqrt{2}} + 2.352, - \frac{5}{\sqrt{2}} + 3.92, - 2 \sqrt{2} + 3.136) \end{aligned}

1.7.4.8.

Solution.

Equation 1.7.2 tells us the normal force exerted by the track is

W \hat{N},

where

W = m κ | v |^{2} + m g \hat{ȷ ȷ} \cdot \hat{N} .

So, we need to find

κ

and

\hat{ȷ ȷ} \cdot \hat{N}

at the point

θ = \frac{13 π}{3} .

Note that

θ

is the parameter used to describe the track, but it is not time. So

| v (θ) | = | \frac{d r}{d θ} |

is not the same as

| v | = | \frac{d r}{d t} |,

the speed of the bead.

\begin{aligned} r (θ) & = (\sin θ, \sin θ - θ) \\ v (θ) & = (\cos θ, \cos θ - 1) \\ | v (θ) | = \frac{d s}{d θ} & = \sqrt{2 \cos^{2} θ - 2 \cos θ + 1} \\ a (θ) & = (- \sin θ, - \sin θ) \\ | v \times a | & = | \sin θ | \\ κ (θ) & = \frac{| v \times a |}{{(\frac{d s}{d θ})}^{3}} = \frac{| \sin θ |}{(2 \cos^{2} θ - 2 \cos θ + 1)^{3 / 2}} \\ \frac{d^{2} s}{d θ^{2}} & = \frac{\sin θ (1 - 2 \cos θ)}{\sqrt{2 \cos^{2} θ - 2 \cos θ + 1}} \end{aligned}

Equation 1.3.3 part (c) gives us the relation

a (θ) = \frac{d^{2} s}{d θ^{2}} \hat{T} + κ {(\frac{d s}{d θ})}^{2} \hat{N} .

We use this to find

\hat{ȷ ȷ} \cdot \hat{N}

θ = 13 π / 3

without differentiating (actually, without even finding)

\hat{T} .

\begin{aligned} a (13 π / 3) & = (- \sqrt{3} / 2, - \sqrt{3} / 2) \\ \frac{d^{2} s}{d θ^{2}} (13 π / 3) & = 0 \\ κ (13 π / 3) & = \sqrt{6} \\ \frac{d s}{d θ} (13 π / 3) & = 1 / \sqrt{2} \\ a (θ) \cdot \hat{ȷ ȷ} & = (\frac{d^{2} s}{d θ^{2}} \hat{T} + κ {(\frac{d s}{d θ})}^{2} \hat{N}) \cdot \hat{ȷ ȷ} \\ - \frac{\sqrt{3}}{2} & = 0 + \sqrt{6} (1 / 2) \hat{N} \cdot \hat{ȷ ȷ} \\ \hat{N} \cdot \hat{ȷ ȷ} & = - \frac{1}{\sqrt{2}} \end{aligned}

Now we can find the speed

| v |

of the bead when

| W | = 100

and it breaks off the track.

\begin{aligned} W & = m κ | v |^{2} + m g \hat{ȷ ȷ} \cdot \hat{N} \\ \pm 100 & = (\frac{1}{9.8}) \sqrt{6} | v |^{2} + \frac{9.8}{9.8} (- \frac{1}{\sqrt{2}}) \\ | v | & = \sqrt{\frac{9.8}{\sqrt{6}} (100 + \frac{1}{\sqrt{2}})} \approx 20 m / s \approx 72 kph \end{aligned}

(Because

| v | > 0,

the equation above has no solution for

W = - 100 .

)

Quite fast! 100 N is a lot of force for such a light object.

1.7.4.9.

Solution.

According to the equation in §1.7.2, the skier will become airborne when:

| v | > \sqrt{\frac{g}{κ} | \hat{ȷ ȷ} \cdot \hat{N} |}

We’ll use the equation of the curve to find

κ

and

\hat{N} .

Note that

g

is given in metres per second, while the other quantities are in kilometres and hours. Converting,

9.8

m/s

^{2}

is the same as

(\frac{9.8 m}{1 s^{2}}) (\frac{1 km}{1000 m}) {(\frac{3600 s}{1 hr})}^{2} = 98 \cdot 6^{4} \frac{km}{h^{2}} = 2^{5} \cdot 3^{4} \cdot 7^{2} \frac{km}{h^{2}} .

\begin{aligned} r (t) & = (\ln t, 1 - t) \\ r^{'} (t) & = v (t) = (t^{- 1}, - 1) \frac{d s}{d t} = | v (t) | = \sqrt{1 + t^{- 2}} \\ r^{″} (t) & = a (t) = (- t^{- 2}, 0) \\ κ (t) & = \frac{| v \times a |}{{(\frac{d s}{d t})}^{3}} = \frac{t^{- 2}}{{\sqrt{1 + t^{- 2}}}^{3}} = \frac{| t |}{(1 + t^{2})^{3 / 2}} = \frac{t}{(1 + t^{2})^{3 / 2}} \end{aligned}

Note $t$ is positive in the interval in question.

\begin{aligned} \hat{T} (t) & = \frac{v (t)}{| v (t) |} = \frac{1}{\sqrt{1 + t^{- 2}}} (t^{- 1}, - 1) = (\frac{1}{\sqrt{1 + t^{2}}}, \frac{- t}{\sqrt{1 + t^{2}}}) \\ {\hat{T}}^{'} (t) & = (\frac{- t}{(1 + t^{2})^{3 / 2}}, \frac{- 1}{(1 + t^{2})^{3 / 2}}) | {\hat{T}}^{'} (t) | = \frac{1}{t^{2} + 1} \\ \hat{N} (t) & = \frac{{\hat{T}}^{'} (t)}{| {\hat{T}}^{'} (t) |} = (\frac{- t}{\sqrt{1 + t^{2}}}, \frac{- 1}{\sqrt{1 + t^{2}}}) \\ | \hat{N} \cdot \hat{ȷ ȷ} | & = \frac{1}{\sqrt{1 + t^{2}}} \end{aligned}

Now, we have all the pieces we need to find the “escape velocity” of the ground.

\begin{aligned} | v | & = \sqrt{\frac{g}{κ} | \hat{N} \cdot \hat{ȷ ȷ} |} = \sqrt{\frac{g \cdot (1 + t^{2})^{3 / 2}}{t (1 + t^{2})^{1 / 2}}} = \sqrt{\frac{g (1 + t^{2})}{t}} \end{aligned}

Since the skier can take off anywhere on the hill, we just need their velocity to be larger than the smallest value of

\sqrt{\frac{g (1 + t^{2})}{t}}

when

1 / e \leq t \leq e .

To find that minimum, we find the location of the minimum of the simpler function

g (t) = \frac{1 + t^{2}}{t} .

Using first-semester calculus, we find it to occur when

t = 1 .

So, the minimum value of

\sqrt{\frac{g (1 + t^{2})}{t}}

(that is, smallest speed to achieve lift-off) occurs at

t = 1 .

We therefore need a minimum speed greater than:

\begin{aligned} \sqrt{\frac{g (1 + t^{2})}{| t |}} |_{t = 1} & = \sqrt{2 g} = \sqrt{2^{6} \cdot 3^{4} \cdot 7^{2}} = 2^{3} \cdot 3^{2} \cdot 7 = 504 kph \end{aligned}

(It seems unlikely that one could reach this speed on skis. The skier is probably earth-bound until they find a curvier hill.)

1.7.4.10.

Solution.

We now have three forces acting on the bead, rather than the two in §1.7.1. The wire still exerts a normal force

W \hat{N}

on the bead to keep it on the wire; gravity still exerts a force

- m g \hat{ȷ ȷ}

straight down. Now our jet-pack force also exerts a force parallel to the direction of the bead’s motion, i.e. parallel to

\hat{T} .

This force is

U \hat{T} .

The net force acting on the bead is the sum of these three forces:

\begin{aligned} F = m a & = U \hat{T} + W \hat{N} - m g \hat{ȷ ȷ} \end{aligned}

To focus on the force in the direction of $\hat{T},$ we dot both sides of the equation with $\hat{T} (s) = (\frac{d x}{d s}, \frac{d y}{d s}) .$ (Recall $r (s)$ was parametrized with respect to arclength, so $\hat{T} (s) = \frac{d r}{d s}$ everywhere.) Since the speed of the bead is constant, the tangential component of its acceleration, $a \cdot \hat{T},$ is 0 (see Theorem 1.3.3.c).

\begin{aligned} 0 & = (U \hat{T} + W \hat{N} - m g \hat{ȷ ȷ}) \cdot \hat{T} \\ = (U \hat{T} \cdot \hat{T}) + (W \hat{N} \cdot \hat{T}) - m g \hat{ȷ ȷ} \cdot \hat{T} \\ = U + 0 - m g \frac{d y}{d s} \\ U & = m g \frac{d y}{d s} \end{aligned}

1.7.4.11.

Solution.

(a) There are three forces acting on the snowmachine. If it’s not accelerating, then

F = m a = 0 :

that is, the forces all cancel out.

So, we have the equation

\begin{aligned} m a & = W \hat{N} + M \hat{T} - m g \hat{ȷ ȷ} \end{aligned}

To isolate $M,$ we dot both sides of the equation with $\hat{T} .$ Remember $\hat{T}$ is a unit vector, and it is perpendicular to $\hat{N} .$

\begin{aligned} m a \cdot \hat{T} & = W \hat{N} \cdot \hat{T} + M \hat{T} \cdot \hat{T} - m g \hat{ȷ ȷ} \cdot \hat{T} \\ = 0 + M - m g \hat{ȷ ȷ} \cdot \hat{T} \end{aligned}

Since the speed of the snowmachine is constant, the tangential component of its acceleration, $a \cdot \hat{T},$ is 0 (see Theorem 1.3.3.c).

\begin{aligned} 0 & = M - m g \hat{ȷ ȷ} \cdot \hat{T} \\ M & = m g \hat{ȷ ȷ} \cdot \hat{T} \end{aligned}

(b) We would expect, from looking at the situation, that the engine would have to provide a “backwards” force to slow the acceleration due to gravity. So, we would expect

M < 0 .

Indeed, if

\hat{T}

points downhill, then the

y

-component of

\hat{T}

is negative, so

M = m g \hat{ȷ ȷ} \cdot \hat{T}

is negative.

(This is the purpose of driving downhill in a low gear: the friction inside the motor provides a force opposing the direction of motion, slowing the vehicle.)

M = m g \hat{ȷ ȷ} \cdot \hat{T},

we’ll need to find

\hat{ȷ ȷ} \cdot \hat{T} .

\begin{aligned} r (x) & = (x, 1 + \cos x) & r^{'} (x) & = (1, - \sin x) \\ | r^{'} (x) | & = \sqrt{1 + \sin^{2} x} & \hat{T} (x) & = \frac{1}{\sqrt{1 + \sin^{2} x}} (1, - \sin x) \\ \hat{T} (3 π / 4) & = (\sqrt{\frac{2}{3}}, - \frac{1}{\sqrt{3}}) \end{aligned}

So,

M = (200 kg) (9.8 m / s^{2}) (- \frac{1}{\sqrt{3}} m) = - \frac{1960}{\sqrt{3}} N \approx - 1131.6 N

1.7.4.12.

Solution.

We begin with the usual computations.

\begin{aligned} r (θ) & = (4 \cos θ, 3 (1 + \sin θ)) \\ v (θ) = r^{'} (θ) & = (- 4 \sin θ, 3 \cos θ) \\ | v (θ) | = \frac{d s}{d θ} & = \sqrt{16 \sin^{2} θ + 9 \cos^{2} θ} = \sqrt{9 + 7 \sin^{2} θ} \\ a (θ) & = (- 4 \cos θ, - 3 \sin θ) \\ | v (θ) \times a (θ) | & = 12 \\ κ (θ) & = \frac{| v (θ) \times a (θ) |}{{(\frac{d s}{d θ})}^{3}} = \frac{12}{(9 + 7 \sin^{2} θ)^{3 / 2}} \\ \hat{T} (θ) & = \frac{(- 4 \sin θ, 3 \cos θ)}{\sqrt{9 + 7 \sin^{2} θ}} \\ {\hat{T}}^{'} (θ) & = \frac{(36 \cos θ, 48 \sin θ)}{- (9 \cos^{2} θ + 16 \sin^{2} θ)^{3 / 2}} \\ | {\hat{T}}^{'} (θ) | & = \frac{12}{9 \cos^{2} θ + 16 \sin^{2} θ} \\ \hat{N} (θ) & = \frac{(3 \cos θ, 4 \sin θ)}{- \sqrt{9 \cos^{2} θ + 16 \sin^{2} θ}} \end{aligned}

We want to find the height

y_{S}

where

| v | = 0,

and the height

y_{A}

where

W = 0 .

Remember that

v

in these equations is the derivative of position with respect to time, and is not the same as

v (θ) .

\begin{aligned} Equation 1.7.1 : E & = \frac{1}{2} m | v |^{2} + m g y \\ If | v | = 0 : E & = m g y_{S} ⟹ y_{S} = \frac{E}{m g} \end{aligned}

This answers part a.

\begin{aligned} Equation 1.7.2 : W & = 2 κ (E - m g y) + m g \hat{ȷ ȷ} \cdot \hat{N} \\ If W = 0 : 0 & = 2 κ (E - m g y_{A}) + m g \hat{ȷ ȷ} \cdot \hat{N} \\ = \frac{24 (E - m g y_{A})}{(9 + 7 \sin^{2} θ)^{3 / 2}} - m g (4 \frac{\sin θ}{\sqrt{9 + 7 \sin^{2} θ}}) \end{aligned}

Using $y = 3 + 3 \sin θ :$

\begin{aligned} = \frac{24 (E - m g y_{A})}{{(9 + 7 {(\frac{y_{A} - 3}{3})}^{2})}^{3 / 2}} - 4 m g (\frac{\frac{y_{A} - 3}{3}}{\sqrt{9 + 7 {(\frac{y_{A} - 3}{3})}^{2}}}) \end{aligned}

So, for part b., we can write (say)

\frac{24 (E - m g y_{A})}{{(9 + 7 {(\frac{y_{A} - 3}{3})}^{2})}^{3 / 2}} = 4 m g (\frac{\frac{y_{A} - 3}{3}}{\sqrt{9 + 7 {(\frac{y_{A} - 3}{3})}^{2}}})

Now, suppose the skater’s speed at the bottom of the culvert (

y = 0

) is 11 m/s. Then their energy is

E = \frac{1}{2} m (11^{2}) + 0,

\frac{121 m}{2}

joules, where

m

is their mass. Then

y_{S} = \frac{E}{m g} = \frac{121}{2 \cdot 9.8} \approx 6.2 .

Since the half-way height of the culvert is at height

y = 3,

the skater makes it onto the ceiling of the culvert. Now the question is: did they make it all around, or fall off the ceiling?

For this, we need to find

y_{A} .

If they go airborne on the ceiling, they fall; but if

y_{A} > 6,

then they never lose contact with the culvert, and they go all the way around.

\begin{aligned} \frac{24 (E - m g y_{A})}{{(9 + 7 {(\frac{y_{A} - 3}{3})}^{2})}^{3 / 2}} & = 4 m g (\frac{\frac{y_{A} - 3}{3}}{\sqrt{9 + 7 {(\frac{y_{A} - 3}{3})}^{2}}}) \\ \Leftrightarrow \frac{6 (\frac{E}{m g} - y_{A})}{{(9 + 7 {(\frac{y_{A} - 3}{3})}^{2})}^{3 / 2}} & = \frac{\frac{y_{A} - 3}{3}}{\sqrt{9 + 7 {(\frac{y_{A} - 3}{3})}^{2}}} \\ \Leftrightarrow \frac{6 (\frac{11^{2}}{2 \cdot 9.8} - y_{A})}{{(9 + 7 {(\frac{y_{A} - 3}{3})}^{2})}^{3 / 2}} & = \frac{\frac{y_{A} - 3}{3}}{\sqrt{9 + 7 {(\frac{y_{A} - 3}{3})}^{2}}} \end{aligned}

To simplify to a more standard form, we multiply both sides by ${(9 + 7 {(\frac{y_{A} - 3}{3})}^{2})}^{3 / 2} :$

\begin{aligned} 6 (\frac{11^{2}}{2 \cdot 9.8} - y_{A}) & = (\frac{y_{A} - 3}{3}) (9 + 7 {(\frac{y_{A} - 3}{3})}^{2}) \end{aligned}

Now, we simplify to

\begin{aligned} 0 & = \frac{7}{9} y_{A}^{3} - 7 y_{A}^{2} + 48 y_{A} - \frac{7797}{49} \end{aligned}

Now, solving for

y_{A}

involves solving a cubic function, which is no small task. We could ask a computer, but we can also get an idea of its root(s) by plugging in numbers and using the intermediate value theorem. In particular, we need to know whether

y_{A}

is greater than 6 (the skater makes it!) or between 3 and 6 (they fall off the ceiling).

Let

f (y) = \frac{7}{9} y^{3} - 7 y^{2} + 48 y - \frac{7797}{49} .

Note

f (4) = - \frac{12941}{441},

which is negative, and

f (5) = \frac{1367}{441},

which is positive. So, by the intermediate value theorem, there is a root of

f (y)

between

y = 4

and

y = 5 .

That is,

y_{A}

is between 4 and 5, so the skater falls off the ceiling somewhere between these heights, rather than making it all the way around.

1.7.4.13.

Solution.

(a) By Newton’s law of motion

\begin{aligned} E^{'} (t) & = \frac{d}{d t} [\frac{1}{2} m | v (t) |^{2} + m g r (t) \cdot \hat{k}] = m v (t) \cdot v^{'} (t) + m g v (t) \cdot \hat{k} \\ = v (t) \cdot [N (r (t)) - m g \hat{k}] + m g v (t) \cdot \hat{k} \\ = 0 \end{aligned}

since

v (t) \cdot N (r (t)) = 0 .

E (t)

is a constant, independent of

t .

(b) By part (a),

\begin{aligned} E (t) = E (0) & ⟹ \frac{1}{2} m | v (t) |^{2} + m g b θ (t) = m g (2 π b) \\ ⟹ | v (t) |^{2} = 2 g b (2 π - θ (t)) \end{aligned}

θ = 2 π

θ = 0 .

We’ll first determine

\frac{d θ}{d t} .

\begin{aligned} v = \frac{d r}{d t} = \frac{d r}{d θ} \frac{d θ}{d t} = (- a \sin θ, a \cos θ, b) \frac{d θ}{d t} \\ ⟹ | v |^{2} = [a^{2} + b^{2}] {(\frac{d θ}{d t})}^{2} \\ ⟹ \frac{d θ}{d t} = - {[\frac{| v |^{2}}{a^{2} + b^{2}}]}^{1 / 2} = - {[\frac{2 g b (2 π - θ)}{a^{2} + b^{2}}]}^{1 / 2} \end{aligned}

We have chosen the negative sign because

θ

must decrease from

2 π

to 0. The time required to do so is

\begin{aligned} \int d t & = \int_{2 π}^{0} \frac{d t}{d θ} d θ = - {[\frac{a^{2} + b^{2}}{2 g b}]}^{1 / 2} \int_{2 π}^{0} \frac{1}{(2 π - θ)^{1 / 2}} d θ \\ = {[\frac{a^{2} + b^{2}}{2 g b}]}^{1 / 2} \int_{0}^{2 π} \frac{1}{(2 π - θ)^{1 / 2}} d θ \\ = {[\frac{a^{2} + b^{2}}{2 g b}]}^{1 / 2} {[- 2 (2 π - θ)^{1 / 2}]}_{0}^{2 π} = 2 {[\frac{a^{2} + b^{2}}{g b} π]}^{1 / 2} \end{aligned}

1.8 Optional — Polar Coordinates

Exercises

1.8.1.

Solution.

The upper sketch below contains the points,

(x_{1}, y_{1}),

(x_{3}, y_{3}),

(x_{5}, y_{5}),

that are on the axes. The lower sketch below contains the points,

(x_{2}, y_{2}),

(x_{4}, y_{4}),

that are not on the axes.

Recall that the polar coordinates

r,

θ

are related to the cartesian coordinates

x,

y,

x = r \cos θ,

y = r \sin θ .

r = \sqrt{x^{2} + y^{2}}

and

\tan θ = \frac{y}{x}

(assuming that

x \neq 0

) and

\begin{aligned} (x_{1}, y_{1}) & = (3, 0) & ⟹ r_{1} = 3, \tan θ_{1} = 0 \\ ⟹ θ_{1} = 0 as (x_{1}, y_{1}) is on the positive x -axis \\ (x_{2}, y_{2}) & = (1, 1) & ⟹ r_{2} = \sqrt{2}, \tan θ_{2} = 1 \\ ⟹ θ_{2} = \frac{π}{4} as (x_{2}, y_{2}) is in the first octant \\ (x_{3}, y_{3}) & = (0, 1) & ⟹ r_{3} = 1, \cos θ_{3} = 0 \\ ⟹ θ_{3} = \frac{π}{2} as (x_{3}, y_{3}) is on the positive y -axis \\ (x_{4}, y_{4}) & = (- 1, 1) & ⟹ r_{4} = \sqrt{2}, \tan θ_{4} = - 1 \\ ⟹ θ_{4} = \frac{3 π}{4} as (x_{4}, y_{4}) is in the third octant \\ (x_{5}, y_{5}) & = (- 2, 0) & ⟹ r_{5} = 2, \tan θ_{5} = 0 \\ ⟹ θ_{5} = π as (x_{5}, y_{5}) is on the negative x -axis \end{aligned}

1.8.2.

Solution.

Note that the distance from the point

(r \cos θ, r \sin θ)

to the origin is

\sqrt{r^{2} \cos^{2} θ + r^{2} \sin^{2} θ} = \sqrt{r^{2}} = | r |

Thus

r

can be either the distance to the origin or minus the distance to the origin.

(a) The distance from

(- 2, 0)

to the origin is

2 .

So either

r = 2

r = - 2 .

If $r = 2,$ then $θ$ must obey
$\begin{aligned} (- 2, 0) = (2 \cos θ, 2 \sin θ) & ⟺ \sin θ = 0, \cos θ = - 1 \\ ⟺ θ = n π, n integer, \cos θ = - 1 \\ ⟺ θ = n π, n odd integer \end{aligned}$
If $r = - 2,$ then $θ$ must obey
$\begin{aligned} (- 2, 0) = (- 2 \cos θ, - 2 \sin θ) & ⟺ \sin θ = 0, \cos θ = 1 \\ ⟺ θ = n π, n integer, \cos θ = 1 \\ ⟺ θ = n π, n even integer \end{aligned}$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates

θ = π,

r > 0

and the pink half-line is the set of all points with polar coordinates

θ = π,

r < 0 .

In the figure on the right below, the blue half-line is the set of all points with polar coordinates

θ = 0,

r > 0

and the pink half-line is the set of all points with polar coordinates

θ = 0,

r < 0 .

(b) The distance from

(1, 1)

to the origin is

\sqrt{2} .

So either

r = \sqrt{2}

r = - \sqrt{2} .

If $r = \sqrt{2},$ then $θ$ must obey
$\begin{aligned} (1, 1) = (\sqrt{2} \cos θ, \sqrt{2} \sin θ) & ⟺ \sin θ = \cos θ = \frac{1}{\sqrt{2}} \\ ⟺ θ = \frac{π}{4} + 2 n π, n integer \end{aligned}$
If $r = - \sqrt{2},$ then $θ$ must obey
$\begin{aligned} (1, 1) = (- \sqrt{2} \cos θ, - \sqrt{2} \sin θ) & ⟺ \sin θ = \cos θ = - \frac{1}{\sqrt{2}} \\ ⟺ θ = \frac{5 π}{4} + 2 n π, n integer \end{aligned}$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates

θ = \frac{π}{4},

r > 0

and the pink half-line is the set of all points with polar coordinates

θ = \frac{π}{4},

r < 0 .

In the figure on the right below, the blue half-line is the set of all points with polar coordinates

θ = \frac{5 π}{4},

r > 0

and the pink half-line is the set of all points with polar coordinates

θ = \frac{5 π}{4},

r < 0 .

(- 1, - 1)

to the origin is

\sqrt{2} .

So either

r = \sqrt{2}

r = - \sqrt{2} .

If $r = \sqrt{2},$ then $θ$ must obey
$\begin{aligned} (- 1, - 1) = (\sqrt{2} \cos θ, \sqrt{2} \sin θ) & ⟺ \sin θ = \cos θ = - \frac{1}{\sqrt{2}} \\ ⟺ θ = \frac{5 π}{4} + 2 n π, n integer \end{aligned}$
If $r = - \sqrt{2},$ then $θ$ must obey
$\begin{aligned} (- 1, - 1) = (- \sqrt{2} \cos θ, - \sqrt{2} \sin θ) & ⟺ \sin θ = \cos θ = \frac{1}{\sqrt{2}} \\ ⟺ θ = \frac{π}{4} + 2 n π, n integer \end{aligned}$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates

θ = \frac{5 π}{4},

r > 0

and the pink half-line is the set of all points with polar coordinates

θ = \frac{5 π}{4},

r < 0 .

In the figure on the right below, the blue half-line is the set of all points with polar coordinates

θ = \frac{π}{4},

r > 0

and the pink half-line is the set of all points with polar coordinates

θ = \frac{π}{4},

r < 0 .

1.8.3.

Solution.

(a) The lengths are

\begin{aligned} | {\hat{e}}_{r} (θ) | & = \sqrt{\cos^{2} θ + \sin^{2} θ} = 1 \\ | {\hat{e}}_{θ} (θ) | & = \sqrt{(- \sin θ)^{2} + \cos^{2} θ} = 1 \end{aligned}

{\hat{e}}_{r} (θ) \cdot {\hat{e}}_{θ} (θ) = (\cos θ) (- \sin θ) + (\sin θ) (\cos θ) = 0

the two vectors are perpendicular and the angle between them is

\frac{π}{2} .

The cross product is

\begin{aligned} {\hat{e}}_{r} (θ) \times {\hat{e}}_{θ} (θ) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \cos θ & \sin θ & 0 \\ - \sin θ & \cos θ & 0 \end{array}] = \hat{k} \end{aligned}

(b) Note that for

θ

determined by

x = r \cos θ,

y = r \sin θ,

the vector ${\hat{e}}_{r} (θ)$ is a unit vector in the same direction as the vector from $(0, 0)$ to $(x, y)$ and
the vector ${\hat{e}}_{θ} (θ)$ is a unit vector that is perpendicular to ${\hat{e}}_{r} (θ) .$
The $y$ -component of ${\hat{e}}_{θ} (θ)$ has the same sign as the $x$ -component of ${\hat{e}}_{r} (θ) .$ The $x$ -component of ${\hat{e}}_{θ} (θ)$ has opposite sign to that of the $y$ -component of ${\hat{e}}_{r} (θ) .$

Here is a sketch of

(x_{i}, y_{i}),

{\hat{e}}_{r} (θ_{i}),

{\hat{e}}_{θ} (θ_{i})

for

i = 1, 3, 5

(the points on the axes)

and here is a sketch (to a different scale) of

(x_{i}, y_{i}),

{\hat{e}}_{r} (θ_{i}),

{\hat{e}}_{θ} (θ_{i})

for

i = 2, 4

(the points off the axes).

1.8.4. (✳).

Solution.

(a) Since

- 1 \leq \sin (4 θ) \leq 1,

the coordinate

r = 2 + \sin (4 θ)

oscillates between

r = 1

and

r = 3

θ

runs from

0

2 π .

The maximum value

r = 3

is achieved when

\sin (4 θ) = 1,

i.e when

4 θ = \frac{π}{2} + 2 n π,

i.e. when

θ = \frac{π}{8} + \frac{n π}{2} .

That matches figure (E).

(b) Since

- 1 \leq \sin (4 θ) \leq 1,

the coordinate

r = 1 + 2 \sin (4 θ)

takes its maximum value

r = 3

when

\sin (4 θ) = 1,

i.e. when

θ = \frac{π}{8} + \frac{n π}{2},

just as the case with

(a) .

But now

r

can also take the value

0 .

That matches figure (B).

(c)

r = 1

is completely indepedent of

θ .

All points on the curve

r = 1

are a distance

1

from the origin. That is,

r = 1

is the circle of radius

1

centred on the origin. That’s figure (F).

(d) In this case,

θ

is subject to the restriction

- \frac{π}{2} \leq θ \leq \frac{π}{2},

like figure (C). Figure (C) looks like a circle. We can verify that

r = 2 \cos (θ)

is indeed a circle by converting to Cartesian coordinates. We can convert the right hand side to exactly

2 x = 2 r \cos (θ)

by multiplying the whole equation by

r .

\begin{aligned} r = 2 \cos (θ) & ⟺ r^{2} = 2 r \cos (θ) ⟺ x^{2} + y^{2} = 2 x \\ ⟺ (x - 1)^{2} + y^{2} = 1 \end{aligned}

r = 2 \cos (θ)

is the circle of radius

1

centred on

x = 1,

y = 0,

which indeed matches figure (C).

(e) When

θ = 0,

r = e^{θ / 10} + e^{- θ / 10} = 2 .

\frac{d}{d θ} (e^{θ / 10} + e^{- θ / 10}) = \frac{1}{10} (e^{θ / 10} - e^{- θ / 10}) > 0 for all θ > 0

r = e^{θ / 10} + e^{- θ / 10}

increases as

θ

increases for all

θ \geq 0 .

Furthermore the rate of increase gets bigger and bigger as

θ

gets bigger and bigger. So

r

starts at

r = 2

when

θ = 0

and increases faster and faster as

θ

increases. That matches figure (A).

(f) When

θ = 0,

r = θ = 0 .

\frac{d}{d θ} θ = 1 for all θ

r = θ

increases as

θ

increases for all

θ \geq 0 .

Furthermore the rate of increase is independent of

θ .

r

starts at

r = 0

when

θ = 0

and increases at a constant rate as

θ

increases. That matches figure (D).

1.8.5.

Solution.

Think of

θ

as a time parameter and recall that

κ (θ) = \frac{| v (θ) \times a (θ) |}{| v (θ) |^{3}} .

The given curve has

\begin{aligned} x (θ) & = f (θ) \cos θ \\ y (θ) & = f (θ) \sin θ \\ r (θ) & = f (θ) [\cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ}] \\ v (θ) = r^{'} (θ) & = f^{'} (θ) [\cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ}] + f (θ) [- \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ}] \\ a (θ) = r^{″} (θ) & = {f^{″} (θ) - f (θ)} [\cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ}] + 2 f^{'} (θ) [- \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ}] \end{aligned}

The efficient way to compute

| v (θ) |

and the cross product

v (θ) \times a (θ)

is to observe that

\begin{aligned} v (θ) & = f^{'} (θ) {\hat{e}}_{r} (θ) + f (θ) {\hat{e}}_{θ} (θ) \\ a (θ) & = {f^{″} (θ) - f (θ)} {\hat{e}}_{r} (θ) + 2 f^{'} (θ) {\hat{e}}_{θ} (θ) \end{aligned}

where

{\hat{e}}_{r} (θ)

and

{\hat{e}}_{θ} (θ)

are the vectors of Q[1.8.3]. As

{\hat{e}}_{r} (θ)

and

{\hat{e}}_{θ} (θ)

are mutually perpendicular unit vectors obeying

{\hat{e}}_{r} (θ) \times {\hat{e}}_{θ} (θ) = \hat{k}

and

{\hat{e}}_{r} (θ) \times {\hat{e}}_{r} (θ) = {\hat{e}}_{θ} (θ) \times {\hat{e}}_{θ} (θ) = 0,

\begin{aligned} | v (θ) |^{2} & = v (θ) \cdot v (θ) = [f^{'} (θ) {\hat{e}}_{r} (θ) + f (θ) {\hat{e}}_{θ} (θ)] \cdot [f^{'} (θ) {\hat{e}}_{r} (θ) + f (θ) {\hat{e}}_{θ} (θ)] \\ = f^{'} (θ)^{2} {\hat{e}}_{r} (θ) \cdot {\hat{e}}_{r} (θ) + f (θ)^{2} {\hat{e}}_{θ} (θ) \cdot {\hat{e}}_{θ} (θ) + 2 f^{'} (θ) f (θ) {\hat{e}}_{r} (θ) \cdot {\hat{e}}_{θ} (θ) \\ = f^{'} (θ)^{2} + f (θ)^{2} \\ | v (θ) | & = \sqrt{f^{'} (θ)^{2} + f (θ)^{2}} \\ v (θ) \times a (θ) & = [f^{'} (θ) {\hat{e}}_{r} (θ) + f (θ) {\hat{e}}_{θ} (θ)] \times [{f^{″} (θ) - f (θ)} {\hat{e}}_{r} (θ) + 2 f^{'} (θ) {\hat{e}}_{θ} (θ)] \\ = 2 f^{'} (θ)^{2} {\hat{e}}_{r} (θ) \times {\hat{e}}_{θ} (θ) + f (θ) [f^{″} (θ) - f (θ)] {\hat{e}}_{θ} (θ) \times {\hat{e}}_{r} (θ) \\ = {2 f^{'} (θ)^{2} - f (θ) [f^{″} (θ) - f (θ)]} \hat{k} \end{aligned}

\begin{aligned} κ (θ) & = \frac{| v (θ) \times a (θ) |}{| v (θ) |^{3}} = \frac{| f (θ)^{2} + 2 f^{'} (θ)^{2} - f (θ) f^{″} (θ) |}{{[f (θ)^{2} + f^{'} (θ)^{2}]}^{3 / 2}} \end{aligned}

1.8.6.

Solution.

By the Q[1.8.5] with

\begin{matrix} f (θ) = a (1 - \cos θ) f^{'} (θ) = a \sin θ f^{″} (θ) = a \cos θ \end{matrix}

we have

\begin{aligned} κ (θ) & = \frac{| f (θ)^{2} + 2 f^{'} (θ)^{2} - f (θ) f^{″} (θ) |}{[f (θ)^{2} + f^{'} (θ)^{2}]^{3 / 2}} \\ = \frac{| a^{2} - 2 a^{2} \cos θ + a^{2} \cos^{2} θ + 2 a^{2} \sin^{2} θ - a^{2} \cos θ + a^{2} \cos^{2} θ |}{[a^{2} - 2 a^{2} \cos θ + a^{2} \cos^{2} θ + a^{2} \sin^{2} θ]^{3 / 2}} \\ = \frac{3 a^{2} - 3 a^{2} \cos θ}{[2 a^{2} - 2 a^{2} \cos θ]^{3 / 2}} = \frac{3}{2^{3 / 2} a \sqrt{1 - \cos θ}} = \frac{3}{2 \sqrt{2 a r (θ)}} \end{aligned}

1.9 Optional — Central Forces

Exercises

1.9.1. (✳).

Solution.

(a) Both (i) and (ii) were proven in §1.9. Here are those arguments.

Define

Ω Ω (t) = r (t) \times v (t)

By the product rule,

\begin{aligned} \frac{d Ω Ω}{d t} (t) & = \frac{d}{d t} (r (t) \times v (t)) = v (t) \times v (t) + r (t) \times a (t) \\ = m r (t) \times f (r (t)) r (t)) = 0 \end{aligned}

Ω Ω (t)

is in fact independent of

t .

It is a constant vector that we’ll just denote

Ω Ω .

r (t) \times v (t) = Ω Ω,

we have that

r (t)

is always perpendicular to

Ω Ω

and

r (t) \cdot Ω Ω = 0

If $Ω Ω \neq 0,$ this is exactly the statement that $r (t)$ always lies in the plane through the origin with normal vector $Ω Ω .$
If $Ω Ω = 0,$ then $r (t)$ is always parallel to $v (t)$ and there is some function $α (t)$ such that
$\frac{d r}{d t} (t) = v (t) = α (t) r (t)$
This is a first order, linear, ordinary differential equation that we can solve by using an integrating factor. Set
$β (t) = \int_{0}^{t} α (t) d t$
Then
$\begin{aligned} \frac{d r}{d t} (t) = α (t) r (t) & ⟺ e^{- β (t)} \frac{d r}{d t} (t) - α (t) e^{- β (t)} r (t) = 0 \\ ⟺ \frac{d}{d t} [e^{- β (t)} r (t)] = 0 \\ ⟺ e^{- β (t)} r (t) = r (0) \\ ⟺ r (t) = e^{β (t)} r (0) \end{aligned}$
so that $r (t)$ lies on a line through the origin. This makes sense — the particle is always moving parallel to its radius vector.

This completes the verification that

r (t)

lies in a plane.

Now we show that the radius vector

r (t)

sweeps out equal areas in equal times. In other words, we now verify that the rate at which

r (t)

sweeps out area is independent of time. To do so we rewrite the statement that

| r (t) \times v (t) |

is constant in polar coordinates. Writing

r (t) = r (t) \hat{r} (θ (t))

and then applying Lemma 1.8.2.b gives that

\begin{aligned} constant = | r \times v | = | r \hat{r} \times (\frac{d r}{d t} \hat{r} + r \frac{d θ}{d t} \hat{θ θ}) | = r^{2} \frac{d θ}{d t} \\ since | \hat{r} \times \hat{r} | = 0, | \hat{r} \times \hat{θ θ} | = 1 \end{aligned}

is constant. It now suffices to observe that

r (t)^{2} \frac{d θ}{d t} (t)

is exactly twice the rate at which

r (t)

sweeps out area. To see this, just look at the figure below. The shaded area is essentially a wedge of a circular disk of radius

r .

(If

r (t)

were independent of

t,

it would be exactly a wedge of a circular disk.) Its area is the fraction

\frac{d θ}{2 π}

of the area of the full disk, which is

\frac{d θ}{2 π} π r^{2} = \frac{1}{2} r^{2} d θ

(b) If

f (r)

is identically zero, then

r^{″} (t) = 0,

so that

r^{'} (t)

is a constant, say

v_{0},

and

r (t) = r_{0} + v_{0} t,

for some constant

r_{0} .

That’s a straight line.

We’ll now show that if the motion of the particle always lies on a straight line, then

f (r)

must be identically zero. Suppose that

r (t)

is a straight line. Then there are constant vectors

r_{0}

and

\hat{T}

such that

r (t) = r_{0} + g (t) \hat{T}

for some scalar valued function

g (t) .

Then

\begin{aligned} r^{″} (t) & = f (r (t)) r (t) \end{aligned}

becomes

\begin{aligned} g^{″} (t) \hat{T} & = f (r (t)) [r_{0} + g (t) \hat{T}] = f (r (t)) r_{0} + f (r (t)) g (t) \hat{T} \end{aligned}

We may always choose the initial conditions so that, for example,

r_{0} = \hat{ı ı}

and

\hat{T} = \hat{ȷ ȷ} .

\begin{aligned} g^{″} (t) \hat{ȷ ȷ} & = f (r (t)) \hat{ı ı} + f (r (t)) g (t) \hat{ȷ ȷ} \end{aligned}

Taking the dot product of both sides with

\hat{ı ı}

gives

f (r (t)) = 0,

as desired.

- \frac{G M m}{r^{3}} r

can produce elliptical orbits. So any

f (r)

which is a positive constant times

- \frac{1}{r^{3}}

does the job.

1.9.2. (✳).

Solution.

(a) Our object is subject to a central force. So the acceleration

a (t)

is parallel to

r (t)

and

\frac{d}{d t} (r (t) \times v (t)) = v (t) \times v (t) + r (t) \times a (t) = 0 + 0 = 0

By Lemma 1.8.2,

v (t) = \frac{d r}{d t} (t) \hat{r} (θ (t)) + r (t) \frac{d θ}{d t} (t) \hat{θ θ} (θ (t)) .

Because

r (t) = r (t) \hat{r} (θ (t))

is parallel to

\hat{r} (θ (t))

and is perpendicular to

\hat{θ θ} (θ (t)),

r (t) \times v (t) = r (t)^{2} \dot{θ} (t) \hat{r} (θ (t)) \times \hat{θ θ} (θ (t))

and, in particular,

| r (t) \times v (t) | = r (t)^{2} | \dot{θ} (t) |

is a constant. As

\dot{θ} (t)

is continuous,

r (t)^{2} \dot{θ} (t)

is also constant.

(b) By Lemma 1.8.2, the acceleration

\begin{aligned} a (t) & = (\frac{d^{2} r}{d t^{2}} (t) - r (t) (\frac{d θ}{d t} (t))^{2}) \hat{r} (θ (t)) \\ + (r (t) \frac{d^{2} θ}{d t^{2}} (t) + 2 \frac{d r}{d t} (t) \frac{d θ}{d t} (t)) \hat{θ θ} (θ (t)) \end{aligned}

Because our object is subject to a central force, the acceleration

a (t)

is parallel to

\hat{r} (θ (t)) .

So the

\hat{θ θ} (θ (t))

component of the acceleration is zero and

a (t) = (\frac{d^{2} r}{d t^{2}} (t) - r (t) (\frac{d θ}{d t} (t))^{2}) \hat{r} (θ (t))

so that

| a (t) | = | \frac{d^{2} r}{d t^{2}} (t) - r (t) (\frac{d θ}{d t} (t))^{2} |

Since

r (t) = \frac{1}{θ (t) + α}

\begin{matrix} \dot{r} (t) = - \frac{1}{[θ (t) + α]^{2}} \dot{θ} (t) = - r (t)^{2} \dot{θ} (t) = - h \end{matrix}

\frac{d^{2} r}{d t^{2}} (t) = 0

and

\begin{matrix} | a (t) | = r (t) \dot{θ} (t)^{2} = \frac{r (t)^{4} \dot{θ} (t)^{2}}{r (t)^{3}} = \frac{h^{2}}{r (t)^{3}} \end{matrix}

2 Vector Fields
2.1 Definitions and First Examples

Exercises

2.1.1.

Solution.

The vectors are pointing to the right when

x > 0,

to the left when

x < 0,

and are vertical when

x = 0 .

So, at least for

(x, y)

shown in the sketch,

v (x, y) \cdot \hat{ı ı} {\begin{cases} > 0 & when x > 0 \\ = 0 & when x = 0 \\ < 0 & when x < 0 \end{cases}

The behaviour of the

y

-values is more complicated. Vectors in one vertical line seem to be all pointing up, or all pointing down. So, the sign of

v \cdot \hat{ȷ ȷ}

depends only on

x,

not on

y

(although the magnitude of

v \cdot \hat{ȷ ȷ}

depends on both). Roughly, the vectors are pointing

Down when $x < - 2;$
horizontally when $x = - 2$ (remember the vector is positioned with the base of $v (x, y)$ at $(x, y);$
up when $- 2 < x < 2;$
horizontally when $x = 2;$
up when $2 < x .$

Since we’re assuming there’s nothing surprising happening between the samples pictured, at least for

(x, y)

shown in the sketch,

v (x, y) \cdot \hat{ȷ ȷ} {\begin{cases} > 0 & when - 2 < x < 2 \\ = 0 & when x \in {- 2, 2} \\ < 0 & when x < - 2 or x > 2 \end{cases}

2.1.2.

Solution.

To start out, we find the places where

v (x, y) \cdot \hat{ı ı} = 0

(vertical vectors) or

v (x, y) \cdot \hat{ȷ ȷ} = 0

(horizontal vectors). Remember the vector

v (x, y)

has its tail at

(x, y) .

We see the vertical vectors (those with

v (x, y) \cdot \hat{ı ı} = 0

) occur at every point along the line

y = - x,

while horizontal vectors (those with

v (x, y) \cdot \hat{ȷ ȷ} = 0

) occur at every point along the line

y = x .

Indeed, below the line

y = - x,

vectors point to the left, while above the line

y = - x

they point to the right. Similarly, vectors point down when they’re above the line

y = x,

and the point up when they’re below the line

y = x .

So, at least for

(x, y)

shown in the sketch,

v (x, y) \cdot \hat{ı ı} {\begin{cases} > 0 & when y > - x \\ = 0 & when y = - x \\ < 0 & when y < - x \end{cases} and v (x, y) \cdot \hat{ȷ ȷ} {\begin{cases} > 0 & when y < x \\ = 0 & when y = x \\ < 0 & when y > x \end{cases}

2.1.3.

Solution.

Since all conveyors point towards the origin, the direction of motion of an object at location

(x, y)

\frac{(- x, - y)}{\sqrt{x^{2} + y^{2}}} .

Its magnitude is

| y |,

v (x, y) = \frac{- | y |}{\sqrt{x^{2} + y^{2}}} (x, y) .

2.1.4.

Solution.

The arrows near the point

A

are pointing to the right, indicating that

P > 0,

and upward, indicating that

Q > 0 .

Moving from left to right near

A,

the vertical component of the arrows is decreasing, indicating that

\frac{\partial Q}{\partial x} < 0 .

Moving vertically upwards near

A,

the vertical component of the arrows is increasing, indicating that

\frac{\partial Q}{\partial y} > 0 .

2.1.5.

Solution.

(a) At time

0

the velocity of the twig is

v (1, 1) = \hat{ı ı} + \hat{ȷ ȷ} .

So at time

t = 0.01,

the position of the twig is approximately

(1, 1) + 0.01 (1, 1) = (1.01, 1.01)

(b) At time

0

the velocity of the twig is

v (0, 0) = 0 .

So at time

t = 0.01,

the position of the twig is

(0, 0) + 0.01 (0, 0) = (0, 0)

0

the velocity of the twig is

v (0, 0) = 0 .

So it is stationary and its velocity remains zero for all time. The position of the twig at time

10,

and in fact at all times, is

(0, 0) .

2.1.6.

Solution.

The velocity of the fluid at all points of the

y

-axis is

- \hat{ȷ ȷ} .

So the twig will remain on the

y

-axis and will consequently have velocity

- \hat{ȷ ȷ}

for all time. The position of the twig at time

10

will be

(0, 0) + 10 (0, - 1) = (0, - 10)

2.1.7.

Solution.

Set your face to be at the origin of our coordinate system,

(0, 0, 0) .

A bee at position

(x, y, z)

is a distance of

\sqrt{x^{2} + y^{2} + z^{2}}

from your face, heading in the direction

(- x, - y, - z) .

So, the unit vector indicating the direction of travel of one bee is

\frac{- 1}{\sqrt{x^{2} + y^{2} + z^{2}}} (x, y, z) .

Now all we need to find is the length of the velocity vector, i.e. the speed of the bee.

The speed of the friendly bee is inversely proportional to

\sqrt{x^{2} + y^{2} + z^{2}},

its distance from your face. (Bees that are closer to you are buzzing towards you more excitedly.) So, the speed is given by

\frac{α}{\sqrt{x^{2} + y^{2} + z^{2}}}

for some constant

α .

The bee velocity has the direction of the unit vector

\frac{- 1}{\sqrt{x^{2} + y^{2} + z^{2}}} (x, y, z)

with length

\frac{α}{\sqrt{x^{2} + y^{2} + z^{2}}}

for some positive constant

α .

That is,

v (x, y, z) = - \frac{α}{x^{2} + y^{2} + z^{2}} (x, y, z)

2.1.8.

Solution.

Beginning as in Example 2.1.4, we note

v (x, y) \cdot \hat{ı ı} = x^{2} {\begin{cases} > 0 & x \neq 0 \\ = 0 & x = 0 \end{cases}

and

v (x, y) \cdot \hat{ȷ ȷ} = y {\begin{cases} > 0 & y > 0 \\ = 0 & y = 0 \\ < 0 & y < 0 \end{cases} .

That leads to the following picture:

This gives us a general idea to start with. Refining, we notice that when

x^{2} > | y |,

then the vector

v (x, y)

will be more horizontal than vertical. As we move away from the

y

-axis in a horizontal line, the difference between

x^{2}

and

| y |

grows, so the vectors get more and more horizontal. However, for a fixed value of

x,

vectors farther from the axis will be more vertical than vectors closer to it.

2.1.9.

Solution.

Although ultimately we’ll sketch only unit-length vectors, we can still find the direction of

v (x, y)

by finding its

x

- and

y

components.

Note

v (x, y) \cdot \hat{ı ı}

is the distance from

(x, y)

to the origin, while

v (x, y) \cdot \hat{ȷ ȷ}

is the distance from

(x, y)

to the point

(1, 1) .

Both these numbers are always nonnegative. This leads to the following sketch:

When

(x, y)

is far from the origin, its distance from

(0, 0)

is almost the same as its distance from

(1, 0) .

So, we expect

v (x, y)

to be approximately a scalar multiple of

(1, 1) .

(0, 0),

v (0, 0) \cdot \hat{ı ı} = 0,

so our vector is horizontal; similarly,

v (1, 1) \cdot \hat{ȷ ȷ} = 0

so this vector is horizontal. Vectors very near to

(0, 0)

are nearly horizontal, while vectors near to

(1, 1)

are nearly vertical.

For the direction field, we normalize our vectors to have unit length.

2.1.10.

Solution.

The sign of

v (x, y) \cdot \hat{ı ı} = x (x + y)

depends on the signs of

x

and

x + y .

When they have the same signs,

v (x, y) \cdot \hat{ı ı}

is positive, so

v (x, y)

points to the right; when they have different signs,

v (x, y)

points to the left.

Similarly, the sign of

v (x, y) \cdot \hat{ȷ ȷ} = y (y - x)

depends on the signs of

y

and

y - x .

All together:

Refining, we notice that as we move straight up or down,

| v (x, y) \cdot \hat{ı ı} |

has its minimum along the lines

y = - x

and

x = 0 .

So, the vectors become more strongly vertical as we approach

y = - x

and

x = 0

from above or below.

Similarly,

| v (x, y) \cdot \hat{ȷ ȷ} |

has its minima along the lines

y = x

and

y = 0,

so the vectors become more strongly horizontal as we approach

y = x

horizontally.

2.1.11.

Solution.

The field

v (x, y)

is the sum, scaled by 1/3, of the unit vector pointing away from the origin and the unit vector pointing away from

(1, 0) .

This tells us about a few regions:

Along the $x$ axis between $(0, 0)$ and $(1, 0),$ the vectors away from these points are pointing in opposite directions (and have the same length), so they cancel each other out. That is, $v (x, 0) = 0$ for all $x \in (0, 1) .$
$v (0, 0)$ and $v (1, 0)$ are not defined.
Along the $x$ -axis outside of $[0, 1],$ the vector pointing away from the point $(0, 0)$ is the same as the vector pointing away from the point $(1, 0) .$ So, $v (x, 0) = (- 2 / 3, 0)$ for $x < 0$ and $v (x, 0) = (2 / 3, 0)$ for $x > 1 .$
As the distance from $(x, y)$ to the origin grows, the vector pointing away from $(0, 0)$ looks more and more like the vector pointing away from $(1, 0) .$ So, our vectors far away from the origin look like vectors of length about 2/3, pointing away from the origin.

2.1.12.

Solution.

(a) The vector field

v (x, y) = x \hat{ı ı} + y \hat{ȷ ȷ}

is the same as the radius vector. It points radially outward and has length growing linearly with the distance from the origin.

(b) The vertical component of

v (x, y) = 2 x \hat{ı ı} - \hat{ȷ ȷ}

is always

- 1 .

Its horizontal component is

2 x,

so that

$v (x, y)$ is rightward pointing when $x > 0$ and leftward pointing when $x < 0,$ and
the magnitude of the horizontal component grows linearly with the distance from the $y$ -axis.

It is sketched in the figure on the left below.

(x, y)

the vector

v (x, y) = \frac{y \hat{ı ı} - x \hat{ȷ ȷ}}{\sqrt{x^{2} + y^{2}}}

is of length $1$ and
is perpendicular to the radius vector $x \hat{ı ı} + y \hat{ȷ ȷ} .$
$v (x, y)$ is rightward pointing when $y > 0$ and leftward pointing when $y < 0,$ and
$v (x, y)$ is downward pointing when $x > 0$ and upward pointing when $x < 0 .$

It is sketched in the figure on the right above.

2.1.13.

Solution.

A particle of unit mass at position

(x, y)

has distance

D_{1} = \sqrt{x^{2} + y^{2}}

from the 5kg mass, so that mass exerts a force of magnitude

\frac{G (5)}{x^{2} + y^{2}}

on the particle. This force has direction

(- x, - y) .

So, the force exerted by the 5kg mass is

f_{1} (x, y) = \frac{- 5 G}{(x^{2} + y^{2})^{3 / 2}} (x, y) .

Similarly, the 3 kg mass at

(2, 3)

exerts a force of

f_{2} (x, y) = \frac{3 G}{((x - 2)^{2} + (y - 3)^{2})^{3 / 2}} (2 - x, 3 - y);

and the 7 kg mass at

(4, 0)

exerts a force of

f_{3} (x, y) = \frac{7 G}{((x - 4)^{2} + y^{2})^{3 / 2}} (4 - x, - y) .

The net force on a unit mass is therefore

\begin{aligned} f (x, y) & = f_{1} (x, y) + f_{2} (x, y) + f_{3} (x, y) \\ = \frac{- 5 G (x, y)}{(x^{2} + y^{2})^{3 / 2}} + \frac{3 G (2 - x, 3 - y)}{((x - 2)^{2} + (y - 3)^{2})^{3 / 2}} + \frac{7 G (4 - x, - y)}{((x - 4)^{2} + y^{2})^{3 / 2}} \end{aligned}

2.1.14.

Solution.

Consider a point $P$ on the pole that is a distance $p$ away from the bottom end. Use this point to make a smaller right triangle, as in the picture below.

Using similar triangles:

$\begin{aligned} h & = \frac{p}{2} H & b & = \frac{p}{2} \sqrt{4 - H^{2}} \end{aligned}$
If $P$ is at position $(x, y),$ then:
$\begin{aligned} y & = h = \frac{p}{2} H & x & = \sqrt{4 - H^{2}} - b = (1 - \frac{p}{2}) \sqrt{4 - H^{2}} \\ \frac{d y}{d t} & = \frac{p}{2} \frac{d H}{d t} = - \frac{p}{4} & \frac{d x}{d t} & = (1 - \frac{p}{2}) \frac{- H}{\sqrt{4 - H^{2}}} \frac{d H}{d t} \\ = (1 - \frac{p}{2}) \frac{H}{2 \sqrt{4 - H^{2}}} \end{aligned}$

When $H = 1 :$

$\begin{aligned} {\frac{d y}{d t} |}_{H = 1} & = - \frac{p}{4} & {\frac{d x}{d t} |}_{H = 1} & = (1 - \frac{p}{2}) \frac{1}{2 \sqrt{3}} \end{aligned}$

Therefore,

$v (p) = {(\frac{d x}{d t}, \frac{d y}{d t}) |}_{H = 1} = ((1 - \frac{p}{2}) \frac{1}{2 \sqrt{3}}, - \frac{p}{4})$

For our model, we set the domain of this function to be $[0, 2] .$
Let’s start by seeing what we can salvage from our work on part a. As in part a., consider a point $P$ on one of the poles, $p$ metres from the bottom end.

Let $P$ have position $(x, y, z) .$ Noting that $\frac{d H}{d t}$ is now positive, not negative, if we stick to this two-dimensional slice,

$V (p) = ((1 - \frac{p}{2}) \frac{- 1}{2 \sqrt{3}}, \frac{p}{4})$

where the second coordinate is $z$ and the first coordinate refers to the (horizontal) line in the direction of the vector $(x, y, 0) .$

So, we know ${\frac{d z}{d t} |}_{H = 1} = \frac{p}{4},$ and we know ${(\frac{d x}{d t}, \frac{d y}{d t}) |}_{H = 1} = (x, y) c$ for some negative constant $c$ with $| (x, y) c | = (1 - \frac{p}{2}) \frac{1}{2 \sqrt{3}} .$ Since we have the direction and the magnitude of the vector, we can find the vector:

${(\frac{d x}{d t}, \frac{d y}{d t}) |}_{H = 1} = (x, y) c = - \frac{(1 - \frac{p}{2})}{2 \sqrt{3} \sqrt{x^{2} + y^{2}}} (x, y)$

We want our equation to be in terms of $x,$ $y,$ and $z,$ so we need to get rid of $p .$ Using similar triangles, $\frac{p}{2} = \frac{\sqrt{4 - H^{2}} - \sqrt{x^{2} + y^{2}}}{\sqrt{4 - H^{2}}} .$ When $H = 1,$ then $1 - \frac{p}{2} = \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{3}} .$ So:

${(\frac{d x}{d t}, \frac{d y}{d t}) |}_{H = 1} = - \frac{1}{6} (x, y)$

Finally:

$V (x, y, z) = {(\frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t}) |}_{H = 1} = (- \frac{1}{6} x, - \frac{1}{6} y, \frac{1}{2} z)$

Not all values of $(x, y, z)$ are on the frame. But, for those values of $(x, y, z)$ that are on the frame, this equation holds.

2.2 Optional — Field Lines
2.2.2 Exercises

2.2.2.1.

Solution.

(a) At every point of the positive

y

-axis, the velocity vector

v (0, y)

points straight down. So a rubber ducky placed in the water at

(0, 2)

just floats straight down the positive

y

-axis towards the origin.

(b) At every point of the positive

x

-axis, the velocity vector

v (x, 0)

points straight to the right. So a rubber ducky placed in the water at

(1, 0)

just floats rightward along the positive

x

-axis.

v (x, y)

points downwards and towards the right. So a rubber ducky placed in the water at

(1, 2)

always floats down and to the right. The closer the ducky gets to the

x

--axis the more rightward its motion becomes.

2.2.2.2.

Solution.

The derivatives

\begin{aligned} x^{'} (t) & = - e^{- t} \cos t - e^{- t} \sin t & = - x (t) - y (t) \\ y^{'} (t) & = - e^{- t} \sin t + e^{- t} \cos t & = - y (t) + x (t) \end{aligned}

(x (t), y (t))

is a solution of the system of differential equations

\begin{aligned} \frac{d x}{d t} & = v_{1} (x, y) = - x - y \\ \frac{d y}{d t} & = v_{2} (x, y) = x - y \end{aligned}

So the vector field is

v (x, y) = (v_{1} (x, y), v_{2} (x, y)) = (- x - y, x - y) .

2.2.2.3. (✳).

Solution.

(a) The field lines of

F (x, y) = \nabla \nabla f = y \hat{ı ı} + x \hat{ȷ ȷ}

obey

\begin{matrix} \frac{d x}{y} = \frac{d y}{x} ⟺ x d x = y d y ⟺ \frac{x^{2}}{2} = \frac{y^{2}}{2} + C \end{matrix}

for any constant

C .

(b) The sign data

\begin{array}{r} \hat{ı ı} \cdot F (x, y) = y {\begin{cases} > 0 & if y > 0 \\ = 0 & if y = 0 \\ < 0 & if y < 0 \end{cases}} \\ \hat{ȷ ȷ} \cdot F (x, y) = x {\begin{cases} > 0 & if x > 0 \\ = 0 & if x = 0 \\ < 0 & if x < 0 \end{cases}} \end{array}

is visually displayed in the figure on the left below. The arrows in the figure on the left gives us the direction of motion along the field lines

\frac{x^{2}}{2} = \frac{y^{2}}{2} + C

(in red) in the figure on the right below. Some equipotential curves

x y = C

are also sketched (in blue) in the figure on the right below.

2.2.2.4. (✳).

Solution.

The field lines obey

\frac{d x}{2 y} = \frac{d y}{x / y^{2}} = \frac{d z}{e^{y}} if x, y \neq 0

In particular

\begin{matrix} \frac{d x}{2 y} = \frac{y^{2} d y}{x} ⟹ x d x = 2 y^{3} d y ⟹ \frac{1}{2} x^{2} = \frac{1}{2} y^{4} + C \end{matrix}

Since

y = 1

when

x = 1,

C = 0 .

x = y^{2}

and

\frac{d y}{x / y^{2}} = \frac{d z}{e^{y}} ⟹ e^{y} d y = d z ⟹ z = e^{y} + D

Since

z = e

when

y = 1,

D = 0 .

So the field line is

x = y^{2} z = e^{y}

2.2.2.5. (✳).

Solution.

The field lines obey

\begin{aligned} \frac{d x}{x} = \frac{d y}{3 y} if x, y \neq 0 \\ ⟹ 3 \ln | x | = \ln | y | + C \\ ⟹ | x |^{3} = e^{C} | y | \\ ⟹ y = \pm e^{- C} x^{3} \\ ⟹ y = C^{'} x^{3} \end{aligned}

with

C^{'}

a nonzero constant.

x = 0

and

y = 0

are also field lines, since on the

y

-axis

F ∥ \hat{ȷ ȷ}

and on the

x

-axis

F ∥ \hat{ı ı} .

2.3 Conservative Vector Fields

Exercises

2.3.1.

Solution.

False, in general.

In the context of Equation 1.7.1, the only forces acting on the particle are gravity,

- m g \hat{ȷ ȷ},

and the normal force,

W \hat{N} .

We make no such constraints on the force in Example 2.3.3. Certainly

F

could arise from gravity and the normal force of a track, but there’s nothing saying it has to. For example, suppose

φ

is an equation that does not depend on

m

and/or

g .

Alternately, suppose the

y

-coordinate of our three-dimensional system is not “up.”

2.3.2.

Solution.

Remember that the screening test can only rule out conservativity — it can never, by itself, guarantee conservativity. So, A is never the case.

\begin{aligned} F & = x \hat{ı ı} + z \hat{ȷ ȷ} + y \hat{k} \\ \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \\ = (1 - 1) \hat{ı ı} + (0 - 0) \hat{ȷ ȷ} + (0 - 0) \hat{k} = 0 \end{aligned}

This field passes the screening test. That means the screening test doesn’t rule out the possibility of

F

being conservative. So, we have option C.

\begin{aligned} F & = y^{2} z \hat{ı ı} + x^{2} z \hat{ȷ ȷ} + x^{2} y \hat{k} \\ \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \\ = (x^{2} - x^{2}) \hat{ı ı} + (y^{2} - 2 x y) \hat{ȷ ȷ} + (2 x z - 2 y z) \hat{k} \neq 0 \end{aligned}

So,

F

fails the screening test — it’s not conservative. That’s option B.

\begin{aligned} F & = (y e^{x y} + 1) \hat{ı ı} + (x e^{x y} + z) \hat{ȷ ȷ} + (\frac{1}{z} + y) \hat{k} \\ \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \\ = (1 - 1) \hat{ı ı} + (0 - 0) \hat{ȷ ȷ} + (e^{x y} (x y + 1) - e^{x y} (x y + 1)) \hat{k} = 0 \end{aligned}

F

passes the screening test, so it may or may not be conservative. That is Option C.

\begin{aligned} F & = y \cos (x y) \hat{ı ı} + x \sin (x y) \hat{ȷ ȷ} \\ \frac{\partial F_{2}}{\partial x} & = x y \cos (x y) + \sin (x y) \\ \frac{\partial F_{1}}{\partial y} & = - x y \sin (x y) + \cos (x y) \\ \frac{\partial F_{2}}{\partial x} & \neq \frac{\partial F_{1}}{\partial y} \end{aligned}

F

fails the screening test, so it is not conservative. That is Option B.

2.3.3.

Solution.

Let

φ

be a potential for

F .

Define

ϕ = φ + a x + b y + c z .

Then

\nabla \nabla ϕ = \nabla \nabla φ + (a, b, c) = F + (a, b, c) .

So,

F + (a, b, c)

is also conservative.

2.3.4.

Solution.

If $F + G$ is conservative for any particular $F$ and $G,$ then by definition, there exists a potential $φ$ with $F + G = \nabla \nabla φ .$

Since $F$ is conservative, there also exists a potential $ψ$ with $F = \nabla \nabla ψ .$

But now $G = (F + G) - F = \nabla \nabla φ - \nabla \nabla ψ = \nabla \nabla (φ - ψ) .$ That means the function $(φ - ψ)$ is a potential for $G .$ However, this is impossible: since $G$ is non-conservative, no function with this property exists.

So it is not possible that $F + G$ is conservative. It must be non-conservative.
Counterexample: if $F = - G,$ then $F + G = 0 = \nabla \nabla c$ for any constant $c .$
Since both fields are conservative, they both have potentials, say $F = \nabla \nabla φ$ and $G = \nabla \nabla ψ .$ Then $F + G = \nabla \nabla φ + \nabla \nabla ψ = \nabla \nabla (φ + ψ) .$ That is, $(φ + ψ)$ is a potential for $F + G,$ so $F + G$ is conservative.

2.3.5. (✳).

Solution.

Set

φ (x, y) = \arctan \frac{y}{x}

(using the standard

\arctan

that takes values between

- \frac{π}{2}

and

\frac{π}{2}

). Note that

φ (x, y)

is well-defined, with all partial derivatives continuous, on

D

since

x > 1

there. Then

\begin{aligned} \frac{\partial φ}{\partial x} (x, y) & = \frac{- \frac{y}{x^{2}}}{1 + (\frac{y}{x})^{2}} & = - \frac{y}{x^{2} + y^{2}} \\ \frac{\partial φ}{\partial y} (x, y) & = \frac{\frac{1}{x}}{1 + (\frac{y}{x})^{2}} & = \frac{x}{x^{2} + y^{2}} \end{aligned}

so that

F = \nabla \nabla φ .

2.3.6.

Solution.

φ

is a potential for

F,

then:

$\frac{\partial φ}{\partial x} = x + y,$ so $φ = \frac{1}{2} x^{2} + x y + ψ_{1} (y)$
$\frac{\partial φ}{\partial y} = x - y,$ so $φ = x y - \frac{1}{2} y^{2} + ψ_{2} (x)$

So, for instance,

φ = \frac{1}{2} x^{2} + x y - \frac{1}{2} y^{2}

is a potential for

F .

2.3.7.

Solution.

φ

is a potential for

F,

then:

$\frac{\partial φ}{\partial x} = \frac{1}{x} - \frac{1}{y},$ so $φ = \ln | x | - \frac{x}{y} + ψ_{1} (y)$
$\frac{\partial φ}{\partial y} = \frac{x}{y^{2}},$ so $φ = - \frac{x}{y} + ψ_{2} (x)$

So, for instance,

φ = \ln | x | - \frac{x}{y}

is a potential for

F .

2.3.8.

Solution.

None exists:

\frac{\partial F_{2}}{\partial z} = \frac{1}{3} x^{3},

while

\frac{\partial F_{3}}{\partial y} = \frac{1}{3} x^{3} + 1,

F

fails the screening test, Theorem 2.3.9.

2.3.9.

Solution.

φ (x, y, z)

is a potential for

F (x, y, z),

then:

$\frac{\partial φ}{\partial x} (x, y, z) = \frac{x}{x^{2} + y^{2} + z^{2}},$ so $φ (x, y, z) = \frac{1}{2} \ln (x^{2} + y^{2} + z^{2}) + ψ_{1} (y, z)$
$\frac{\partial φ}{\partial y} (x, y, z) = \frac{y}{x^{2} + y^{2} + z^{2}},$ so $φ (x, y, z) = \frac{1}{2} \ln (x^{2} + y^{2} + z^{2}) + ψ_{2} (x, z)$
$\frac{\partial φ}{\partial z} (x, y, z) = \frac{z}{x^{2} + y^{2} + z^{2}},$ so $φ (x, y, z) = \frac{1}{2} \ln (x^{2} + y^{2} + z^{2}) + ψ_{2} (x, y)$

So, for instance,

φ (x, y, z) = \frac{1}{2} \ln (x^{2} + y^{2} + z^{2})

is a potential for

F (x, y, z) .

2.3.10.

Solution.

(a) We shall show that

F (x, y, z)

is conservative by finding a potential for it.

φ (x, y, z)

is a potential for this

F

if and only if

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = x \\ \frac{\partial φ}{\partial y} (x, y, z) & = - 2 y \\ \frac{\partial φ}{\partial z} (x, y, z) & = 3 z \end{aligned}

Integrating the first of these equations gives

φ (x, y, z) = \frac{x^{2}}{2} + f (y, z)

Substituting this into the second equation gives

\frac{\partial f}{\partial y} (y, z) = - 2 y

which integrates to

f (y, z) = - y^{2} + g (z)

Finally, substituting

φ (x, y, z) = \frac{x^{2}}{2} - y^{2} + g (z)

into the last equation gives

g^{'} (z) = 3 z

which integrates to

g (z) = \frac{3}{2} z^{2} + C

with

C

being an arbitrary constant. So,

F (x, y, z)

is conservative and

φ (x, y, z) = \frac{1}{2} x^{2} - y^{2} + \frac{3}{2} z^{2}

is one allowed potential.

(b) The field

F = F_{1} \hat{ı ı} + F_{2} \hat{ȷ ȷ}

can be conservative only if it passes the screening test

\frac{\partial F_{1}}{\partial y} = \frac{\partial F_{2}}{\partial x}

In this case

\frac{\partial F_{1}}{\partial y} = \frac{\partial}{\partial y} (\frac{x}{x^{2} + y^{2}}) = - \frac{2 x y}{{(x^{2} + y^{2})}^{2}}

is different from

\frac{\partial F_{2}}{\partial x} = \frac{\partial}{\partial x} (\frac{- y}{x^{2} + y^{2}}) = \frac{2 x y}{{(x^{2} + y^{2})}^{2}}

for all

(x, y)

with

x

and

y

both nonzero. So

F

is not conservative.

2.3.11.

Solution.

By Theorem 2.4.8, the field

F = F_{1} \hat{ı ı} + F_{2} \hat{ȷ ȷ} + F_{3} \hat{k}

is conservative only if it passes the screening test

\nabla \nabla \times F = 0 .

That is, if and only if

\begin{matrix} \frac{\partial F_{1}}{\partial y} = \frac{\partial F_{2}}{\partial x} \frac{\partial F_{1}}{\partial z} = \frac{\partial F_{3}}{\partial x} \frac{\partial F_{2}}{\partial z} = \frac{\partial F_{3}}{\partial y} \end{matrix}

or,

\begin{aligned} \frac{\partial}{\partial y} (e^{(z^{2})}) & = \frac{\partial}{\partial x} (2 B y z^{3}) & ⟺ & 0 & = 0 \\ \frac{\partial}{\partial z} (e^{(z^{2})}) & = \frac{\partial}{\partial x} (A x z e^{(z^{2})} + 3 B y^{2} z^{2}) & ⟺ & 2 z e^{(z^{2})} & = A z e^{(z^{2})} \\ \frac{\partial}{\partial z} (2 B y z^{3}) & = \frac{\partial}{\partial y} (A x z e^{(z^{2})} + 3 B y^{2} z^{2}) & ⟺ & 6 B y e^{(z^{2})} & = 6 B y e^{(z^{2})} \end{aligned}

Hence only

A = 2

works. We shall see in part (b) that any

B

works.

(b) When

A = 2,

and

B

is any real number.

F = e^{(z^{2})} \hat{ı ı} + 2 B y z^{3} \hat{ȷ ȷ} + (2 x z e^{(z^{2})} + 3 B y^{2} z^{2}) \hat{k}

φ (x, y, z)

is a potential for this

F

if and only if

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = e^{(z^{2})} \\ \frac{\partial φ}{\partial y} (x, y, z) & = 2 B y z^{3} \\ \frac{\partial φ}{\partial z} (x, y, z) & = 2 x z e^{(z^{2})} + 3 B y^{2} z^{2} \end{aligned}

Integrating the first of these equations gives

φ (x, y, z) = x e^{(z^{2})} + f (y, z)

Substituting this into the second equation gives

\frac{\partial f}{\partial y} (y, z) = 2 B y z^{3}

which integrates to

f (y, z) = B y^{2} z^{3} + g (z)

Finally, substituting

φ (x, y, z) = x e^{(z^{2})} + B y^{2} z^{3} + g (z)

into the last equation gives

2 x z e^{(z^{2})} + 3 B y^{2} z^{2} + g^{'} (z) = 2 x z e^{(z^{2})} + 3 B y^{2} z^{2} or g^{'} (z) = 0

which integrates to

g (z) = C

with

C

being an arbitrary constant. So, for each real number

B,

φ (x, y, z) = x e^{(z^{2})} + B y^{2} z^{3}

is one allowed potential.

2.3.12.

Solution.

In each second

2 π m

{c m}^{2}

of fluid crosses each circle of radius

r

(and hence circumference

2 π r

) centred on the origin. So the speed of flow at radius

r

\frac{m}{r} .

As the direction of flow is radially outward

\begin{matrix} v = m \frac{x \hat{ı ı} + y \hat{ȷ ȷ}}{x^{2} + y^{2}} \end{matrix}

φ (x, y)

is a potential for this

F

if and only if

\begin{aligned} \frac{\partial φ}{\partial x} (x, y) & = m \frac{x}{x^{2} + y^{2}} \\ \frac{\partial φ}{\partial y} (x, y) & = m \frac{y}{x^{2} + y^{2}} \end{aligned}

Integrating the first of these equations gives

φ (x, y) = \frac{1}{2} m \ln (x^{2} + y^{2}) + f (y)

Substituting this into the second equation gives

m \frac{y}{x^{2} + y^{2}} + f^{'} (y) = m \frac{y}{x^{2} + y^{2}} or f^{'} (y) = 0

which integrates to

f (y) = C

with

C

an arbitrary constant. So one possible potential is

φ = \frac{1}{2} m \ln (x^{2} + y^{2})

2.3.13.

Solution.

Following Example 2.3.3, the particle can never escape the region

{(x, y, z) : φ (x, y, z) \geq - E} .

So, we should find

E,

then figure out the region.

The kinetic energy of the particle is

\frac{1}{2} m | v |^{2},

so the total energy of the system (also the kinetic energy when the potential energy is 0) is

\frac{1}{2} (10) (2^{2}) = 20

Therefore, a region it can never escape is

{(x, y, z) | φ (x, y, z) \geq - 20}

that is,

{(x, y, z) | x^{2} + y^{2} + z^{2} \leq 20}

So, it can never escape the sphere centred at the origin with radius

\sqrt{20} .

2.3.14.

Solution.

Example 2.3.3 tells us

\frac{1}{2} m | v (t) |^{2} - φ (x (t), y (t), z (t)) = E

is a constant quantity, provided

F

is conservative with potential

φ .

So, it would be nice if

F

were conservative.

F = \nabla \nabla φ,

then

$\frac{\partial φ}{\partial x} = 0,$ so $φ = ψ_{1} (y, z)$
$\frac{\partial φ}{\partial y} = 1,$ so $φ = y + ψ_{2} (x, z)$
$\frac{\partial φ}{\partial z} = 3 z^{1 / 3},$ so $φ = \frac{9}{4} z^{4 / 3} + ψ_{3} (x, y)$

We can choose

φ (x, y, z) = y + \frac{9}{4} z^{4 / 3} .

So,

\frac{1}{2} m | v (t) |^{2} - φ (x (t), y (t), z (t)) = E

is a constant quantity, as desired. Using the information that the particle has mass

1 / 2,

and speed

1

when it is at the origin:

\begin{aligned} E & = \frac{1}{2} \cdot \frac{1}{2} | 1 |^{2} - φ (0, 0, 0) = \frac{1}{4} \end{aligned}

When the particle is at $(1, 1, 1) :$

\begin{aligned} \frac{1}{4} & = \frac{1}{2} \cdot \frac{1}{2} | v |^{2} - φ (1, 1, 1) = \frac{| v |^{2}}{4} - (1 + \frac{9}{4}) \\ | v | & = \sqrt{14} \end{aligned}

So, at the point

(1, 1, 1),

the particle has speed

\sqrt{14} .

2.3.15.

Solution.

We can start with the screening test, Theorem 2.3.9.

\begin{aligned} \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \\ = (g^{'} (y) h^{'} (z) - g^{'} (y) h^{'} (z)) \hat{ı ı} + (0 - 0) \hat{ȷ ȷ} + (0 - 0) \hat{k} = 0 \end{aligned}

So, it’s possible that the field is conservative. Remember, this test alone isn’t enough to tell us it’s conservative. (Had the test come out differently, though, we’d be done.)

Suppose

F = \nabla \nabla φ (x, y, z) .

Then:

$\frac{\partial φ}{\partial x} = 2 f (x) f^{'} (x) .$ By inspection, we see $φ = f^{2} (x) + ψ_{1} (y, z) .$ (We could also find this by evaluating $\int 2 f (x) f^{'} (x) d x$ with the substitution $u = f (x) .$ )
$\frac{\partial φ}{\partial y} = g^{'} (y) h (z),$ so $φ = g (y) h (z) + ψ_{2} (x, z) .$
$\frac{\partial φ}{\partial z} = g (y) h^{'} (z),$ so $φ = g (y) h (z) + ψ_{2} (x, y) .$

All together, we can choose

φ (x, y, z) = f^{2} (x) + g (y) h (z) .

2.3.16.

Solution.

Following Definition 2.3.8, The curl of a vector field is defined by

\begin{aligned} \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \end{aligned}

When $F = ⟨ x y, x z, y^{2} + z ⟩,$

\begin{aligned} \nabla \nabla \times F & = (2 y - x) \hat{ı ı} + (0 - 0) \hat{ȷ ȷ} + (z - x) \hat{k} \end{aligned}

When the curl is

0 \hat{ı ı} + 0 \hat{ȷ ȷ} + 0 \hat{k},

we have

x = 2 y

and

x = z .

That is, our points are of the form

(2 c, c, 2 c)

for any constant

c .

So, the region in question is the line through the origin in the direction of the vector

(2, 1, 2) .

2.4 Line Integrals
2.4.2 Exercises

2.4.2.1.

Solution.

The square has four sides, each of which is a line segment.

On the first side, $y = 0$ and $d y = 0 .$ That is, we may parametrize the first side by $r (x) = x \hat{ı ı}$ with $0 \leq x \leq 1 .$
On the second side, $x = 1$ and $d x = 0 .$ We may parametrize the second side by $r (y) = \hat{ı ı} + y \hat{ȷ ȷ}$ with $0 \leq y \leq 1 .$
On the third side, $y = 1$ and $d y = 0 .$ We may parametrize the third side by $r (x) = x \hat{ı ı} + \hat{ȷ ȷ}$ with $x$ running from $1$ to $0 .$
On the final side, $x = 0$ and $d x = 0 .$ We may parametrize the fourth side by $r (y) = y \hat{ȷ ȷ}$ with $y$ running from $1$ to $0 .$

\begin{aligned} \int_{C} x^{2} y^{2} d x + x^{3} y d y & = \int_{0}^{1} x^{2} \times 0^{2} d x + \int_{0}^{1} 1^{3} \times y d y + \int_{1}^{0} x^{2} \times 1^{2} d x \\ + \int_{1}^{0} 0^{3} \times y d y \\ = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \end{aligned}

2.4.2.2.

Solution.

Every

F

in this problem is defined and has continuous first-order partial derivatives on all of

R^{2}

R^{3} .

The characterization in Theorem 2.4.8 tells us that our fields will be conservative if and only if they pass the screening test, i.e. have curl 0.

\begin{aligned} F & = x \hat{ı ı} + z \hat{ȷ ȷ} + y \hat{k} \\ \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \\ = (1 - 1) \hat{ı ı} + (0 - 0) \hat{ȷ ȷ} + (0 - 0) \hat{k} = 0 \end{aligned}

This field passes the screening test. Since

F

is defined and has continuous first-order partial derivatives on all of

R^{3},

it is conservative. So, we have option A.

\begin{aligned} F & = y^{2} z \hat{ı ı} + x^{2} z \hat{ȷ ȷ} + x^{2} y \hat{k} \\ \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \\ = (x^{2} - x^{2}) \hat{ı ı} + (y^{2} - 2 x y) \hat{ȷ ȷ} + (2 x z - 2 y z) \hat{k} \neq 0 \end{aligned}

So,

F

fails the screening test. So, it’s not conservative. That’s option B.

\begin{aligned} F & = (y e^{x y} + 1) \hat{ı ı} + (x e^{x y} + z) \hat{ȷ ȷ} + (z + y) \hat{k} \\ \nabla \nabla \times F & = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{ı ı} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k} \\ = (1 - 1) \hat{ı ı} + (0 - 0) \hat{ȷ ȷ} + {e^{x y} (x y + 1) - e^{x y} (x y + 1)} \hat{k} = 0 \end{aligned}

F

passes the screening test. Since

F

is defined and has continuous first-order partial derivatives on all of

R^{3},

it is conservative. So, we have option A.

\begin{aligned} F & = y \cos (x y) \hat{ı ı} + x \sin (x y) \hat{ȷ ȷ} \\ \frac{\partial F_{2}}{\partial x} & = x y \cos (x y) + \sin (x y) \\ \frac{\partial F_{1}}{\partial y} & = - x y \sin (x y) + \cos (x y) \\ \frac{\partial F_{2}}{\partial x} & \neq \frac{\partial F_{1}}{\partial y} \end{aligned}

F

fails the screening test, so it is not conservative. That is Option B.

2.4.2.3.

Solution.

Since

F

is conservative,

\int_{C} F \cdot d r = 0

over any closed curve

C .

The given curve is closed, so the integral is simply zero.

2.4.2.4.

Solution.

Since

F

is conservative, and

A

and

B

start and end at the same points, by path-independence

\int_{B} F \cdot d r = \int_{A} F \cdot d r = 5.

2.4.2.5. (✳).

Solution.

Note

F

is defined and continuous on all of

R^{3} .

By Theorem 2.4.7, the integral

\int_{C} F \cdot d r = 0

for all closed paths

C

if and only if

F

is conservative. Furthermore,

F

has continuous first-order partial derivatives on all of

R^{3} .

Using Theorem 2.4.8,

F

is conservative if and only if it passes the scsreening test

\nabla \nabla \times F = 0 :

\begin{aligned} 0 = \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{x} \sin y & a e^{x} \cos y + b z & c x \end{array}] \\ = (0 - b) \hat{ı ı} - (c - 0) \hat{ȷ ȷ} + (a e^{x} \cos y - e^{x} \cos y) \hat{k} \end{aligned}

This is the case if and only if

a = 1, b = c = 0 .

2.4.2.6.

Solution.

(a) Consider the circle

C

in the figure (a) on the left below, oriented clockwise. The vector field

F

is in the same direction as

\frac{d r}{d t}

at every point of the curve. So

F \cdot \frac{d r}{d t} > 0

at every point of

C

and

C

is a closed curve with

\oint_{C} F \cdot d r > 0 .

As a consequence

F

is not conservative.

(a)

(b)

(b) Consider the square in the figure (b) on the right above, oriented counterclockwise. It consists of the four line segments

L_{1},

L_{2},

L_{3}

and

L_{4} .

On all of

L_{1},

L_{2},

L_{3}

we have that

F (r (t)) \cdot r^{'} (t) = 0

because the vector field is perpendicular to the line segment. On

L_{4}

we have

F (r (t)) \cdot r^{'} (t) > 0 .

\begin{aligned} \oint_{C} F \cdot d r & = \int_{L_{1}} F \cdot d r + \int_{L_{2}} F \cdot d r + \int_{L_{3}} F \cdot d r + \int_{L_{4}} F \cdot d r \\ = 0 + 0 + 0 + \int_{L_{4}} F \cdot d r > 0 \end{aligned}

C

is a closed curve with

\oint_{C} F \cdot d r > 0

and

F

is not conservative.

(c) Consider the square in the figure (c) on the left below, oriented counterclockwise. It consists of the four line segments

L_{1},

L_{2},

L_{3}

and

L_{4} .

L_{1}

and

L_{3}

we have that the dot product

F (r (t)) \cdot r^{'} (t) = 0

because the vector field is perpendicular to the line segment. On

L_{2}

we have

F (r (t)) \cdot r^{'} (t) < 0

while on

L_{4}

we have

F (r (t)) \cdot r^{'} (t) > 0 .

The vector field

F

is longer on

L_{4}

than on

L_{2} .

F (r (t)) \cdot r^{'} (t)

has a larger magnitude on

L_{4}

than

L_{2}

and

\begin{aligned} \oint_{C} F \cdot d r & = \int_{L_{1}} F \cdot d r + \int_{L_{2}} F \cdot d r + \int_{L_{3}} F \cdot d r + \int_{L_{4}} F \cdot d r \\ = 0 + \int_{L_{2}} F \cdot d r + 0 + \int_{L_{4}} F \cdot d r > 0 \end{aligned}

C

is a closed curve with

\oint_{C} F \cdot d r > 0

and

F

is not conservative.

(c)

(d)

(d) We are told that one of the four vector fields is conservative. Only the vector field in (d) is left, so it is conservative.

Remark: We can verify that vector field (d) is indeed conservative by observing (look at the figure (d) on the right above) that the

\hat{ı ı}

component of the vector field is exactly zero and that the

\hat{ȷ ȷ}

component depends only on

y .

So the vector field is of the form

F (x, y) = a (y) \hat{ȷ ȷ}

for some function

a (y) .

A (y)

is any antiderivative of

a (y),

we have

F = \nabla \nabla A,

so that

F

is conservative with potential

A (y) .

2.4.2.7. (✳).

Solution.

(a) The (largest possible) domain is

D = {(x, y, z) | x^{2} + y^{2} \neq 0} .

That is, all of

R^{3}

except the points lying along the

z

-axis.

(b) As preliminary computations, let’s find

\begin{aligned} \frac{\partial}{\partial y} (\frac{x - 2 y}{x^{2} + y^{2}}) & = \frac{- 2}{x^{2} + y^{2}} - \frac{2 y (x - 2 y)}{{(x^{2} + y^{2})}^{2}} = \frac{- 2 x^{2} + 2 y^{2} - 2 x y}{{(x^{2} + y^{2})}^{2}} \\ \frac{\partial}{\partial x} (\frac{2 x + y}{x^{2} + y^{2}}) & = \frac{2}{x^{2} + y^{2}} - \frac{2 x (2 x + y)}{{(x^{2} + y^{2})}^{2}} = \frac{- 2 x^{2} + 2 y^{2} - 2 x y}{{(x^{2} + y^{2})}^{2}} \end{aligned}

So the curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{x - 2 y}{x^{2} + y^{2}} & \frac{2 x + y}{x^{2} + y^{2}} & z \end{array}] \\ = (\frac{- 2 x^{2} + 2 y^{2} - 2 x y}{{(x^{2} + y^{2})}^{2}} - \frac{- 2 x^{2} + 2 y^{2} - 2 x y}{{(x^{2} + y^{2})}^{2}}) \hat{k} \\ = 0 \end{aligned}

on the domain of $F$ .

r (t) = 2 \cos t \hat{ı ı} + 2 \sin t \hat{ȷ ȷ} + 3 \hat{k} r^{'} (t) = - 2 \sin t \hat{ı ı} + 2 \cos t \hat{ȷ ȷ}

with

0 \leq θ \leq 2 π .

So the integral is

\begin{aligned} \int_{C} F \cdot d r & = \int_{0}^{2 π} {\overset{\frac{x - 2 y}{x^{2} + y^{2}}}{\overset{⏞}{\frac{2 \cos t - 4 \sin t}{4}}} \hat{ı ı} + \overset{\frac{2 x + y}{x^{2} + y^{2}}}{\overset{⏞}{\frac{4 \cos t + 2 \sin t}{4}}} \hat{ȷ ȷ} + \overset{z}{\overset{⏞}{3}} \hat{k}} \cdot \\ {\overset{r^{'} (t)}{\overset{⏞}{- 2 \sin t \hat{ı ı} + 2 \cos t \hat{ȷ ȷ}}}} d t \\ = \int_{0}^{2 π} \frac{- 4 \sin t \cos t + 8 \sin^{2} t + 8 \cos^{2} t + 4 \sin t \cos t}{4} d t \\ = 2 \int_{0}^{2 π} d t = 4 π \end{aligned}

(d) As the integral of

F

around the simple closed curve

C

is not zero,

F

cannot be conservative. See Theorem 2.4.7 and Examples 2.3.14 and 4.3.8.

2.4.2.8.

Solution.

The point here is that

F

is conservative, as

F = \nabla ϕ

with

ϕ = \frac{x^{2}}{2} + y x - y z + \frac{z^{2}}{2}

So, for all paths from

r (t_{0}) = (1, 0, - 1)

r (t_{1}) = (0, - 2, 3),

\begin{aligned} \int_{C} F \cdot d r = ϕ (r (t_{1})) - ϕ (r (t_{0})) & = ϕ (0, - 2, 3) - ϕ (1, 0, - 1) \\ = [0 + 0 + 6 + \frac{9}{2}] - [\frac{1}{2} + 0 - 0 + \frac{1}{2}] \\ = 9 \frac{1}{2} \end{aligned}

2.4.2.9.

Solution.

Note that:

Along the line segment from $(0, 0)$ to $(1, 0),$ $x$ increases from $0$ to $1,$ while $y$ is held fixed at $y = 0 .$ So we may parametrize this segment by $r (x) = x \hat{ı ı},$ $0 \leq x \leq 1 .$
Along the line segment from $(1, 0)$ to $(1, π),$ $y$ increases from $0$ to $π,$ while $x$ is held fixed at $x = 1 .$ So we may parametrize this segment by $r (x) = \hat{ı ı} + y \hat{ȷ ȷ},$ $0 \leq y \leq π .$
Along the line segment from $(1, π)$ to $(0, π),$ $x$ decreases from $1$ to $0,$ while $y$ is held fixed at $y = π .$ So we may parametrize this segment by $r (x) = x \hat{ı ı} + π \hat{ȷ ȷ}$ with $x$ running from $1$ to $0 .$

Hence

\begin{aligned} \int_{C} V \cdot d r & = \int_{0}^{1} V (x, 0) \cdot \hat{ı ı} d x + \int_{0}^{π} V (1, y) \cdot \hat{ȷ ȷ} d y + \int_{1}^{0} V (x, π) \cdot \hat{ı ı} d x \\ = \int_{0}^{1} (e^{x} + x^{2}) d x + \int_{0}^{π} (y + 3) d y + \int_{1}^{0} (- e^{x} + x^{2}) d x \\ = 2 \int_{0}^{1} e^{x} d x + \int_{0}^{π} (y + 3) d y \\ = 2 (e - 1) + \frac{π^{2}}{2} + 3 π \end{aligned}

2.4.2.10.

Solution.

(a) We may parametrize the curve by

r (t) = t \hat{ı ı} + t^{2} \hat{ȷ ȷ}

with

0 \leq t \leq 1 .

Then

v (t) = \frac{d r}{d t} (t) = \hat{ı ı} + 2 t \hat{ȷ ȷ}

and

F (x (t), y (t)) = t^{3} \hat{ı ı} - t^{2} \hat{ȷ ȷ}

\begin{aligned} \int_{C} F \cdot d r & = \int_{0}^{1} F (x (t), y (t)) \cdot \frac{d r}{d t} (t) d t = \int_{0}^{1} [t^{3} \hat{ı ı} - t^{2} \hat{ȷ ȷ}] \cdot [\hat{ı ı} + 2 t \hat{ȷ ȷ}] d t \\ = \int_{0}^{1} [- t^{3}] d t \\ = - \frac{1}{4} \end{aligned}

(b) The path is the union of three line segments.

On the first segment of the path $y = z = 0$ so $F$ simplifies to $x \hat{ı ı} - x \hat{k}$ and $d r = \hat{ı ı} d x$ (i.e. we can parametrize the first segment of the path by $r (x) = x \hat{ı ı}$ with $0 \leq x \leq 1$ ), so $F \cdot d r = x d x .$
On the second segment of the path $x = 1,$ $z = 0$ so $F$ simplifies to $\hat{ı ı} + y \hat{ȷ ȷ} - (1 + y) \hat{k}$ and $d r = \hat{ȷ ȷ} d y$ (parametrize the second segment of the path by $r (y) = \hat{ı ı} + y \hat{ȷ ȷ}$ with $0 \leq y \leq 1$ ), so $F \cdot d r = y d y .$
On the final segment of the path $x = y = 1$ so $F$ simplifies to $(1 - z) \hat{ı ı} + (1 - z) \hat{ȷ ȷ} - 2 \hat{k}$ and $d r = \hat{k} d z$ (parametrize the third segment of the path by $r (z) = \hat{ı ı} + \hat{ȷ ȷ} + z \hat{k}$ with $0 \leq z \leq 1$ ), so $F \cdot d r = - 2 d z .$

\int_{C} F \cdot d r = \int_{0}^{1} x d x + \int_{0}^{1} y d y + \int_{0}^{1} (- 2) d z = \frac{1}{2} + \frac{1}{2} - 2 = - 1

2.4.2.11. (✳).

Solution.

Parametrize the curve using

y

as a parameter. Then

y = t,

x = 2 y = 2 t

and

z = \frac{8}{x y} = \frac{8}{2 t^{2}}

so that:

\begin{aligned} r (t) & = 2 t \hat{ı ı} + t \hat{ȷ ȷ} + \frac{4}{t^{2}} \hat{k}, 1 \leq t \leq 2 \\ r^{'} (t) & = 2 \hat{ı ı} + \hat{ȷ ȷ} - \frac{8}{t^{3}} \hat{k} \\ F (r (t)) & = 4 t^{2} \hat{ı ı} + 4 t^{3} \hat{k} \\ F (r (t)) \cdot r^{'} (t) & = 8 t^{2} - 32 \end{aligned}

Then

\begin{aligned} \int_{C} F \cdot d r & = \int_{1}^{2} F (r (t)) \cdot r^{'} (t) d t = \int_{1}^{2} (8 t^{2} - 32) d t = {[\frac{8}{3} t^{3} - 32 t]}_{1}^{2} \\ = - \frac{40}{3} \end{aligned}

2.4.2.12.

Solution.

(a), (b) The curls of

F

and

G

are

\begin{aligned} \nabla \nabla \times F = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 6 x^{2} y z^{2} & 2 x^{3} z^{2} + 2 y - x z & 4 x^{3} y z \end{array}] \\ = (4 x^{3} z - 4 x^{3} z + x) \hat{ı ı} - (12 x^{2} y z - 12 x^{2} y z) \hat{ȷ ȷ} + (6 x^{2} z^{2} - z - 6 x^{2} z^{2}) \hat{k} \\ = x \hat{ı ı} - z \hat{k} \\ \nabla \nabla \times G = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y z & 0 & x y \end{array}] \\ = x \hat{ı ı} - z \hat{k} \end{aligned}

Hence the screening test for

\nabla \nabla \times (F + λ G) = (x + λ x) \hat{ı ı} - (z + λ z) \hat{k}

passes for

λ = - 1 .

Furthermore

\begin{aligned} F - G & = (6 x^{2} y z^{2} - y z) \hat{ı ı} + (2 x^{3} z^{2} + 2 y - x z) \hat{ȷ ȷ} + (4 x^{3} y z - x y) \hat{k} \\ = \nabla \nabla (2 x^{3} y z^{2} - x y z + y^{2}) \end{aligned}

The potential was found by guessing. Alternatively, we can find it by using that

ϕ (x, y, z)

is a potential for

F - G

if and only if

\begin{aligned} \frac{\partial ϕ}{\partial x} (x, y, z) & = 6 x^{2} y z^{2} - y z \\ \frac{\partial ϕ}{\partial y} (x, y, z) & = 2 x^{3} z^{2} + 2 y - x z \\ \frac{\partial ϕ}{\partial z} (x, y, z) & = 4 x^{3} y z - x y \end{aligned}

Integrating the first of these equations gives

ϕ (x, y, z) = 2 x^{3} y z^{2} - x y z + f (y, z)

Substituting this into the second equation gives

2 x^{3} z^{2} - x z + \frac{\partial f}{\partial y} (y, z) = 2 x^{3} z^{2} + 2 y - x z or \frac{\partial f}{\partial y} (y, z) = 2 y

which integrates to

f (y, z) = y^{2} + g (z)

Finally, substituting

ϕ (x, y, z) = 2 x^{3} y z^{2} - x y z + y^{2} + g (z)

into the last equation gives

4 x^{3} y z - x y + g^{'} (z) = 4 x^{3} y z - x y or g^{'} (z) = 0

which integrates to

g (z) = K

with

K

being an arbitrary constant. Choosing

K = 0

gives the potential

ϕ (x, y, z) = 2 x^{3} y z^{2} - x y z + y^{2}

as in the guess above.

(x, y, z)

on the curve must have

z = x

and

y = e^{x z} = e^{x^{2}} .

So we may parametrize the curve by

r (x) = x \hat{ı ı} + e^{x^{2}} \hat{ȷ ȷ} + x \hat{k},

0 \leq x \leq 1 .

Hence

\begin{aligned} \int_{C} F \cdot d r = \int_{C} (F - G) \cdot d r + \int_{C} G \cdot d r \\ = [2 x^{3} y z^{2} - x y z + y^{2}]_{(0, 1, 0)}^{(1, e, 1)} + \int_{0}^{1} \overset{G (r (x))}{\overset{⏞}{[x e^{x^{2}} \hat{ı ı} + x e^{x^{2}} \hat{k}]}} \cdot \overset{\frac{d r}{d x}}{\overset{⏞}{[\hat{ı ı} + 2 x e^{x^{2}} \hat{ȷ ȷ} + \hat{k}]}} d x \\ = e + e^{2} - 1 + \int_{0}^{1} 2 x e^{x^{2}} d x = e + e^{2} - 1 + [e^{x^{2}}]_{0}^{1} = e^{2} + 2 e - 2 \end{aligned}

2.4.2.13. (✳).

Solution.

Parametrize the line segment by

r (t) = (0, 0, 1) + t {(2, 1, 0) - (0, 0, 1)} = (2 t, t, 1 - t) 0 \leq t \leq 1

so that

r (0) = (0, 0, 1)

is the initial point of the line segment and

r (1) = (2, 1, 0)

is the final point of the segment. Then

r^{'} (t) = (2, 1, - 1)

and the work is

\begin{aligned} \int F \cdot d r & = \int_{0}^{1} F (r (t)) \cdot r^{'} (t) d t \\ = \int_{0}^{1} (2 t - t^{2}, t - (1 - t)^{2}, (1 - t) - 4 t^{2}) \cdot (2, 1, - 1) d t \\ = \int_{0}^{1} (4 t - 2 t^{2} + t - 1 + 2 t - t^{2} - 1 + t + 4 t^{2}) d t \\ = \int_{0}^{1} (t^{2} + 8 t - 2) d t \\ = \frac{1}{3} + 4 - 2 = \frac{7}{3} \end{aligned}

2.4.2.14. (✳).

Solution.

P,

z = \ln \frac{1}{x} = - \ln (x) .

So parametrize the curve

P

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} - \ln (\cos θ) \hat{k} 0 \leq θ \leq \frac{π}{4}

Then

\begin{aligned} r^{'} (θ) & = - \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ} + \tan θ \hat{k} \\ F (r (θ)) & = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + \cos^{3} θ \hat{k} \\ F (r (θ)) \cdot r^{'} (θ) & = \sin θ \cos^{2} θ \end{aligned}

so that

\begin{aligned} Work & = \int_{P} F \cdot d r = \int_{0}^{π / 4} F (r (θ)) \cdot r^{'} (θ) d θ = \int_{0}^{π / 4} \sin θ \cos^{2} θ d θ \\ = - \frac{1}{3} \cos^{3} θ |_{0}^{π / 4} \\ = \frac{1}{3} [1 - \frac{1}{2^{3 / 2}}] \approx 0.2155 \end{aligned}

2.4.2.15. (✳).

Solution.

Hmmm.

F

looks suspiciously complicated. Let’s guess that

F

is conservative and look for a potential for it.

ϕ (x, y, z)

is a potential for this

F

if and only if

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = y z \cos x \\ \frac{\partial φ}{\partial y} (x, y, z) & = z \sin x + 2 y z \\ \frac{\partial φ}{\partial z} (x, y, z) & = y \sin x + y^{2} - \sin z \end{aligned}

Integrating the first of these equations gives

φ (x, y, z) = y z \sin x + f (y, z)

Substituting this into the second equation gives

z \sin x + \frac{\partial f}{\partial y} (y, z) = z \sin x + 2 y z or \frac{\partial f}{\partial y} (y, z) = 2 y z

which integrates to

f (y, z) = y^{2} z + g (z)

Finally, substituting

φ (x, y, z) = y z \sin x + y^{2} z + g (z)

into the last equation gives

y \sin x + y^{2} + g^{'} (z) = y \sin x + y^{2} - \sin z or g^{'} (z) = - \sin z

which integrates to

g (z) = \cos z + C

with

C

being an arbitrary constant. So

ϕ (x, y, z) = y z \sin x + y^{2} z + \cos z

is one allowed scalar potential and the specified integral is

\begin{matrix} \int_{C} F \cdot d r = φ (r) |_{r (0)}^{r (π / 2)} = φ (π / 2, π / 2, π / 2) - φ (0, 0, 0) = \frac{π^{3}}{8} + \frac{π^{2}}{4} - 1 \end{matrix}

2.4.2.16. (✳).

Solution 1.

We are being asked to evaluate the line integral

\int_{C} F \cdot d r

with

C

being the specified semi-circle and

F = x y \hat{ȷ ȷ} .

\nabla \nabla \times F \neq 0,

the vector field

F

is not conservative. So we’ll evaluate the integral directly. First, using the figure,

we parametrize

C

r (θ) = (x (θ), y (θ)) = (1 - \cos θ) \hat{ı ı} + \sin θ \hat{ȷ ȷ} 0 \leq θ \leq π

So the integral is

\begin{matrix} \int_{C} x y d y = \int_{0}^{π} x (θ) y (θ) y^{'} (θ) d θ = \int_{0}^{π} (1 - \cos θ) \sin θ \cos θ d θ \end{matrix}

Making the substitution

u = \cos θ,

d u = - \sin θ d θ,

u (0) = 1,

u (π) = - 1,

\begin{aligned} \int_{C} x y d y & = \int_{1}^{- 1} (1 - u) u (- d u) = \int_{- 1}^{1} (u - u^{2}) d u \\ = - 2 \int_{0}^{1} u^{2} d u = - 2 \frac{1^{3}}{3} = - \frac{2}{3} \end{aligned}

Solution 2.

We can write

x

in terms of

y

over

C

in two pieces:

Let $C_{1}$ be the quarter-circle $x = 1 - \sqrt{1 - y^{2}}$ as $y$ goes from 0 to 1, and
Let $C_{2}$ be the quarter-circle $x = 1 + \sqrt{1 - y^{2}}$ as $y$ goes from 1 to 0.

Then:

\begin{aligned} \int_{C} x y d y & = \int_{C_{1}} x y d y + \int_{c_{2}} x y d y \\ = \int_{0}^{1} (1 - \sqrt{1 - y^{2}}) y d y + \int_{1}^{0} (1 + \sqrt{1 - y^{2}}) y d y \\ = \int_{0}^{1} y d y - \int_{0}^{1} y \sqrt{1 - y^{2}} d y + \int_{1}^{0} y d y + \int_{1}^{0} y \sqrt{1 - y^{2}} d y \\ = - 2 \int_{0}^{1} y \sqrt{1 - y^{2}} d y \end{aligned}

Using the substitution $u = 1 - y^{2},$ $d u = - 2 y d y :$

\begin{aligned} = \int_{1}^{0} u^{1 / 2} d u = - \frac{2}{3} \end{aligned}

2.4.2.17. (✳).

Solution.

The line integral is

\int_{C} F \cdot d r

with

F = (y e^{x} + \sin y) \hat{ı ı} + (e^{x} + \sin y + x \cos y) \hat{ȷ ȷ} .

We are to show that it is independent of path. That is the case if and only if

F

is conservative. So let’s look for a potential

φ

for

F .

That is, let’s look for a function

φ

that obeys

\begin{aligned} \frac{\partial φ}{\partial x} (x, y) & = y e^{x} + \sin y \\ \frac{\partial φ}{\partial y} (x, y) & = e^{x} + \sin y + x \cos y \end{aligned}

Integrating the first of these equations gives

φ (x, y) = y e^{x} + x \sin y + f (y)

Substituting this into the second equation gives

e^{x} + x \cos y + f^{'} (y) = e^{x} + \sin y + x \cos y or f^{'} (y) = \sin y

which integrates to

f (y) = - \cos y + C

F

is indeed conservative with one potential being

φ (x, y) = y e^{x} + x \sin y - \cos y

and the line integral is

\begin{aligned} \int_{C} (y e^{x} + \sin y) d x + (e^{x} + \sin y + x \cos y) d y = \int_{C} F \cdot d r = φ (x, y) |_{(1, 0)}^{(0, π / 2)} \\ = [y e^{x} + x \sin y - \cos y]_{(1, 0)}^{(0, π / 2)} \\ = 1 + \frac{π}{2} \end{aligned}

2.4.2.18. (✳).

Solution.

Here is a sketch of

C .

Note that

$y = 0$ on the line segment from $(1, 0, 0)$ to $(0, 0, 1)$ so that the integral reduces to $\int z x d z$ on that line segment and
$x = 0$ on the line segment from $(0, 0, 1)$ to $(0, 1, 0)$ so that the integral reduces to $\int y z d y$ on that line segment and
$z = 0$ on the line segment from $(0, 1, 0)$ to $(1, 0, 0)$ so that the integral reduces to $\int x y d x$ on that line segment.

So it looks feasible to evaluate the integral directly. Label the sides of the triangle

C_{1},

C_{2}

and

C_{3}

as in the sketch above.

We parametrize $C_{1}$ by $r (t) = (1, 0, 0) + t [(0, 0, 1) - (1, 0, 0)] = (1 - t, 0, t),$ $0 \leq t \leq 1 .$ So
$\begin{aligned} \int_{C_{1}} x y d x + y z d y + z x d z & = \int_{C_{1}} z x d z = \int_{0}^{1} \overset{x}{\overset{⏞}{(1 - t)}} \overset{z}{\overset{⏞}{(t)}} \overset{z^{'} (t)}{\overset{⏞}{(1)}} d t \\ = \int_{0}^{1} (t - t^{2}) d t \\ = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \end{aligned}$
We parametrize $C_{2}$ by $r (t) = (0, 0, 1) + t [(0, 1, 0) - (0, 0, 1)] = (0, t, 1 - t),$ $0 \leq t \leq 1 .$ So
$\begin{aligned} \int_{C_{2}} x y d x + y z d y + z x d z & = \int_{C_{2}} y z d y = \int_{0}^{1} \overset{y}{\overset{⏞}{(t)}} \overset{z}{\overset{⏞}{(1 - t)}} \overset{y^{'} (t)}{\overset{⏞}{(1)}} d t \\ = \int_{0}^{1} (t - t^{2}) d t \\ = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \end{aligned}$
We parametrize $C_{3}$ by $r (t) = (0, 1, 0) + t [(1, 0, 0) - (0, 1, 0)] = (t, 1 - t, 0),$ $0 \leq t \leq 1 .$ So
$\begin{aligned} \int_{C_{3}} x y d x + y z d y + z x d z & = \int_{C_{3}} x y d x = \int_{0}^{1} \overset{x}{\overset{⏞}{(t)}} \overset{y}{\overset{⏞}{(1 - t)}} \overset{x^{'} (t)}{\overset{⏞}{(1)}} d t \\ = \int_{0}^{1} (t - t^{2}) d t \\ = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \end{aligned}$

All together

\begin{aligned} \int_{C} x y d x + y z d y + z x d z & = \sum_{ℓ = 1}^{3} \int_{C_{ℓ}} x y d x + y z d y + z x d z = 3 \times \frac{1}{6} = \frac{1}{2} \end{aligned}

2.4.2.19. (✳).

Solution.

We are told that

F

is conservative. Let’s find a potential

φ

obeying

\nabla \nabla φ = F .

That is,

\begin{aligned} \frac{\partial φ}{\partial x} & = y + z e^{x} \\ \frac{\partial φ}{\partial y} & = x + e^{y} \sin z \\ \frac{\partial φ}{\partial z} & = z + e^{x} + e^{y} \cos z \end{aligned}

The first equation forces

φ (x, y, z) = x y + z e^{x} + ψ (y, z) .

Substituting this into the second equation gives

x + \frac{\partial ψ}{\partial y} (y, z) = x + e^{y} \sin z

\frac{\partial ψ}{\partial y} (y, z) = e^{y} \sin z

which forces

ψ (y, z) = e^{y} \sin z + ζ (z) .

So far, we have

φ (x, y, z) = x y + z e^{x} + e^{y} \sin z + ζ (z) .

Substituting this into the third equation gives

e^{x} + e^{y} \cos z + ζ^{'} (z) = z + e^{x} + e^{y} \cos z

ζ^{'} (z) = z

which forces

ζ (z) = \frac{z^{2}}{2} + C,

for some constant

C,

which we take to be zero. So our potential is

φ (x, y, z) = x y + z e^{x} + e^{y} \sin z + \frac{z^{2}}{2}

So the line integral

\int_{C} F \cdot d r = φ (r (π)) - φ (r (0)) = φ (π, e^{π}, 0) - φ (0, 1, 0) = π e^{π}

2.4.2.20. (✳).

Solution.

(a) Note

F

is defined and continuous on all of

R^{3} .

Furthermore,

F

has continuous first-order partial derivatives on all of

R^{3} .

Using Theorem 2.4.8,

F

is conservative if and only if it has zero curl:

\begin{aligned} 0 & = \nabla \nabla \times F = \nabla \nabla \times (α e^{y} \hat{ı ı} + (x e^{y} + β \cos z) \hat{ȷ ȷ} - γ y \sin z \hat{k}) \\ = (- γ \sin z + β \sin z) \hat{ı ı} + (e^{y} - α e^{y}) \hat{k} \end{aligned}

which is the case if and only if

α = 1,

β = γ .

(b) We use Theorem 2.4.2: if

φ

is a potential for

F,

then

\int_{C} F \cdot d r = φ (P_{1}) - φ (P_{0})

where

C

runs from

P_{0}

P_{1} .

So, we find

φ .

Assume that

α = 1,

β = γ .

We find a potential

φ

for

F

by antidifferentiating.

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = e^{y} & ⟹ φ (x, y, z) & = x e^{y} + ψ_{1} (y, z) \\ \frac{\partial φ}{\partial y} (x, y, z) & = x e^{y} + β \cos z & ⟹ φ (x, y, z) & = x e^{y} + β y \cos z + ψ_{2} (x, z) \\ \frac{\partial φ}{\partial z} (x, y, z) & = - β y \sin z & ⟹ φ (x, y, z) & = β y \cos z + ψ_{3} (x, y) \end{aligned}

for some functions

ψ_{1} (y, z),

ψ_{2} (x, z)

and

ψ_{3} (x, y)

to be determined.

We’d like a single function

φ (x, y, z)

that simultaneously obeys all three of these equations, for some

ψ_{j}

’s. An initial guess is simply the sum of all of the distinct terms, other that the

ψ_{j}

’s, that appear in the three equations above. The term

x e^{y}

appears in the

ψ_{1}

and

ψ_{2}

equations and the term

β y \cos z

appears in the

ψ_{2}

and

ψ_{3}

equations. So we guess

φ (x, y, z) \overset{?}{=} x e^{y} + β y \cos z

If we let

ψ_{1} (y, z) = β y \cos z,

ψ_{2} (x, z) = 0,

and

ψ_{3} (x, y) = x e^{y},

then we see this function

φ (x, y, z)

does indeed obey all three equations and so is a potential for

F .

The curve

C

runs from

P_{0} = (0^{2}, e^{0}, π \cdot 0) = (0, 1, 0)

P_{1} = (1^{2}, e^{1}, π \cdot 1) = (1, e, π) .

Using Theorem 2.4.2:

\int_{C} F \cdot d r = φ (1, e, π) - φ (0, 1, 0) = (e^{e} - β e) - β = e^{e} - β (e + 1)

2.4.2.21. (✳).

Solution.

(a) The curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \cos x & 2 + \sin y & e^{z} \end{array}] = 0 \end{aligned}

Because

F_{1}

is a function only of

x,

F_{2}

is a function only of

y,

and

F_{3}

is a function only of

z,

that all partial derivatives used in computing the curl are 0.

(b) The vector field

F

passes the screening test on all of

R^{3}

and so is conservative by Theorem 2.4.8 in the text. Alternatively, we can see that

F = \nabla \nabla (\sin x + 2 y - \cos y + e^{z})

by inspection. Alternatively,

f

can be found by antidifferentiating its partial derivatives:

\begin{aligned} \frac{\partial f}{\partial x} (x, y, z) & = \cos x & ⟹ f (x, y, z) & = \sin x + ψ_{1} (y, z) \\ \frac{\partial f}{\partial y} (x, y, z) & = 2 + \sin y & ⟹ f (x, y, z) & = 2 y - \cos y + ψ_{2} (x, z) \\ \frac{\partial f}{\partial z} (x, y, z) & = e^{z} & ⟹ f (x, y, z) & = e^{z} + ψ_{3} (x, y) \end{aligned}

We’d like a single function

f (x, y, z)

that simultaneously obeys all three of these equations, for some

ψ_{j}

’s. An initial guess is simply the sum of all of the distinct terms, other than the

ψ_{j}

’s, that appear in the three equations. The term

\sin x

appears in the

ψ_{1}

equation, the terms

2 y

and

- \cos y

appears in the

ψ_{2}

equation, and the term

e^{z}

appears in the

ψ_{3}

equation. So we guess

f (x, y, z) \overset{?}{=} \sin x + 2 y - \cos y + e^{z}

If we let

ψ_{1} (y, z) = 2 y - \cos y + e^{z},

ψ_{2} (x, z) = \sin x + e^{z},

and

ψ_{3} (x, y) = \sin x + 2 y - \cos y,

then we see this function

f (x, y, z)

is indeed a potential for

F .

F = \nabla \nabla f,

\begin{aligned} \int_{C} F \cdot d r & = f (r (3 π)) - f (r (0)) \\ = f (3 π, - 1, 0) - f (0, 1, 0) \\ = (0 - 2 - \cos (- 1) + 1) - (0 + 2 - \cos 1 + 1) \end{aligned}

Since cosine is an even function, $\cos (- 1) = \cos 1 .$

\begin{aligned} = - 4 \end{aligned}

2.4.2.22. (✳).

Solution.

(a) The curl is

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z + e^{y} & x e^{y} - e^{z} \sin y & 1 + x + e^{z} \cos y \end{array}] = 0 \end{aligned}

F

passes the screening test. Since its first-order partial derivatives are continuous on all of

R^{3},

it is conservative by Theorem 2.4.8 in the text.

By inspection, the potential is

φ (x, y, z) = x z + x e^{y} + e^{z} \cos y + z

— this is another way to verify that

F

is conservative. Alternatively,

φ

can be found by antidifferentiating its partial derivatives.

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = z + e^{y} \\ ⟹ φ (x, y, z) = z x + x e^{y} + ψ_{1} (y, z) \\ \frac{\partial φ}{\partial y} (x, y, z) & = x e^{y} - e^{z} \sin y \\ ⟹ φ (x, y, z) = x e^{y} + e^{z} \cos y + ψ_{2} (x, z) \\ \frac{\partial φ}{\partial z} (x, y, z) & = 1 + x + e^{z} \cos y \\ ⟹ φ (x, y, z) = z + z x + e^{z} \cos y + ψ_{3} (x, y) \end{aligned}

We’d like a single function

φ (x, y, z)

that simultaneously obeys all three of these equations. An initial guess is simply the sum of the distinct terms (without the

ψ_{j}

’s) that appear in the equations above:

φ (x, y, z) \overset{?}{=} z x + x e^{y} + e^{z} \cos y + z

If we let

ψ_{1} (y, z) = e^{z} \cos y + z,

ψ_{2} (x, z) = z x + z,

and

ψ_{3} (x, y) = x e^{y},

then we see this function

φ (x, y, z)

is indeed a potential for

F .

(b) Since

F = \nabla \nabla φ,

with

φ = x z + x e^{y} + e^{z} \cos y + z,

\begin{aligned} \int_{C} F \cdot d r & = φ (r (π)) - φ (r (0)) = [x z + x e^{y} + e^{z} \cos y + z]_{r (0)}^{r (π)} \\ = [x z + x e^{y} + e^{z} \cos y + z]_{(0, 0, 1)}^{(π^{2}, 0, 1)} \\ = (π^{2} + π^{2} + e + 1) - (0 + 0 + e + 1) = 2 π^{2} \end{aligned}

2.4.2.23. (✳).

Solution.

(a) For

F

to be conservative, it must pass the screening test

\begin{aligned} 0 = \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (x - a) y e^{x} & x e^{x} + z^{3} & b y z^{2} \end{array}] \\ = (b z^{2} - 3 z^{2}) \hat{ı ı} - (0 - 0) \hat{ȷ ȷ} + (e^{x} + x e^{x} - (x - a) e^{x}) \hat{k} \end{aligned}

This is the case if and only if

b = 3

and

a = - 1

(b) Set

a = - 1

and

b = 3 .

For

f

to be a potential for

F,

it must obey

\begin{aligned} \frac{\partial f}{\partial x} (x, y, z) & = (x + 1) y e^{x} \\ \frac{\partial f}{\partial y} (x, y, z) & = x e^{x} + z^{3} \\ \frac{\partial f}{\partial z} (x, y, z) & = 3 y z^{2} \end{aligned}

Integrating the second of these equations gives

f (x, y, z) = x y e^{x} + y z^{3} + g (x, z)

Substituting this into the last equation gives

3 y z^{2} + \frac{\partial g}{\partial z} (x, z) = 3 y z^{2} or \frac{\partial g}{\partial z} (x, z) = 0

which forces

g (x, z) = h (x)

Finally, substituting

f (x, y, z) = x y e^{x} + y z^{3} + h (x)

into the first equation gives

x y e^{x} + y e^{x} + h^{'} (x) = (x + 1) y e^{x} or h^{'} (x) = 0

h (x) = C

and hence

f (x, y, z) = x y e^{x} + y z^{3} + C

works for any constant

C .

F = \nabla \nabla f,

\begin{aligned} \int_{C} F \cdot d r & = \int_{C} \nabla \nabla f \cdot d r = f (r (π)) - f ((r (0)) = f (π, 1, - 1) - f (0, 1, 1) \\ = [π e^{π} - 1] - [1] = π e^{π} - 2 \end{aligned}

(d) Since

\int_{C} F \cdot d r = \int_{C} (x + 1) y e^{x} d x + (x e^{x} + z^{3}) d y + 3 y z^{2} d z

we have

\begin{aligned} I & = \int_{C} F \cdot d r + \int_{C} y z^{2} d z \\ = π e^{π} - 2 + \int_{0}^{π} \overset{y}{\overset{⏞}{(\cos 2 t)}} \overset{z^{2}}{\overset{⏞}{\cos^{2} t}} \overset{d z}{\overset{⏞}{(- \sin t) d t}} \\ = π e^{π} - 2 + \int_{0}^{π} (2 \cos^{2} t - 1) \cos^{2} t (- \sin t) d t \\ = π e^{π} - 2 + \int_{1}^{- 1} (2 u^{2} - 1) u^{2} d u with u = \cos t, d u = - \sin t d t \\ = π e^{π} - 2 + [\frac{2 u^{5}}{5} - \frac{u^{3}}{3}]_{1}^{- 1} \\ = π e^{π} - 2 + [- \frac{4}{5} + \frac{2}{3}] \\ = π e^{π} - \frac{32}{15} \end{aligned}

2.4.2.24. (✳).

Solution.

(a) The vector field

F

is conservative if and only if it passes the screening test

\nabla \nabla \times F = 0 .

That is, if and only if,

\begin{aligned} 0 & = \nabla \nabla \times F = det | \begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^{2} e^{3 z} + A x y^{3} & 2 x y e^{3 z} + 3 x^{2} y^{2} & B x y^{2} e^{3 z} \end{array} | \\ = (2 B x y e^{3 z} - 6 x y e^{3 z}) \hat{ı ı} - (B y^{2} e^{3 z} - 3 y^{2} e^{3 z}) \hat{ȷ ȷ} \\ + (2 y e^{3 z} + 6 x y^{2} - 2 y e^{3 z} - 3 A x y^{2}) \hat{k} \end{aligned}

F

is conservative if and only if

A = 2

and

B = 3 .

(b) Let

A = 2

and

B = 3 .

We find a potential

φ

for

F

by antidifferentiating its partial derivatives.

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = y^{2} e^{3 z} + 2 x y^{3} \\ ⟹ φ (x, y, z) = x y^{2} e^{3 z} + x^{2} y^{3} + ψ_{1} (y, z) \\ \frac{\partial φ}{\partial y} (x, y, z) & = 2 x y e^{3 z} + 3 x^{2} y^{2} \\ ⟹ φ (x, y, z) = x y^{2} e^{3 z} + x^{2} y^{3} + ψ_{2} (x, z) \\ \frac{\partial φ}{\partial z} (x, y, z) & = 3 x y^{2} e^{3 z} \\ ⟹ φ (x, y, z) = x y^{2} e^{3 z} + ψ_{3} (x, y) \end{aligned}

Let’s guess that

φ (x, y, z) = x y^{2} e^{3 z} + x^{2} y^{3}

(This was obtained by summing the distinct terms in the above three equations, without the

ψ_{i}

’s.) If we set

ψ_{1} (y, z) = ψ_{2} (x, z) = 0

and

ψ_{3} (x, y) = x^{2} y^{3},

we see our choice of

φ

is indeed a potential for

F .

A = 2

and

B = 3 .

We are asked the evaluate

\int_{C} G \cdot d r

with

G = (y^{2} e^{3 z} + x y^{3}) \hat{ı ı} + (2 x y e^{3 z} + 3 x^{2} y^{2}) \hat{ȷ ȷ} + 3 x y^{2} e^{3 z} \hat{k} = F - x y^{3} \hat{ı ı}

\begin{aligned} \int_{C} (y^{2} e^{3 z} + x y^{3}) d x + (2 x y e^{3 z} + 3 x^{2} y^{2}) d y + 3 x y^{2} e^{3 z} d z \\ = \int_{C} F \cdot d r - \int_{C} x y^{3} d r \\ = φ (r (1)) - φ (r (0)) - \int_{0}^{1} \overset{x y^{3} \hat{ı ı}}{\overset{⏞}{e^{2 t} (e^{- t})^{3} \hat{ı ı}}} \cdot (\overset{r^{'} (t)}{\overset{⏞}{2 e^{2 t} \hat{ı ı} - e^{- t} \hat{ȷ ȷ} + \frac{1}{1 + t} \hat{k}}}) d t \\ = φ (e^{2}, \frac{1}{e}, \ln 2) - φ (1, 1, 0) - \int_{0}^{1} 2 e^{t} d t \\ = {e^{2} (\frac{1}{e})^{2} e^{3 \ln 2} + e^{4} (\frac{1}{e})^{3}} - (1 + 1) - 2 (e - 1) \\ = 2^{3} + e - 2 - 2 e + 2 \\ = 8 - e \end{aligned}

2.4.2.25. (✳).

Solution.

(a) The field is conservative only if

\frac{\partial F_{1}}{\partial y} = \frac{\partial F_{2}}{\partial x} \frac{\partial F_{1}}{\partial z} = \frac{\partial F_{3}}{\partial x} \frac{\partial F_{2}}{\partial z} = \frac{\partial F_{3}}{\partial y}

That is,

\begin{aligned} \frac{\partial}{\partial y} (2 x \sin (π y) - e^{z}) & = \frac{\partial}{\partial x} (a x^{2} \cos (π y) - 3 e^{z}) \\ ⟺ 2 π x \cos (π y) = 2 a x \cos (π y) \\ \frac{\partial}{\partial z} (2 x \sin (π y) - e^{z}) & = - \frac{\partial}{\partial x} (x + b y) e^{z} \\ ⟺ - e^{z} = - e^{z} \\ \frac{\partial}{\partial z} (a x^{2} \cos (π y) - 3 e^{z}) & = - \frac{\partial}{\partial y} (x + b y) e^{z} \\ ⟺ - 3 e^{z} = - b e^{z} \end{aligned}

Hence only

a = π, b = 3

works.

(b) When

a = π, b = 3

\begin{aligned} F & = (2 x \sin (π y) - e^{z}) \hat{ı ı} + (π x^{2} \cos (π y) - 3 e^{z}) \hat{ȷ ȷ} - (x + 3 y) e^{z} \hat{k} \\ = \nabla \nabla (x^{2} \sin (π y) - x e^{z} - 3 y e^{z} + C) \end{aligned}

φ (x, y, z) = x^{2} \sin (π y) - x e^{z} - 3 y e^{z} + C

for any constant

C .

Here

φ

was guessed. Alternatively, it can be found by antidifferentiating the partial derivatives of

F .

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = 2 x \sin (π y) - e^{z} \\ ⟹ φ (x, y, z) = x^{2} \sin (π y) - x e^{z} + ψ_{1} (y, z) \\ \frac{\partial φ}{\partial y} (x, y, z) & = π x^{2} \cos (π y) - 3 e^{z} \\ ⟹ φ (x, y, z) = x^{2} \sin (π y) - 3 y e^{z} + ψ_{2} (x, z) \\ \frac{\partial φ}{\partial z} (x, y, z) & = - (x + 3 y) e^{z} \\ ⟹ φ (x, y, z) = - x e^{z} - 3 y e^{z} + ψ_{3} (x, y) \end{aligned}

Summing the distinct terms on the right hand sides of the three equations above, we guess

φ (x, y, z) = x^{2} \sin (π y) - x e^{z} - 3 y e^{z}

is a potential for

F .

Setting

ψ_{1} (y, z) = - 3 y e^{z},

ψ_{2} (x, z) = - x e^{z},

and

ψ_{3} (x, y) = x^{2} \sin (π y)

convinces us that our guess is indeed a valid potential.

\begin{aligned} \int_{C} F \cdot d r & = φ (1, 1, \ln 2) - φ (0, 0, 0) \\ = (\sin π - e^{\ln 2} - 3 e^{\ln 2}) - (\sin (0) - 0 - 0) \\ = - 8 \end{aligned}

(d) Observe that

G = F + 3 y e^{z} \hat{k},

with

F

evaluated with

a = π, b = 3 .

Hence

\int_{C} G \cdot d r = \int_{C} F \cdot d r + \int_{C} 3 y e^{z} \hat{k} \cdot d r = - 8 + \int_{C} 3 y e^{z} \hat{k} \cdot d r

To evaluate the remaining integral, parametrize the curve by

r (t) = t \hat{ı ı} + t \hat{ȷ ȷ} + \ln (1 + t) \hat{k}

with

0 \leq t \leq 1 .

Then

r^{'} (t) = \hat{ı ı} + \hat{ȷ ȷ} + \frac{1}{1 + t} \hat{k}

and

3 y e^{z} \hat{k} = 3 t (1 + t) \hat{k}

so that

3 y e^{z} \hat{k} \cdot d r = 3 t d t .

Subbing in

\int_{C} G \cdot d r = - 8 + \int_{0}^{1} 3 t d t = - 8 + \frac{3}{2} = - \frac{13}{2}

2.4.2.26. (✳).

Solution.

(a) The potential

f

must obey

\begin{aligned} \frac{\partial f}{\partial x} (x, y, z) & = - 2 y \cos x \sin x \\ \frac{\partial f}{\partial y} (x, y, z) & = \cos^{2} x + (1 + y z) e^{y z} \\ \frac{\partial f}{\partial z} (x, y, z) & = y^{2} e^{y z} \end{aligned}

Integrating the last of these equations with respect to

z

gives

f (x, y, z) = y e^{y z} + g (x, y)

Substituting this into the second equation gives

e^{y z} + y z e^{y z} + \frac{\partial g}{\partial y} (x, y) = \cos^{2} x + (1 + y z) e^{y z} or \frac{\partial g}{\partial y} (x, y) = \cos^{2} x

which forces

g (x, y) = y \cos^{2} x + h (x)

Finally, substituting

f (x, y, z) = y e^{y z} + y \cos^{2} x + h (x)

into the first equation gives

- 2 y \sin x \cos x + h^{'} (x) = - 2 y \cos x \sin x or h^{'} (x) = 0

h (x) = C

and hence

f (x, y, z) = y e^{y z} + y \cos^{2} x + C

works for any constant

C .

(b) By part (a)

\begin{aligned} \int_{C} F \cdot d r & = \int_{C} \nabla \nabla f \cdot d r = f (π, e^{π}, 0) - f (0, 1, - π^{2}) \\ = [y e^{y z} + y \cos^{2} x]_{(0, 1, - π^{2})}^{(π, e^{π}, 0)} \\ = (2 e^{π}) - (e^{- π^{2}} + 1) \end{aligned}

2.4.2.27. (✳).

Solution.

(a) The curl of

F

is zero because

F_{1}

is a function only of

x,

F_{2}

is a function only of

y,

and

F_{3}

is a function only of

z .

That is:

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2 x & 2 y & 2 z \end{array}] = (0 - 0) \hat{ı ı} + (0 - 0) \hat{ȷ ȷ} + (0 - 0) \hat{k} = 0 \end{aligned}

(b) All first-order partial derivative of

F

are continuous on all of

R^{3} .

By part (a),

F

passes the screening test and is conservative by Theorem 2.4.8 in the text. By inspection, a potential is

φ = x^{2} + y^{2} + z^{2} .

Since

F = \nabla \nabla φ,

\begin{aligned} \int_{C} F \cdot d r & = [x^{2} + y^{2} + z^{2}]_{(0, 0, 0)}^{(a_{1}, a_{2}, a_{3})} = a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = a \cdot a \end{aligned}

2.4.2.28. (✳).

Solution.

(a) The curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{y z} & x z e^{y z} + z e^{y} & x y e^{y z} + e^{y} \end{array}] \\ = [(x e^{y z} + x y z e^{y z} + e^{y}) - (x e^{y z} + x y z e^{y z} + e^{y})] \hat{ı ı} - [y e^{y z} - y e^{y z}] \hat{ȷ ȷ} \\ + [z e^{y z} - z e^{y z}] \hat{k} \\ = 0 \end{aligned}

(b)

F

is defined on all of

R^{3}

and passes the conservative field screening test

\nabla \nabla \times F = 0 .

F

is conservative. We find a potential

φ

for

F

by antidifferentiating its partial derivatives.

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = e^{y z} & ⟹ φ (x, y, z) & = x e^{y z} + ψ_{1} (y, z) \\ \frac{\partial φ}{\partial y} (x, y, z) & = x z e^{y z} + z e^{y} & ⟹ φ (x, y, z) & = x e^{y z} + z e^{y} + ψ_{2} (x, z) \\ \frac{\partial φ}{\partial z} (x, y, z) & = x y e^{y z} + e^{y} & ⟹ φ (x, y, z) & = x e^{y z} + z e^{y} + ψ_{3} (x, y) \end{aligned}

All together,

φ (x, y, z) = x e^{y z} + z e^{y} + C

works for any constant

C .

So the specified work integral is

\begin{matrix} \int_{C} F \cdot d r = φ (r (π / 2)) - φ (r (0)) = φ (0, 1, π / 2) - φ (1, 0, 0) = \frac{π e}{2} - 1 \end{matrix}

2.4.2.29. (✳).

Solution.

(a), (b) The function

f (x, y)

is a potential for

F (x, y)

if and only if it obeys

\begin{aligned} \frac{\partial f}{\partial x} (x, y) & = 2 x y \cos (x^{2}) \\ \frac{\partial f}{\partial y} (x, y) & = \sin (x^{2}) - \sin (y) \end{aligned}

Integrating the first of these equations gives

f (x, y) = y \sin (x^{2}) + g (y)

Substituting this into the second equation gives

\sin (x^{2}) + g^{'} (y) = \sin (x^{2}) - \sin (y) or g^{'} (y) = - \sin (y)

which integrates to

g (y) = \cos (y) + C

with

C

an arbitrary constant. Hence

f (x, y) = y \sin (x^{2}) + \cos (y) + C

is a potential for any constant

C .

Because

F

has a potential, it is conservative.

C

r (t) = \sin (t) \hat{ı ı} + t \hat{ȷ ȷ} \frac{π}{2} \leq t \leq π

f (x, y) = y \sin (x^{2}) + \cos (y)

is a potential for

F

\begin{aligned} \int_{C} F \cdot d r & = f (r (π)) - f (r (\frac{π}{2})) = f (0, π) - f (1, \frac{π}{2}) \\ = (- 1) - (\frac{π}{2} \sin (1)) \\ = - 1 - \frac{π}{2} \sin (1) \end{aligned}

2.4.2.30. (✳).

Solution.

(a) The stated integral property is characteristic of conservative fields (Theorem 2.4.7). Since all partial derivatives of

F

are defined on all of

R^{3},

an equivalent property is

\begin{aligned} 0 & = \nabla \times F = | \begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (m x y z + z^{2} - n y^{2}) & (x^{2} z - 4 x y) & (x^{2} y + p x z + q z^{3}) \end{array} | \\ = \hat{ı ı} (x^{2} - x^{2}) - \hat{ȷ ȷ} (2 x y + p z - m x y - 2 z) + \hat{k} (2 x z - 4 y - m x z + 2 n y) . \end{aligned}

This requires

p = 2,

m = 2,

and

n = 2,

but leaves

q \in R

completely free.

(b) Solution 1: The choices from (a) give

F = (2 x y z + z^{2} - 2 y^{2}) \hat{ı ı} + (x^{2} z - 4 x y) \hat{ȷ ȷ} + (x^{2} y + 2 x z + q z^{3}) \hat{k} .

We find a potential

φ

for

F

by antidifferentiating its partial derivatives.

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = 2 x y z + z^{2} - 2 y^{2} \\ ⟹ φ (x, y, z) = x^{2} y z + x z^{2} - 2 x y^{2} + ψ_{1} (y, z) \\ \frac{\partial φ}{\partial y} (x, y, z) & = x^{2} z - 4 x y \\ ⟹ φ (x, y, z) = x^{2} y z - 2 x y^{2} + ψ_{2} (x, z) \\ \frac{\partial φ}{\partial z} (x, y, z) φ (x, y, z) & = x^{2} y + 2 x z + q z^{3} \\ ⟹ φ (x, y, z) = x^{2} y z + x z^{2} + \frac{q}{4} z^{4} + ψ_{3} (x, y) \end{aligned}

All together,

F = \nabla φ

for

φ (x, y, z) = x^{2} y z + x z^{2} - 2 x y^{2} + \frac{1}{4} q z^{4} + C

where

C

is any constant.

Rearranging the sphere’s equation to

x^{2} + y^{2} + (z - 1)^{2} = 1

reveals that its bottom is at

r_{0} = (0, 0, 0),

and its top is at

r_{1} = (0, 0, 2) .

Hence the work done is

W = \int_{C} F \cdot d r = \int_{C} \nabla φ \cdot d r = φ (0, 0, 2) - φ (0, 0, 0) = 4 q

Solution 2: Since the integral is path-independent, all paths from

r_{0}

r_{1}

produce the same result. A simple choice is

C : r = (0, 0, t), 0 \leq t \leq 2

Here

r^{'} (t) = (0, 0, 1),

so direct calculation gives

\int_{C} F \cdot d r = \int_{t = 0}^{2} F (r (t)) \cdot r^{'} (t) d t = \int_{t = 0}^{2} q t^{3} d t = [\frac{1}{4} q t^{4}]_{t = 0}^{2} = 4 q

2.4.2.31. (✳).

Solution.

(a) Since the particle has mass

m = 1,

Newton’s law of motion

m a = F

simplifies to

r^{″} (t) = \hat{ȷ ȷ} - \sin t \hat{k}

Integrating once gives

r^{'} (t) = t \hat{ȷ ȷ} + \cos t \hat{k} + C

for some constant vector

C .

To satisfy the initial condition that

r^{'} (0) = v_{0} = \hat{ı ı} + \hat{k},

we need

\begin{matrix} \hat{ı ı} + \hat{k} = r^{'} (0) = \hat{k} + C ⟹ C = \hat{ı ı} \end{matrix}

r^{'} (t) = \hat{ı ı} + t \hat{ȷ ȷ} + \cos t \hat{k}

Integrating a second time, and imposing the initial condition that

r (0) = \hat{ȷ ȷ},

gives

r (t) = t \hat{ı ı} + \frac{t^{2}}{2} \hat{ȷ ȷ} + \sin t \hat{k} + \hat{ȷ ȷ} = t \hat{ı ı} + (1 + \frac{t^{2}}{2}) \hat{ȷ ȷ} + \sin t \hat{k}

(b) The particle has

x (t) = π / 2

when

t = π / 2 .

r (π / 2) = \frac{π}{2} \hat{ı ı} + (1 + \frac{π^{2}}{8}) \hat{ȷ ȷ} + \hat{k}

\begin{aligned} Work & = \int_{0}^{π / 2} F (t) \cdot r^{'} (t) d t \\ = \int_{0}^{π / 2} (\hat{ȷ ȷ} - \sin t \hat{k}) \cdot (\hat{ı ı} + t \hat{ȷ ȷ} + \cos t \hat{k}) d t \\ = \int_{0}^{π / 2} (t - \sin t \cos t) d t \\ = [\frac{t^{2}}{2} + \frac{1}{2} \cos^{2} t]_{0}^{π / 2} \\ = \frac{π^{2}}{8} - \frac{1}{2} \end{aligned}

2.4.2.32.

Solution.

(a) Parametrize

C

x .

When the first component of a point on the curve is

x,

then the second component,

y,

must be

x^{2}

and the third component,

z,

must be

x^{3} .

\begin{aligned} r (x) & = x \hat{ı ı} + x^{2} \hat{ȷ ȷ} + x^{3} \hat{k} 0 \leq x \leq 1 \\ r^{'} (x) & = \hat{ı ı} + 2 x \hat{ȷ ȷ} + 3 x^{2} \hat{k} \\ \frac{d s}{d x} (x) & = \sqrt{1 + 4 x^{2} + 9 x^{4}} \end{aligned}

and

\begin{aligned} ρ (x) \frac{d s}{d x} (x) & = (8 x + 36 x^{3}) \sqrt{1 + 4 x^{2} + 9 x^{4}} \\ \int_{C} ρ d s & = \int_{0}^{1} (8 x + 36 x^{3}) \sqrt{1 + 4 x^{2} + 9 x^{4}} d x \end{aligned}

Substituting

u = 1 + 4 x^{2} + 9 x^{4},

d u = (8 x + 36 x^{3}) d x,

u (0) = 1,

u (1) = 14,

\begin{aligned} \int_{C} ρ d s & = \int_{1}^{14} \sqrt{u} d u = \frac{2}{3} u^{3 / 2} |_{1}^{14} \\ = \frac{2}{3} [14^{3 / 2} - 1] \approx 34.26 \end{aligned}

(b) Since

F (x, y, z) = \nabla f (x, y, z)

with

f (x, y, z) = x \sin y + y z + \frac{1}{2} z^{2},

\int_{C} F \cdot d r = f (1, 1, 1) - f (0, 0, 0) = \sin 1 + \frac{3}{2} \approx 2.3415

The potential

f

was just guessed. Alternatively, it can be found by solving

\begin{aligned} \frac{\partial f}{\partial x} (x, y, z) & = \sin y \\ \frac{\partial f}{\partial y} (x, y, z) & = x \cos y + z \\ \frac{\partial f}{\partial z} (x, y, z) & = y + z \end{aligned}

Integrating the first of these equations gives

f (x, y, z) = x \sin y + g (y, z)

Substituting this into the second equation gives

x \cos y + \frac{\partial g}{\partial y} (y, z) = x \cos y + z or \frac{\partial g}{\partial y} (x, z) = z

which forces

g (y, z) = y z + h (z)

Finally, substituting

f (x, y, z) = x \sin y + y z + h (z)

into the last equation gives

y + h^{'} (z) = y + z or h^{'} (z) = z

h (x) = \frac{z^{2}}{2} + C

and hence

f (x, y, z) = x \sin y + y z + \frac{z^{2}}{2} + C

for any constant

C .

2.4.2.33. (✳).

Solution.

First, we’ll parametrize

(x, y),

which wraps once, counterclockwise, aroung the circle

x^{2} + y^{2} = 1 .

x (t) = \cos t,

y (t) = \sin t,

0 \leq t \leq 2 π

works. As

(x, y)

wraps around the circle,

z

has to start at

0

(when

t = 0

) and end at

1

(when

t = 2 π

). So

z (t) = \frac{t}{2 π}

works and our parametrization is

r (t) = \cos t \hat{ı ı} + \sin t \hat{ȷ ȷ} + \frac{t}{2 π} \hat{k}

(Compare to Example 1.4.4.) With this parametrization

\begin{aligned} r^{'} (t) & = - \sin t \hat{ı ı} + \cos t \hat{ȷ ȷ} + \frac{1}{2 π} \hat{k} \\ F (x (t), y (t), z (t)) & = - \sin t \hat{ı ı} + \cos t \hat{ȷ ȷ} + \frac{t^{2}}{4 π^{2}} \hat{k} \\ F (x (t), y (t), z (t)) \cdot r^{'} (t) & = 1 + \frac{t^{2}}{8 π^{3}} \end{aligned}

and

\begin{aligned} \int_{C} F \cdot d r & = \int_{0}^{2 π} F (x (t), y (t), z (t)) \cdot r^{'} (t) d t = \int_{0}^{2 π} (1 + \frac{t^{2}}{8 π^{3}}) d t \\ = 2 π + \frac{1}{3} \end{aligned}

2.4.2.34. (✳).

Solution.

(a) Let’s evaluate the integral directly using the parametrization

r (x) = x \hat{ı ı} + (9 - x^{2}) \hat{ȷ ȷ}

with

- 3 \leq x \leq 3 .

Since

r^{'} (x) = \hat{ı ı} - 2 x \hat{ȷ ȷ},

\begin{aligned} \int_{C} (x^{2} + y) d x + x d y & = \int_{- 3}^{3} (x^{2} + \overset{y}{\overset{⏞}{9 - x^{2}}} + x \overset{\frac{d y}{d x}}{\overset{⏞}{(- 2 x)}}) d x \\ = \int_{- 3}^{3} (9 - 2 x^{2}) d x \\ = 2 \int_{0}^{3} (9 - 2 x^{2}) d x = 2 (27 - 2 \frac{3^{3}}{3}) \\ = 18 \end{aligned}

(b) In this solution, we’ll evaluate the integral directly. Label the four sides of the square

L_{1},

L_{2},

L_{3}

and

L_{4}

as in the figure

The parametrization of

L_{1}

by arc length is

r (s) = s \hat{ı ı},

0 \leq s \leq 1 .

As the outward pointing normal to

L_{1}

- \hat{ȷ ȷ},

\begin{matrix} \int_{L_{1}} F \cdot \hat{n} d s = \int_{0}^{1} F (s, 0) \cdot (- \hat{ȷ ȷ}) d s = \int_{0}^{1} (- 0) d s = 0 \end{matrix}

The parametrization of

L_{2}

by arc length is

r (s) = \hat{ı ı} + s \hat{ȷ ȷ},

0 \leq s \leq 1 .

As the outward pointing normal to

L_{2}

\hat{ı ı},

\begin{matrix} \int_{L_{2}} F \cdot \hat{n} d s = \int_{0}^{1} F (1, s) \cdot \hat{ı ı} d s = \int_{0}^{1} 2 d s = 2 \end{matrix}

The parametrization of

L_{3}

by arc length (starting at

(1, 1)

) is

r (s) = (1 - s) \hat{ı ı} + \hat{ȷ ȷ},

0 \leq s \leq 1 .

As the outward pointing normal to

L_{3}

\hat{ȷ ȷ},

\begin{matrix} \int_{L_{3}} F \cdot \hat{n} d s = \int_{0}^{1} F (1 - s, 1) \cdot \hat{ȷ ȷ} d s = \int_{0}^{1} e^{1 - s} d s = [- e^{1 - s}]_{0}^{1} = e - 1 \end{matrix}

The parametrization of

L_{4}

by arc length (starting at

(0, 1)

) is

r (s) = (1 - s) \hat{ȷ ȷ},

0 \leq s \leq 1 .

As the outward pointing normal to

L_{4}

- \hat{ı ı},

\begin{matrix} \int_{L_{2}} F \cdot \hat{n} d s = \int_{0}^{1} F (0, 1 - s) \cdot (- \hat{ı ı}) d s = \int_{0}^{1} (- 0) d s = 0 \end{matrix}

All together

\begin{aligned} \int_{C} F \cdot \hat{n} d s & = \int_{L_{1}} F \cdot \hat{n} d s + \int_{L_{2}} F \cdot \hat{n} d s + \int_{L_{3}} F \cdot \hat{n} d s + \int_{L_{4}} F \cdot \hat{n} d s \\ = 0 + 2 + (e - 1) + 0 = e + 1 \end{aligned}

2.4.2.35. (✳).

Solution.

(a) Since

m = 1,

Newton’s law of motion gives

\begin{matrix} a (t) = v^{'} (t) = F (t) = \hat{ȷ ȷ} - \sin t \hat{k} \end{matrix}

Integating gives

v (t) = t \hat{ȷ ȷ} + \cos t \hat{k} + c

for some constant vector

c .

Since

v (0) = \hat{ı ı} + \hat{k},

we have

c = \hat{ı ı}

so that

\begin{matrix} r^{'} (t) = v (t) = \hat{ı ı} + t \hat{ȷ ȷ} + \cos t \hat{k} \end{matrix}

Integating again gives

r (t) = t \hat{ı ı} + \frac{t^{2}}{2} \hat{ȷ ȷ} + \sin t \hat{k} + c

for some (new) constant vector

c .

Since

r (0) = \hat{ȷ ȷ},

we have

c = \hat{ȷ ȷ}

so that

\begin{matrix} r (t) = t \hat{ı ı} + (1 + \frac{t^{2}}{2}) \hat{ȷ ȷ} + \sin t \hat{k} \end{matrix}

(b) The particle has

x = π / 2

when

t = π / 2

and then

\begin{matrix} r_{1} = r (π / 2) = \frac{π}{2} \hat{ı ı} + (1 + \frac{π^{2}}{8}) \hat{ȷ ȷ} + \hat{k} \end{matrix}

t = 0

and time

t = π / 2

\begin{aligned} \int_{0}^{π / 2} F (t) \cdot d r & = \int_{0}^{π / 2} F (t) \cdot \frac{d r}{d t} (t) d t \\ = \int_{0}^{π / 2} [\hat{ȷ ȷ} - \sin t \hat{k}] \cdot [\hat{ı ı} + t \hat{ȷ ȷ} + \cos t \hat{k}] d t \\ = \int_{0}^{π / 2} [t - \sin t \cos t] d t = [\frac{t^{2}}{2} + \frac{1}{2} \cos^{2} t]_{0}^{π / 2} = \frac{π^{2}}{8} - \frac{1}{2} \end{aligned}

2.4.2.36. (✳).

Solution.

(a) We can parametrize

L

r (t) = (x (t), y (t)) = (t, t),

with

t

running from 2 to 1. Using this parametrization,

\begin{aligned} \int_{L} F \cdot d r & = \int_{2}^{1} F (x (t), y (t)) \cdot (x^{'} (t), y^{'} (t)) d t = \int_{2}^{1} (3 t, t - 1) \cdot (1, 1) d t \\ = \int_{2}^{1} (4 t - 1) d t = - 5 \end{aligned}

(b) First, we note that such a choice of path is even possible: if

F

were conservative, then

\int_{c} F \cdot d r

would be

- 5

for every path starting at

(2, 2)

and ending at

(1, 1),

because it would be path independent. Since

\frac{\partial F_{1}}{\partial y} = 3

and

\frac{\partial F_{2}}{\partial x} = 1 \neq \frac{\partial F_{1}}{\partial y},

by Theorem 2.4.7,

F

is not path-independent.

Solution 1:

Let’s try a family of polygonal paths

C_{Y}

that consist of

the line segment $L_{1}$ from $(2, 2)$ to $(2, Y)$ followed by
the line segment $L_{2}$ from $(2, Y)$ to $(1, Y)$ followed by
the line segment $L_{3}$ from $(1, Y)$ to $(1, 1) .$

This is a way of characterizing a family of alternate paths with only one parameter,

Y .

We are hoping that the value of the integral

\int_{C_{Y}} F \cdot d r

depends on

Y

and that we can choose a specific value of

Y

so as to make the value of the integral

\int_{C_{Y}} F \cdot d r

exactly

4 .

Note that

On $L_{1},$ $x = 2$ is a constant (so that $d x = 0$ ) and $y$ runs from $2$ to $Y .$
On $L_{2},$ $y = Y$ is a constant (so that $d y = 0$ ) and $x$ runs from $2$ to $1 .$
On $L_{3},$ $x = 1$ is a constant (so that $d x = 0$ ) and $y$ runs from $Y$ to $1$

So,

\begin{aligned} \int_{C_{Y}} F \cdot d r & = \int_{L_{1}} {(3 y d x + (x - 1) d y} + \int_{L_{2}} {(3 y d x + (x - 1) d y} \\ + \int_{L_{3}} {(3 y d x + (x - 1) d y} \\ = \int_{2}^{Y} d y + \int_{2}^{1} 3 Y d x + \int_{Y}^{1} 0 d y \\ = (Y - 2) + 3 Y (1 - 2) = - 2 Y - 2 \end{aligned}

Since we want our integral to be 4, we set

4 = - 2 Y - 2,

and find

Y = - 3 .

That is, the path

D

consisting of line segments from

(2, 2)

(2, - 3)

(1, - 3)

(1, 1)

gives us

\int_{D} F \cdot d r = 4 .

Solution 2: Choosing three straight line segments was a convenient way to solve this, but not the only way. To emphasize this point, we show that we also could have considered (for example) the family of parabolas that pass through

(2, 2)

and

(1, 1) .

That is, we consider the family of functions

y = a x^{2} + b x + c

with

2 = 4 a + 2 b + c

and

1 = a + b + c .

Subtracting the equation

a + b + c = 1

from the equation

4 a + 2 b + c = 2

(in order to eliminate

c

) gives

\begin{aligned} (4 a + 2 b + c) - (a + b + c) & = (2) - (1) \\ ⟹ & 3 a + b & = 1 \\ ⟹ & b & = 1 - 3 a \end{aligned}

Using $b = 1 - 3 a,$

\begin{aligned} a + b + c & = 1 \\ ⟹ & a + (1 - 3 a) + c & = 1 \\ ⟹ & c & = 2 a \end{aligned}

So, the class of functions described by

y = a x^{2} + (1 - 3 a) x + 2 a

for some constant

a

are parabolas that pass through

(1, 1)

and

(2, 2) .

So, we consider paths of the form:

\begin{aligned} r (x) & = (x, a x^{2} + (1 - 3 a) x + 2 a) \\ F (r (x)) & = (3 a x^{2} + 3 (1 - 3 a) x + 6 a, x - 1) \\ r^{'} (x) & = (1, 2 a x + 1 - 3 a) \\ F (r (x)) \cdot r^{'} (x) & = (3 a x^{2} + 3 (1 - 3 a) x + 6 a) \\ + (2 a x^{2} + (1 - 3 a) x - 2 a x + (3 a - 1)) \\ = 5 a x^{2} + (4 - 14 a) x + (9 a - 1) \end{aligned}

So, if $C$ is a portion of this parabola from $(2, 2)$ to $(1, 1),$ then

\begin{aligned} \int_{C} F \cdot d r & = \int_{2}^{1} (5 a x^{2} + (4 - 14 a) x + (9 a - 1)) d x \\ = {[\frac{5 a}{3} x^{3} + (2 - 7 a) x^{2} + (9 a - 1) x]}_{2}^{1} \\ = \frac{a}{3} - 5 \end{aligned}

Since we want our integral to have value 4, we set

4 = \frac{a}{3} - 5,

which yields

a = 27 .

If we choose

C

to be the path from

(2, 2)

(1, 1)

along the parabola

27 x^{2} - 80 x + 54,

then

\int_{C} F \cdot d r = 4,

as desired.

2.4.2.37. (✳).

Solution 1.

Let’s try a family of polygonal paths

C_{Y}

(sketched below) that consist of

the line segment $L_{1}$ from $(0, 0)$ to $(0, Y)$ followed by
the line segment $L_{2}$ from $(0, Y)$ to $(2, Y)$ followed by
the line segment $L_{3}$ from $(2, Y)$ to $(2, 0) .$

Here

Y

is a parameter.

We are hoping that the value of the integral

\int_{C_{Y}} F \cdot d r

depends on

Y

and that we can choose a specific value of

Y

so as to make the value of the integral

\int_{C_{Y}} F \cdot d r

exactly

8 .

Note that

on $L_{1},$ $x = 0$ is a constant (so that $d x = 0$ ) and $y$ runs from $0$ to $Y$ and
on $L_{2},$ $y = Y$ is a constant (so that $d y = 0$ ) and $x$ runs from $0$ to $2$ and
on $L_{3},$ $x = 2$ is a constant (so that $d x = 0$ ) and $y$ runs from $Y$ to $0$

Since

F \cdot d r = (2 y + 2) d x,

\begin{aligned} \int_{C_{Y}} F \cdot d r & = \int_{L_{1}} (2 y + 2) d x + \int_{L_{2}} (2 y + 2) d x + \int_{L_{3}} (2 y + 2) d x \\ = 0 + \int_{0}^{2} (2 Y + 2) d x + 0 \\ = 2 (2 Y + 2) \end{aligned}

Y = 1

does the job.

Solution 2.

There’s nothing magical about the form of the path from Solution 1. It’s just a path that’s relatively easy to describe using one constant

Y .

To emphasize this point, we provide a solution with an alternate path based on an ellipse.

A partial ellipse running from

(0, 0)

(2, 2)

can be described by

r (t) = (\cos t + 1, A \sin t)

for a constant

A,

with

t

running from

π

to 0. (To find this: we centre a circle of radius 1 at the point

(1, 0),

then multiply its

y

-coordinate by

A .

)

In this case,

F (r (t)) = (2 A \sin t + 2, 0)

and

r^{'} (t) = (- \sin t, A \cos t),

\begin{aligned} F (r (t)) \cdot r^{'} (t) & = - \sin t (2 A \sin 2 + 2) = - A (2 \sin^{2} t) - 2 \sin t \\ = - A (1 - \cos 2 t) - 2 \sin t \\ \int F \cdot d r & = \int_{π}^{0} (A (\cos 2 t - 1) - 2 \sin t) d t \\ = {[A (\frac{1}{2} \sin (2 t) - t) + 2 \cos t]}_{π}^{0} \\ = A π + 4 \end{aligned}

Setting

A π + 4 = 8,

we find

A = \frac{4}{π} .

So, the half-ellipse

r (t) = (\cos t + 1, \frac{4}{π} \sin t),

with

t

running from

π

to 0, is another path that gives

\int_{C} F \cdot d r = 8 .

2.4.2.38. (✳).

Solution.

The vector field

F

is conservative, with

F = \nabla \nabla φ φ (x, y) = x + \int_{0}^{y} \tilde{y} g (\tilde{y}) d \tilde{y}

Consquently, for

P = (x_{0}, 0)

and

Q = (x_{1}, 0),

\begin{aligned} \int_{C} F \cdot d r & = φ (Q) - φ (P) = x_{1} + \int_{0}^{0} \tilde{y} g (\tilde{y}) d \tilde{y} - x_{0} - \int_{0}^{0} \tilde{y} g (\tilde{y}) d \tilde{y} \\ = x_{1} - x_{0} \end{aligned}

Thus

| \int_{C} F \cdot d r | = | x (Q) - x (P) | = distance between P and Q

2.4.2.39. (✳).

Solution.

First notice that the vector field $\tilde{F} (x, y, z) = z^{2} \hat{k}$ is conservative (with potential $\frac{1}{3} z^{3}$ ), so $\int_{C_{1}} \tilde{F} \cdot d r = \int_{C_{2}} \tilde{F} \cdot d r$ for any two curves $C_{1}$ and $C_{2}$ from $P_{1}$ to $P_{2}$ (whether or not they are on the surface $S$ ). Consequently, the statement “ $\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r$ ” is true if and only if the statement “ $\int_{C_{1}} (F - \tilde{F}) \cdot d r = \int_{C_{2}} (F - \tilde{F}) \cdot d r$ ” is true. So we may replace the vector field $F$ with the vector field
$G (x, y, z) = F (x, y, z) - \tilde{F} (x, y, z) = (x z + a x y^{2}) \hat{ı ı} + y z \hat{ȷ ȷ}$
We are to consider only curves on the surface $S .$ For any such curve $C,$ say parametrized by $r (t)$ with $a \leq t \leq b,$ the integral
$\int_{C} G \cdot d r = \int_{a}^{b} G (r (t)) \cdot \frac{d r}{d t} (t) d t$
depends only on the values of $G$ on the surface $S .$ In particular, if another vector field $H$ obeys $H (x, y, z) = G (x, y, z),$ for all points $(x, y, z)$ on $S,$ we have
$\begin{aligned} \int_{C} G \cdot d r & = \int_{a}^{b} G (r (t)) \cdot \frac{d r}{d t} (t) d t = \int_{a}^{b} H (r (t)) \cdot \frac{d r}{d t} (t) d t \\ = \int_{C} H \cdot d r \end{aligned}$
So we may replace $G$ with
$\begin{aligned} H (x, y, z) & = G (x, y, 2 + x^{2} - 3 y^{2}) \\ = [x (2 + x^{2} - 3 y^{2}) + a x y^{2}] \hat{ı ı} + y (2 + x^{2} - 3 y^{2}) \hat{ȷ ȷ} \\ = (2 x + x^{3} - 3 x y^{2} + a x y^{2}) \hat{ı ı} + (2 y + y x^{2} - 3 y^{3}) \hat{ȷ ȷ} \end{aligned}$
Note that $H (x, y, z)$ is defined on all of $R^{3} .$ It just happens to not depend on $z .$
The curl of $H$ is
$\begin{aligned} \nabla \nabla \times H & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2 x + x^{3} - 3 x y^{2} + a x y^{2} & 2 y + y x^{2} - 3 y^{3} & 0 \end{array}] \\ = (2 x y - [- 6 x y + 2 a x y]) \hat{k} = (8 - 2 a) x y \hat{k} \end{aligned}$
This is zero if $a = 4 .$ As $H$ has continuous first order partial derivatives on all of $R^{3},$ Theorem 2.4.8 tells us that, when $a = 4,$ $H$ is conservative and that $\int_{C_{1}} H \cdot d r = \int_{C_{2}} H \cdot d r$ for any two curves $C_{1}$ and $C_{2}$ from $P_{1}$ to $P_{2}$

a = 4

does the job.

2.4.2.40. (✳).

Solution.

(a) The curl of

F

\begin{aligned} \nabla \nabla \times F = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (1 + a x^{2}) y e^{3 x^{2}} - b x z \cos (x^{2} z) & x e^{3 x^{2}} & x^{2} \cos (x^{2} z) \end{array}] \\ = 0 \hat{ı ı} + [- b x \cos (x^{2} z) + b x^{3} z \sin (x^{2} z) - 2 x \cos (x^{2} z) + 2 x^{3} z \sin (x^{2} z)] \hat{ȷ ȷ} \\ + [e^{3 x^{2}} + 6 x^{2} e^{3 x^{2}} - (1 + a x^{2}) e^{3 x^{2}}] \hat{k} \\ = [- (b + 2) x \cos (x^{2} z) + (b + 2) x^{3} z \sin (x^{2} z)] \hat{ȷ ȷ} + (6 - a) x^{2} e^{3 x^{2}} \hat{k} \end{aligned}

(b) For

F

to be conservative it is necessary that

\nabla \nabla \times F = 0 .

This is the case when

b = - 2

and

a = 6 .

f

to be a potential, when

b = - 2

and

a = 6,

we need

\begin{aligned} \frac{\partial f}{\partial x} (x, y, z) & = (1 + 6 x^{2}) y e^{3 x^{2}} + 2 x z \cos (x^{2} z) \\ \frac{\partial f}{\partial y} (x, y, z) & = x e^{3 x^{2}} \\ \frac{\partial f}{\partial z} (x, y, z) & = x^{2} \cos (x^{2} z) \end{aligned}

Integrating the second of these equations gives

f (x, y, z) = x y e^{3 x^{2}} + g (x, z)

Substituting this into the last equation gives

\frac{\partial g}{\partial z} (x, z) = x^{2} \cos (x^{2} z)

which integrates to

g (x, z) = \sin (x^{2} z) + h (x)

Finally, substituting

f (x, y, z) = x y e^{3 x^{2}} + \sin (x^{2} z) + h (x)

into the first equation gives

\begin{aligned} (1 + 6 x^{2}) y e^{3 x^{2}} + 2 x z \cos (x^{2} z) + h^{'} (x) & = (1 + 6 x^{2}) y e^{3 x^{2}} + 2 x z \cos (x^{2} z) \\ or h^{'} (x) = 0 \end{aligned}

h (x) = C

and hence

f (x, y, z) = x y e^{3 x^{2}} + \sin (x^{2} z) + C

works for any constant

C .

(d) Note that the integral is

\int_{C} (F_{a = 6, b = - 2} - 6 x^{2} y e^{3 x^{2}} \hat{ı ı}) \cdot d r .

\begin{aligned} \int_{C} (y e^{3 x^{2}} + 2 x z \cos (x^{2} z)) d x + x e^{3 x^{2}} d y + x^{2} \cos (x^{2} z) d z \\ = \int_{C} \nabla \nabla f \cdot d r - 6 \int_{C} x^{2} y e^{3 x^{2}} d x \\ = f (1, 1, 1) - f (0, 0, 0) - 6 \int_{0}^{1} t^{3} e^{3 t^{2}} d t \\ = e^{3} + \sin 1 - \frac{1}{3} \int_{0}^{3} u e^{u} d u with u = 3 t^{2}, d u = 6 t d t \\ = e^{3} + \sin 1 - \frac{1}{3} [u e^{u} - e^{u}]_{0}^{3} integration by parts \\ = \frac{1}{3} e^{3} + \sin 1 - \frac{1}{3} \end{aligned}

2.4.2.41. (✳).

Solution.

(a) Parametrize

C

x .

Then

\begin{aligned} r (x) & = x \hat{ı ı} + x^{2} \hat{ȷ ȷ} + x^{3} \hat{k} 0 \leq x \leq 1 \\ r^{'} (x) & = \hat{ı ı} + 2 x \hat{ȷ ȷ} + 3 x^{2} \hat{k} \\ F (r (x)) \cdot r^{'} (x) & = ((x^{4} - x^{2}) \hat{ı ı} + (x + x^{3}) \hat{ȷ ȷ} + x^{2} \hat{k}) \cdot (\hat{ı ı} + 2 x \hat{ȷ ȷ} + 3 x^{2} \hat{k}) \\ = x^{4} - x^{2} + 2 x^{2} + 2 x^{4} + 3 x^{4} = x^{2} + 6 x^{4} \\ \int_{C} F \cdot d r & = \int_{0}^{1} [x^{2} + 6 x^{4}] d x = [\frac{x^{3}}{3} + \frac{6 x^{5}}{5}]_{0}^{1} = \frac{23}{15} = 1.5 \dot{3} \end{aligned}

(b) Parametrize

C

x

as in part (a). Then

\begin{aligned} \frac{d s}{d x} & = | \frac{d r}{d x} | = \sqrt{1 + 4 x^{2} + 9 x^{4}} \\ ρ (x, x^{2}, x^{3}) \frac{d s}{d x} & = (8 x + 36 x^{3}) \sqrt{1 + 4 x^{2} + 9 x^{4}} \\ \int_{C} ρ d s & = \int_{0}^{1} (8 x + 36 x^{3}) \sqrt{1 + 4 x^{2} + 9 x^{4}} d x \end{aligned}

Using the substitution $u = 1 + 4 x^{2} + 9 x^{4},$ $d u = (8 x + 36 x^{3}) d x :$

\begin{aligned} = \frac{2}{3} [1 + 4 x^{2} + 9 x^{4}]^{3 / 2} |_{0}^{1} \\ = \frac{2}{3} [14^{3 / 2} - 1] \approx 34.26 \end{aligned}

F = \nabla \nabla f

with

f = x \sin y + y z + \frac{1}{2} z^{2},

\int_{C} F \cdot d r = f (1, 1, 1) - f (0, 0, 0) = \sin 1 + \frac{3}{2} \approx 2.3415

The potential

f

was just guessed. Alternatively, it can be found by antidifferentiating:

\begin{aligned} \frac{\partial f}{\partial x} (x, y, z) & = \sin y & ⟹ f (x, y, z) & = x \sin y + ψ_{1} (y, z) \\ \frac{\partial f}{\partial y} (x, y, z) & = x \cos y + z & ⟹ f (x, y, z) & = x \sin y + y z + ψ_{2} (x, z) \\ \frac{\partial f}{\partial z} (x, y, z) & = y + z & ⟹ f (x, y, z) & = y z + \frac{1}{2} z^{2} + ψ_{3} (x, y) \end{aligned}

All together,

f (x, y, z) = x \sin y + y z + \frac{z^{2}}{2} + C

works for any constant

C .

2.4.2.42. (✳).

Solution.

(a) This field is conservative if and only if it passes the screening test

\nabla \nabla \times F = 0 .

That is, if and only if,

\frac{\partial F_{1}}{\partial y} = \frac{\partial F_{2}}{\partial x} \frac{\partial F_{1}}{\partial z} = \frac{\partial F_{3}}{\partial x} \frac{\partial F_{2}}{\partial z} = \frac{\partial F_{3}}{\partial y}

That is,

\begin{aligned} \frac{\partial}{\partial y} (A x^{3} y^{2} z) & = \frac{\partial}{\partial x} (z^{3} + B x^{4} y z) & ⟺ & 2 A x^{3} y z & = 4 B x^{3} y z \\ \frac{\partial}{\partial z} (A x^{3} y^{2} z) & = \frac{\partial}{\partial x} (3 y z^{2} - x^{4} y^{2}) & ⟺ & A x^{3} y^{2} & = - 4 x^{3} y^{2} \\ \frac{\partial}{\partial z} (z^{3} + B x^{4} y z) & = \frac{\partial}{\partial y} (3 y z^{2} - x^{4} y^{2}) & ⟺ & 3 z^{2} + B x^{4} y & = 3 z^{2} - 2 x^{4} y \end{aligned}

Hence only

A = - 4,

B = - 2

works.

(b) When

A = - 4, B = - 2

F = - 4 x^{3} y^{2} z \hat{ı ı} + (z^{3} - 2 x^{4} y z) \hat{ȷ ȷ} + (3 y z^{2} - x^{4} y^{2}) \hat{k}

We find a potential function

φ (x, y, z)

for this

F

by antidifferentiating.

\begin{aligned} \frac{\partial φ}{\partial x} (x, y, z) & = - 4 x^{3} y^{2} z & ⟹ φ (x, y, z) & = - x^{4} y^{2} z + ψ_{1} (y, z) \\ \frac{\partial φ}{\partial y} (x, y, z) & = z^{3} - 2 x^{4} y z & ⟹ φ (x, y, z) & = y z^{3} - x^{4} y^{2} z + ψ_{2} (x, z) \\ \frac{\partial φ}{\partial z} (x, y, z) & = 3 y z^{2} - x^{4} y^{2} & ⟹ φ (x, y, z) & = y z^{3} - x^{4} y^{2} z + ψ_{3} (x, y) \end{aligned}

All together,

φ (x, y, z) = - x^{4} y^{2} z + y z^{3} + C

with

C

being an arbitrary constant.

(c)

I = φ (1, - 1, 1) - φ (0, 0, 0) = - 2 .

(d) Note that

J = \int_{C} G \cdot d r

with

\begin{aligned} G & = (z - 4 x^{3} y^{2} z) \hat{ı ı} + (z^{3} - x^{4} y z) \hat{ȷ ȷ} + (3 y z^{2} - x^{4} y^{2}) \hat{k} \\ = F + z \hat{ı ı} + x^{4} y z \hat{ȷ ȷ} \end{aligned}

so that

\begin{aligned} J & = \int_{C} (z \hat{ı ı} + x^{4} y z \hat{ȷ ȷ} + F) \cdot d r = - 2 + \int_{C} (z \hat{ı ı} + x^{4} y z \hat{ȷ ȷ}) \cdot d r \end{aligned}

Parametrize

C

r (x) = x \hat{ı ı} - x \hat{ȷ ȷ} + x^{2} \hat{k}

with

0 \leq x \leq 1 .

\frac{d r}{d x} = \hat{ı ı} - \hat{ȷ ȷ} + 2 x \hat{k}

\begin{aligned} \int_{C} (z \hat{ı ı} + x^{4} y z \hat{ȷ ȷ}) \cdot d r & = \int_{0}^{1} (x^{2} \hat{ı ı} - x^{7} \hat{ȷ ȷ}) \cdot (\hat{ı ı} - \hat{ȷ ȷ} + 2 x \hat{k}) d x \\ = \int_{0}^{1} (x^{2} + x^{7}) d x \\ = \frac{1}{3} + \frac{1}{8} \\ = \frac{11}{24} \\ ⟹ J = - 2 + \frac{11}{24} = - \frac{37}{24} \approx - 1.5417 \end{aligned}

(e)

T

is a closed path and

F

is conservative, so

\int_{T} F \cdot d r = 0 .

Let

T_{1}

be the line segment from

(1, 0, 0)

(0, 1, 0),

T_{2}

be the line segment from

(0, 1, 0)

(0, 0, 1)

and

T_{3}

be the line segment from

(0, 0, 1)

(1, 0, 0) .

T_{1},

z = 0,

\int_{T_{1}} z \hat{ı ı} \cdot d r = 0 .

T_{2},

x = 0,

\hat{ı ı} \cdot d r = d x = 0

and

\int_{T_{2}} z \hat{ı ı} \cdot d r = 0 .

Parametrize

T_{3}

r (t) = t \hat{ı ı} + (1 - t) \hat{k},

0 \leq t \leq 1 .

Then

\frac{d r}{d t} = \hat{ı ı} - \hat{k}

and the

z

-coordinate of the path is parametrized by

1 - t .

So,

\begin{aligned} \int_{T} (z \hat{ı ı} + F) \cdot d r & = \int_{T_{3}} z \hat{ı ı} \cdot d r = \int_{0}^{1} \overset{z \hat{ı ı}}{\overset{⏞}{(1 - t) \hat{ı ı}}} \cdot \overset{d r}{\overset{⏞}{(\hat{ı ı} - \hat{k}) d t}} \\ = \int_{0}^{1} (1 - t) d t = \frac{1}{2} \end{aligned}

2.4.2.43. (✳).

Solution.

(a) By Newton’s law of motion

m a = F ⟹ 2 v^{'} (t) = (4 t, 6 t^{2}, - 4 t) ⟹ v^{'} (t) = (2 t, 3 t^{2}, - 2 t)

\begin{aligned} v (t) & = v (0) + \int_{0}^{t} v^{'} (u) d u = (0, 0, 0) + \int_{0}^{t} (2 u, 3 u^{2}, - 2 u) d u \\ = (t^{2}, t^{3}, - t^{2}) \end{aligned}

(b) From part (a),

r^{'} (t) = v (t) = (t^{2}, t^{3}, - t^{2}) .

\begin{aligned} r (t) & = r (0) + \int_{0}^{t} r^{'} (u) d u = (1, 2, 3) + \int_{0}^{t} (u^{2}, u^{3}, - u^{2}) d u \\ = (1, 2, 3) + (t^{3} / 3, t^{4} / 4, - t^{3} / 3) = (\frac{t^{3}}{3} + 1, \frac{t^{4}}{4} + 2, - \frac{t^{3}}{3} + 3) \end{aligned}

| r^{'} (t) | = | t^{2} (1, t, - 1) | = t^{2} \sqrt{2 + t^{2}}

and

\begin{aligned} r^{'} (t) \times r^{″} (t) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ t^{2} & t^{3} & - t^{2} \\ 2 t & 3 t^{2} & - 2 t \end{array}] \\ = (- 2 t^{4} + 3 t^{4}) \hat{ı ı} - (- 2 t^{3} + 2 t^{3}) \hat{ȷ ȷ} + (3 t^{4} - 2 t^{4}) \hat{k} \\ = t^{4} \hat{ı ı} + t^{4} \hat{k} \\ ⟹ | r^{'} (t) \times r^{″} (t) | & = \sqrt{2} t^{4} \end{aligned}

The curvature is (see \S1.5)

\begin{aligned} κ (t) & = \frac{| r^{'} (t) \times r^{″} (t) |}{| r^{'} (t) |^{3}} = \frac{\sqrt{2} t^{4}}{{(t^{2} \sqrt{2 + t^{2}})}^{3}} \\ = \frac{\sqrt{2}}{t^{2} {(2 + t^{2})}^{3 / 2}} \end{aligned}

(d)

W = \int F \cdot d r :

\begin{aligned} \int_{t = 0}^{t = T} F (t) \cdot d r & = \int_{0}^{T} F (t) \cdot \frac{d r}{d t} (t) d t \\ = \int_{0}^{T} (4 t, 6 t^{2}, - 4 t) \cdot (t^{2}, t^{3}, - t^{2}) d t \\ = \int_{0}^{T} (8 t^{3} + 6 t^{5}) d t = 2 T^{4} + T^{6} \end{aligned}

2.4.2.44. (✳).

Solution.

(a) For the specified curve

\begin{aligned} r (t) & = (\frac{4 \sqrt{2}}{3} t^{3 / 2}, \frac{4 \sqrt{2}}{3} t^{3 / 2}, t (2 - t)) \\ v (t) = r^{'} (t) & = (2 \sqrt{2} t^{1 / 2}, 2 \sqrt{2} t^{1 / 2}, 2 - 2 t) \\ | v | & = \sqrt{8 t + 8 t + 4 - 8 t + 4 t^{2}} = \sqrt{4 (1 + 2 t + t^{2})} \\ = 2 | 1 + t | = 2 (1 + t) \end{aligned}

So the distance travelled is

\int_{0}^{2} | v (t) | d t = \int_{0}^{2} 2 (1 + t) d t = 2 [t + \frac{t^{2}}{2}]_{0}^{2} = 8

(b) As

\begin{aligned} v (t) = r^{'} (t) & = (2 \sqrt{2} t^{1 / 2}, 2 \sqrt{2} t^{1 / 2}, 2 - 2 t) & v (1) & = 2 \sqrt{2} (1, 1, 0) \\ a (t) = v^{'} (t) & = (\sqrt{2} t^{- 1 / 2}, \sqrt{2} t^{- 1 / 2}, - 2) & a (1) & = \sqrt{2} (1, 1, - \sqrt{2}) \\ v (1) \times a (1) & = 4 (- \sqrt{2}, \sqrt{2}, 0) & | v (1) | & = 4 \end{aligned}

the curvature

\begin{aligned} κ (1) & = \frac{| v (1) \times a (1) |}{| v (1) |^{3}} = \frac{8}{4^{3}} = \frac{1}{8} \end{aligned}

(c)

G = \nabla φ

with

φ (x, y, z) = - M g z,

so that gravity is conservative. The work done is

φ (r (2)) - φ (r (0)) = φ (\frac{16}{3}, \frac{16}{3}, 0) - φ (0, 0, 0) = 0

Friction is not conservative, so we have to compute the work long hand.

\begin{aligned} \int_{0}^{2} F \cdot d r & = \int_{0}^{2} F (t) \cdot \frac{d r}{d t} (t) d t = - \int_{0}^{2} | v (t) |^{2} v (t) \cdot v (t) d t \\ = - \int_{0}^{2} | v (t) |^{4} d t \\ = - 2^{4} \int_{0}^{2} (1 + t)^{4} d t = - \frac{16}{5} (1 + t)^{5} |_{0}^{2} \\ = - \frac{16}{5} (3^{5} - 1) \approx - 774.4 \end{aligned}

(d) Solution 1:

We know, from Theorem 1.3.3.c in the text, that

a (t) = \frac{d^{2} s}{d t^{2}} \hat{T} + κ (\frac{d s}{d t})^{2} \hat{N}

We have also been told that, at the apex,

\hat{N} = - \hat{k}

and that

\frac{d s}{d t} (t) = 3

for all

t .

\frac{d^{2} s}{d t^{2}} = 0 .

κ = \frac{1}{8}

at the apex

a (1) = 0 \hat{T} + \frac{1}{8} (3)^{2} (- \hat{k}) = - \frac{9}{8} \hat{k}

Solution 2:

The bird follows the parametrized path

r (u) = (\frac{4 \sqrt{2}}{3} u^{3 / 2}, \frac{4 \sqrt{2}}{3} u^{3 / 2}, u (2 - u))

This is the same path as the plane, but the parameter

u

is not time. Let’s denote by

R (t)

the position of the bird at time

t .

At time

t

the bird is at some point on the parametrized path, so there is some

u (t)

with

R (t) = r (u (t))

We saw in part (a) that

| \frac{d r}{d u} | = 2 (1 + u) .

Since the bird always has speed

3,

\begin{aligned} 3 = | \frac{d R}{d t} (t) | = | \frac{d r}{d u} (u (t)) \frac{d u}{d t} | = 2 (1 + u (t)) \frac{d u}{d t} \\ ⟹ \frac{d u}{d t} = \frac{3}{2 (1 + u (t))} \\ ⟹ \frac{d^{2} u}{d t^{2}} = - \frac{3}{2 (1 + u (t))^{2}} \frac{d u}{d t} = - \frac{9}{4 (1 + u (t))^{3}} \end{aligned}

At the apex

u = 1

so that

\frac{d u}{d t} = \frac{3}{4}

and

\frac{d^{2} u}{d t^{2}} = - \frac{9}{32} .

The bird’s acceleration is

\begin{aligned} \frac{d^{2} R}{d t^{2}} (t) & = \frac{d}{d t} (\frac{d R}{d t} (t)) = \frac{d}{d t} (\frac{d r}{d u} (u (t)) \frac{d u}{d t} (t)) \\ = \frac{d^{2} r}{d u^{2}} (\frac{d u}{d t})^{2} + \frac{d r}{d u} \frac{d^{2} u}{d t^{2}} \end{aligned}

From part (a)

\begin{aligned} \frac{d r}{d u} & = (2 \sqrt{2} u^{1 / 2}, 2 \sqrt{2} u^{1 / 2}, 2 - 2 u) \\ \frac{d^{2} r}{d u^{2}} & = (\sqrt{2} u^{- 1 / 2}, \sqrt{2} u^{- 1 / 2}, - 2) \end{aligned}

At the apex, when

u = 1,

\begin{aligned} \frac{d r}{d u} & = (2 \sqrt{2}, 2 \sqrt{2}, 0) \\ \frac{d^{2} r}{d u^{2}} & = (\sqrt{2}, \sqrt{2}, - 2) \end{aligned}

and the acceleration is

\begin{aligned} \frac{d^{2} R}{d t^{2}} & = \frac{d^{2} r}{d u^{2}} (\frac{d u}{d t})^{2} + \frac{d r}{d u} \frac{d^{2} u}{d t^{2}} \\ = (\sqrt{2}, \sqrt{2}, - 2) (\frac{3}{4})^{2} + (2 \sqrt{2}, 2 \sqrt{2}, 0) (- \frac{9}{32}) \\ = (0, 0, - \frac{9}{8}) \end{aligned}

3 Surface Integrals
3.1 Parametrized Surfaces

Exercises

3.1.1.

Solution.

This parametrization is almost trivial. We know it will have the form

r (x, y) = ψ_{1} (x, y) \hat{ı ı} + ψ_{2} (x, y) \hat{ȷ ȷ} + ψ_{3} (x, y) \hat{k}

where

ψ_{1}

gives the

x

-component (i.e.

x

ψ_{2}

gives the

y

-component (i.e.

y

), and

ψ_{3}

gives the

z

-component (i.e.

e^{x + 1} + x y

). So,

r (x, y) = x \hat{ı ı} + y \hat{ȷ ȷ} + (e^{x + 1} + x y) \hat{k}

3.1.2. (✳).

Solution.

Our parametrization is

\begin{aligned} x (u, v) & = u + v \\ y (u, v) & = u^{2} + v^{2} \\ z (u, v) & = u - v \end{aligned}

Adding $x (u, v)$ and $z (u, v)$ gives $x (u, v) + z (u, v) = 2 u .$
Subtracting $z (u, v)$ from $x (u, v)$ gives $x (u, v) - z (u, v) = 2 v .$

u = \frac{1}{2} (x (u, v) + z (u, v))

and

v = \frac{1}{2} (x (u, v) - z (u, v)) .

So on our surface

\begin{aligned} y (u, v) & = u^{2} + v^{2} = \frac{1}{4} (x (u, v) + z (u, v))^{2} + \frac{1}{4} (x (u, v) - z (u, v))^{2} \\ = \frac{1}{2} x (u, v)^{2} + \frac{1}{2} z (u, v)^{2} \end{aligned}

All points of our surface lie on

2 y = x^{2} + z^{2} .

This is a parabolic bowl:

no points have $y < 0$ and
the $y = Y$ (with $Y > 0$ ) cross-section is the circle $x^{2} + z^{2} = 2 Y,$ $y = Y$
the $x = 0$ cross-section is the parabola $2 y = z^{2},$ $x = 0$
the $z = 0$ cross-section is the parabola $2 y = x^{2},$ $z = 0$

3.1.3. (✳).

Solution.

Note that, since

x^{2} + y^{2} = 1 + 2 z^{2}

S,

the condition

z \geq 1

is equivalent to

x^{2} + y^{2} \geq 3,

z \geq 0 .

So the hyperboloid is

{(x, y, z) | x^{2} + y^{2} = 1 + 2 z^{2}, 3 \leq x^{2} + y^{2} \leq 9, z \geq 0} .

(a) No. Under this parametrization, the condition

3 \leq x^{2} + y^{2} \leq 9

3 \leq u^{2} + v^{2} \leq 9,

not

2 \leq u^{2} + v^{2} \leq 9 .

(b) Yes. Under this parametrization,

x = u \sin v,

y = - u \cos v

and

z = \sqrt{\frac{u^{2}}{2} - \frac{1}{2}} .

$x^{2} + y^{2} - 2 z^{2} = u^{2} - 2 (\frac{u^{2}}{2} - \frac{1}{2}) = 1,$ as desired.
The condition $x^{2} + y^{2} \leq 9$ is equivalent to $u \leq 3,$ since $u \geq 0 .$
The condition $x^{2} + y^{2} \geq 3$ is equivalent to $u \geq \sqrt{3},$ since $u \geq 0 .$
$z = \sqrt{\frac{u^{2}}{2} - \frac{1}{2}} \geq 0$

x = \sqrt{1 + 2 v^{2}} \cos u,

y = \sqrt{1 + 2 v^{2}} \sin u

and

z = v .

$x^{2} + y^{2} - 2 z^{2} = 1 + 2 v^{2} - 2 v^{2} = 1,$ as desired.
The condition $x^{2} + y^{2} \leq 9$ is equivalent to $1 + 2 v^{2} \leq 9,$ which is equivalent to $v \leq 2,$ since $v \geq 0 .$
The condition $x^{2} + y^{2} \geq 3$ is equivalent to $1 + 2 v^{2} \geq 3,$ which is equivalent to $v \geq 1,$ since $v \geq 0 .$
$z = v \geq 0$

(d) Yes. Under this parametrization,

x = \sqrt{1 + u} \sin v,

y = \sqrt{1 + u} \cos v

and

z = \sqrt{u / 2} .

$x^{2} + y^{2} - 2 z^{2} = 1 + u - 2 (u / 2) = 1,$ as desired.
The condition $x^{2} + y^{2} \leq 9$ is equivalent to $1 + u \leq 9,$ which is equivalent to $u \leq 8 .$
The condition $x^{2} + y^{2} \geq 3$ is equivalent to $1 + u \geq 3,$ which is equivalent to $u \geq 2 .$
$z = \sqrt{u / 2} \geq 0$

(e) No. Under this parametrization,

x = \sqrt{u} \cos v,

y = - \sqrt{u} \sin v

and

z = \sqrt{(u + 1) / 2} .

$x^{2} + y^{2} - 2 z^{2} = u - 2 (u + 1) / 2 = - 1,$ not $+ 1$

3.1.4. (✳).

Solution.

(a) No.

z = \sin ϕ \sin θ

is negative when

0 < ϕ \leq \frac{π}{4},

π < θ < 2 π .

(b) Yes. Note that

x^{2} + (- y)^{2} + (\sqrt{2 - x^{2} - y^{2}})^{2} = 2

and that, for

x^{2} + y^{2} \leq 1,

we have both

x^{2} + (- y)^{2} \leq 1

and

\sqrt{2 - x^{2} - y^{2}} \geq 0 .

(u \sin θ)^{2} + (u \cos θ)^{2} = u^{2} > 1

for

1 < u \leq 2 .

Also

\sqrt{2 - u^{2}}

is not defined for

\sqrt{2} < u \leq 2 .

(d) Yes. Note that

$(\sqrt{2} \sin ϕ \cos θ)^{2} + (\sqrt{2} \sin ϕ \sin θ)^{2} + (\sqrt{2} \cos ϕ)^{2} = 2$
For $0 \leq ϕ \leq \frac{π}{4},$ we have $z = \sqrt{2} \cos ϕ > 0 .$
As $ϕ$ runs from $0$ to $\frac{π}{4},$ $r (ϕ) = \sqrt{2} \sin ϕ$ runs from $0$ to $1,$ so that $(x = r (ϕ) \cos θ, y = r (ϕ) \sin θ)$ covers all of $x^{2} + y^{2} \leq 1$ as $ϕ$ runs from $0$ to $\frac{π}{4}$ and $θ$ runs from $0$ to $2 π .$

(e) Yes. Note that

$(- \sqrt{2 - z^{2}} \sin ϕ)^{2} + (\sqrt{2 - z^{2}} \cos ϕ)^{2} + (z)^{2} = 2$
For $1 \leq z \leq \sqrt{2},$ we have obviously have $z > 0 .$
As $z$ runs from $1$ to $\sqrt{2},$ $r (z) = \sqrt{2 - z^{2}}$ runs from $1$ to $0,$ so that $(x = - r (z) \sin ϕ, y = r (z) \cos ϕ)$ covers all of $x^{2} + y^{2} \leq 1$ as $z$ runs from $1$ to $\sqrt{2}$ and $ϕ$ runs from $0$ to $2 π .$

3.1.5. (✳).

Solution.

(a) No. When

u = v = 0,

z = 4

is not between

0

and

1 .

(b) Yes. Note that when

x = \sqrt{4 - u} \cos v,

y = \sqrt{4 - u} \sin v

and

z = u

with

0 \leq u \leq 1,

0 \leq v \leq 2 π,

$z + x^{2} + y^{2} = 4$
$0 \leq z = u \leq 1$
For each fixed $z = u$ between $0$ and $1,$ $(x, y)$ runs once around the circle $x^{2} + y^{2} = 4 - z = 4 - u$ as $v$ runs from $0$ to $2 π .$

x = u \cos v,

y = u \sin v

and

z = 4 - u^{2},

with

\sqrt{3} \leq u \leq 2,

0 \leq v \leq 2 π

$z + x^{2} + y^{2} = 4$
$0 \leq z = 4 - u^{2} \leq 1$
For each fixed $z = 4 - u^{2}$ between $0$ and $1,$ $(x, y)$ runs once around the circle $x^{2} + y^{2} = 4 - z = u^{2}$ as $v$ runs from $0$ to $2 π .$

3.1.6. (✳).

Solution.

First note that,

for A, B and C, $r (θ, ϕ) = x (θ, ϕ) \hat{ı ı} + y (θ, ϕ) \hat{ȷ ȷ} + z (θ, ϕ) \hat{k}$ obeys
$x (θ, ϕ)^{2} + y (θ, ϕ)^{2} + z (θ, ϕ)^{2} = 4$
and so lies on $S_{1}$
for D, E and F, $r (θ, z) = x (θ, z) \hat{ı ı} + y (θ, z) \hat{ȷ ȷ} + z (θ, z) \hat{k}$ obeys
$x (θ, z)^{2} + y (θ, z)^{2} = 4 - z (θ, z)^{2}$
and so lies on $S_{1}$
for G, H and I, $r (θ, z) = x (θ, z) \hat{ı ı} + y (θ, z) \hat{ȷ ȷ} + z (θ, z) \hat{k}$ obeys
$x (θ, z)^{2} + y (θ, z)^{2} = z (θ, z)^{2}$
and so lies on $S_{3}$
for J, K and L, $r (x, y) = x (x, y) \hat{ı ı} + y (x, y) \hat{ȷ ȷ} + z (x, y) \hat{k}$ obeys
$x (x, y)^{2} + y (x, y)^{2} = z (x, y)^{2}$
and so lies on $S_{3}$

(a) To get a part of

S_{1},

we need to use one of the parametrizations A, B, C, D, E, F. In the cases of A, B, C, for

r (θ, ϕ) = x (θ, ϕ) \hat{ı ı} + y (θ, ϕ) \hat{ȷ ȷ} + z (θ, ϕ) \hat{k}

to lie inside

S_{2}

we need (recalling that all points of

S_{1}

have

z (θ, ϕ) \geq 0

and hence

0 \leq ϕ \leq \frac{π}{2}

)

\begin{aligned} x (θ, ϕ)^{2} + y (θ, ϕ)^{2} \leq 1 & ⟺ 4 \sin^{2} ϕ \leq 1 ⟺ \sin ϕ \leq \frac{1}{2} \\ ⟺ 0 \leq ϕ \leq \frac{π}{6} \end{aligned}

In the cases of D, E, F, for

r (θ, z) = x (θ, z) \hat{ı ı} + y (θ, z) \hat{ȷ ȷ} + z (θ, z) \hat{k}

to lie inside

S_{2}

we need (recalling that all points of

S_{1}

have

z (θ, z) \geq 0

and hence

z \geq 0

)

\begin{aligned} x (θ, z)^{2} + y (θ, z)^{2} \leq 1 & ⟺ 4 - z^{2} \leq 1 ⟺ z \geq \sqrt{3} \end{aligned}

So parametrizations A and F work.

(b) To get a part of

S_{1},

we need to use one of the parametrizations A, B, C, D, E, F. In the cases of A, B, C, for

r (θ, ϕ) = x (θ, ϕ) \hat{ı ı} + y (θ, ϕ) \hat{ȷ ȷ} + z (θ, ϕ) \hat{k}

to lie inside

S_{3}

we need (recalling that all points of

S_{1}

have

z (θ, ϕ) \geq 0

and hence

0 \leq ϕ \leq \frac{π}{2}

)

\begin{aligned} x (θ, ϕ)^{2} + y (θ, ϕ)^{2} \leq z (θ, ϕ)^{2} & ⟺ 4 \sin^{2} ϕ \leq 4 \cos^{2} ϕ \\ ⟺ \tan ϕ \leq 1 \\ ⟺ 0 \leq ϕ \leq \frac{π}{4} \end{aligned}

In the cases of D, E, F, for

r (θ, z) = x (θ, z) \hat{ı ı} + y (θ, z) \hat{ȷ ȷ} + z (θ, z) \hat{k}

to lie inside

S_{3}

we need (recalling that all points of

S_{1}

have

z (θ, z) \geq 0

and hence

z \geq 0

)

\begin{aligned} x (θ, z)^{2} + y (θ, z)^{2} \leq z (θ, z)^{2} & ⟺ 4 - z^{2} \leq z^{2} ⟺ z \geq \sqrt{2} \end{aligned}

So parametrizations B and E work.

S_{3},

we need to use one of the parametrizations G, H, I, J, K, L. In the cases of G, H, I, for

r (θ, z) = x (θ, z) \hat{ı ı} + y (θ, z) \hat{ȷ ȷ} + z (θ, z) \hat{k}

to lie inside

S_{2}

we need (recalling that all points of

S_{3}

have

z \geq 0

)

\begin{aligned} x (θ, z)^{2} + y (θ, z)^{2} \leq 1 & ⟺ z^{2} \leq 1 ⟺ 0 \leq z \leq 1 \end{aligned}

In the cases of J, K, L, for

r (x, y) = x (x, y) \hat{ı ı} + y (x, y) \hat{ȷ ȷ} + z (x, y) \hat{k}

to lie inside

S_{3}

we need

\begin{aligned} x (x, y)^{2} + y (x, y)^{2} \leq 1 & ⟺ x^{2} + y^{2} \leq 1 \end{aligned}

So parametrizations G and J work.

(d) To get a part of

S_{3},

we need to use one of the parametrizations G, H, I, J, K, L. In the cases of G, H, I, for

r (θ, z) = x (θ, z) \hat{ı ı} + y (θ, z) \hat{ȷ ȷ} + z (θ, z) \hat{k}

to lie inside

S_{1}

we need (recalling that all points of

S_{3}

have

z \geq 0

)

\begin{aligned} x (θ, z)^{2} + y (θ, z)^{2} + z (θ, z)^{2} \leq 4 & ⟺ 2 z^{2} \leq 4 ⟺ 0 \leq z \leq \sqrt{2} \end{aligned}

In the cases of J, K, L, for

r (x, y) = x (x, y) \hat{ı ı} + y (x, y) \hat{ȷ ȷ} + z (x, y) \hat{k}

to lie inside

S_{3}

we need

\begin{aligned} x (x, y)^{2} + y (x, y)^{2} + z (x, y)^{2} \leq 4 & ⟺ 2 x^{2} + 2 y^{2} \leq 4 \end{aligned}

So parametrizations H and L work.

3.1.7.

Solution.

(a) In the sketch below, the point

(x, y, z)

deviates from the centre

(2, 2, 4)

\sin θ

units in the

\hat{k}

direction, and by

\cos θ

units in the

\sqrt{\frac{1}{\sqrt{2}}} (\hat{ı ı} + \hat{ȷ ȷ})

direction. So,

(x, y, z) = (2 + \frac{1}{\sqrt{2}} \cos θ, 2 + \frac{1}{\sqrt{2}} \cos θ, 4 + \sin θ) .

So, we can parametrize the circle as

(x, y, z) = (2 + \frac{1}{\sqrt{2}} \cos θ, 2 + \frac{1}{\sqrt{2}} \cos θ, 4 + \sin θ),

with

0 \leq θ \leq 2 π .

Remark: it’s easy to check that this equation satisfies the two properties we desire. Since the

x

- and

y

coordinates match, it’s in the plane

x = y .

To check that it’s a circle centred at

(2, 2, 4),

we note the distance from

(x, y, z)

(2, 2, 4)

is:

\begin{aligned} d & = \sqrt{(x - 2)^{2} + (y - 2)^{2} + (z - 4)^{2}} \\ = \sqrt{{(\frac{1}{\sqrt{2}} \cos θ)}^{2} + {(\frac{1}{\sqrt{2}} \cos θ)}^{2} + {(\sin θ)}^{2}} \\ = \sqrt{\frac{1}{2} \cos^{2} θ + \frac{1}{2} \cos^{2} θ + \sin^{2} θ} = \sqrt{\cos^{2} θ + \sin^{2} θ} \\ = 1 \end{aligned}

So, our points all have distance one from the same point — that is, they lie on a circle of radius 1.

(b) Consider a point

(x, y, z) = (2 + \frac{1}{\sqrt{2}} \cos θ, 2 + \frac{1}{\sqrt{2}} \cos θ, 4 + \sin θ),

rotating

ϕ

radians about the line

x = y = 4 .

The new position of the point has the same height,

z = 4 + \sin θ .

Its distance from the line

x = y = 4

is also preserved:

\begin{aligned} R & = \sqrt{(x - 4)^{2} + (y - 4)^{2} + (z - z)^{2}} \\ = \sqrt{(\frac{1}{\sqrt{2}} \cos θ - 2)^{2} + (\frac{1}{\sqrt{2}} \cos θ - 2)^{2} + 0)} \\ = \cos θ - 2 \sqrt{2} \end{aligned}

The circle traced out by a point

(x, y, z) = (2 + \frac{1}{\sqrt{2}} \cos θ, 2 + \frac{1}{\sqrt{2}} \cos θ, 4 + \sin θ)

on the circle is centred at

(4, 4, z)

with radius

\sqrt{2} (4 - x),

so it has equation

x = 4 + \sqrt{2} (2 - \sqrt{2} \cos θ) \cos ϕ,

y = 4 + \sqrt{2} (2 - \sqrt{2} \cos θ) \sin ϕ,

z = 4 \sin θ .

3.2 Tangent Planes

Exercises

3.2.1.

Solution.

Write

F (x, y, z) = x^{2} + y^{2} + (z - 1)^{2} - 1

and

G (x, y, z) = x^{2} + y^{2} + (z + 1)^{2} - 1 .

Let

S_{1}

denote the surface

F (x, y, z) = 0

and

S_{2}

denote the surface

G (x, y, z) = 0 .

First note that

F (0, 0, 0) = G (0, 0, 0) = 0

so that the point

(0, 0, 0)

lies on both

S_{1}

and

S_{2} .

The gradients of

F

and

G

are

\begin{aligned} \nabla \nabla F (x, y, z) & = (\frac{\partial F}{\partial x} (x, y, z), \frac{\partial F}{\partial y} (x, y, z), \frac{\partial F}{\partial z} (x, y, z)) \\ = (2 x, 2 y, 2 (z - 1)) \\ \nabla \nabla G (x, y, z) & = (\frac{\partial G}{\partial x} (x, y, z), \frac{\partial G}{\partial y} (x, y, z), \frac{\partial G}{\partial z} (x, y, z)) \\ = (2 x, 2 y, 2 (z + 1)) \end{aligned}

In particular,

\nabla \nabla F (0, 0, 0) = (0, 0, - 2) \nabla \nabla G (0, 0, 0) = (0, 0, 2)

so that the vector

\hat{k} = - \frac{1}{2} \nabla \nabla F (0, 0, 0) = \frac{1}{2} \nabla \nabla G (0, 0, 0)

is normal to both surfaces at

(0, 0, 0) .

So the tangent plane to both

S_{1}

and

S_{2}

(0, 0, 0)

\hat{k} \cdot (x - 0, y - 0, z - 0) = 0 or z = 0

Denote by

P

the plane

z = 0 .

Thus

S_{1}

is tangent to

P

(0, 0, 0)

and

P

is tangent to

S_{2}

(0, 0, 0) .

So it is reasonable to say that

S_{1}

and

S_{2}

are tangent at

(0, 0, 0) .

3.2.2.

Solution.

Denote by

S

the surface

G (x, y, z) = 0

and by

C

the parametrized curve

r (t) = (x (t), y (t), z (t)) .

To start, we’ll find the tangent plane to

S

r_{0}

and the tangent line to

C

r_{0} .

The tangent vector to $C$ at $r_{0}$ is $(x^{'} (t_{0}), y^{'} (t_{0}), z^{'} (t_{0})),$ so the parametric equations for the tangent line to $C$ at $r_{0}$ are
$\begin{matrix} (E1) & x - x_{0} = t x^{'} (t_{0}) y - y_{0} = t x^{'} (t_{0}) z - z_{0} = t x^{'} (t_{0}) \end{matrix}$
The gradient $(\frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}), \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}), \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0}))$ is a normal vector to the surface $S$ at $(x_{0}, y_{0}, z_{0}) .$ So the tangent plane to the surface $S$ at $(x_{0}, y_{0}, z_{0})$ is
$\begin{aligned} (\frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}), \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}), \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0})) \cdot \\ (x - x_{0}, y - y_{0}, z - z_{0}) = 0 \end{aligned}$
or
$\begin{aligned} \frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}) (x - x_{0}) + \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}) (y - y_{0}) \\ (E2) & + \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0}) (z - z_{0}) = 0 \end{aligned}$

Next, we’ll show that the tangent vector

(x^{'} (t_{0}), y^{'} (t_{0}), z^{'} (t_{0}))

C

r_{0}

and the normal vector

(\frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}), \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}), \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0}))

S

r_{0}

are perpendicular to each other. To do so, we observe that, for every

t,

the point

(x (t), y (t), z (t))

lies on the surface

G (x, y, z) = 0

and so obeys

\begin{matrix} G (x (t), y (t), z (t)) = 0 \end{matrix}

Differentiating this equation with respect to

t

gives, by the chain rule,

\begin{aligned} 0 & = \frac{d}{d t} G (x (t), y (t), z (t)) \\ = \frac{\partial G}{\partial x} (x (t), y (t), z (t)) x^{'} (t) + \frac{\partial G}{\partial y} (x (t), y (t), z (t)) y^{'} (t) \\ + \frac{\partial G}{\partial z} (x (t), y (t), z (t)) z^{'} (t) \end{aligned}

Then setting

t = t_{0}

gives

\begin{aligned} \frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}) x^{'} (t_{0}) + \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}) y^{'} (t_{0}) \\ (E3) & + \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0}) z^{'} (t_{0}) = 0 \end{aligned}

Finally, we are in a position to show that if

(x, y, z)

is any point on the tangent line to

C

r_{0},

then

(x, y, z)

is also on the tangent plane to

S

r_{0} .

(x, y, z)

is on the tangent line to

C

r_{0}

then there is a

t

such that, by (E1),

\begin{aligned} \frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}) {x - x_{0}} + \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}) {y - y_{0}} \\ + \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0}) {z - z_{0}} \\ = \frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}) {t x^{'} (t_{0})} + \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}) {t y^{'} (t_{0})} \\ + \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0}) {t z^{'} (t_{0})} \\ = t [\frac{\partial G}{\partial x} (x_{0}, y_{0}, z_{0}) x^{'} (t_{0}) + \frac{\partial G}{\partial y} (x_{0}, y_{0}, z_{0}) y^{'} (t_{0}) \\ + \frac{\partial G}{\partial z} (x_{0}, y_{0}, z_{0}) z^{'} (t_{0})] \\ = 0 \end{aligned}

by (E3). That is,

(x, y, z)

obeys the equation, (E2), of the tangent plane to

S

r_{0}

and so is on that tangent plane. So the tangent line to

C

r_{0}

is contained in the tangent plane to

S

r_{0} .

3.2.3.

Solution.

By part (b) of Theorem 3.2.1,

\begin{matrix} n = - f_{x} (x_{0}, y_{0}) \hat{ı ı} - f_{y} (x_{0}, y_{0}) \hat{ȷ ȷ} + \hat{k} \end{matrix}

is normal to the surface at

(x_{0}, y_{0}, z_{0}) .

So the parametric equations of the normal line are

\begin{aligned} (x - x_{0}, y - y_{0}, z - z_{0}) = t (- f_{x} (x_{0}, y_{0}), - f_{y} (x_{0}, y_{0}), 1) or \\ x = x_{0} - t f_{x} (x_{0}, y_{0}) y = y_{0} - t f_{y} (x_{0}, y_{0}) z = f (x_{0}, y_{0}) + t \end{aligned}

3.2.4.

Solution.

Use

S_{1}

to denote the surface

F (x, y, z) = 0,

S_{2}

to denote the surface

G (x, y, z) = 0

and

C

to denote the curve of intersection of

S_{1}

and

S_{2} .

Since $C$ is contained in $S_{1},$ the tangent line to $C$ at $(x_{0}, y_{0}, z_{0})$ is contained in the tangent plane to $S_{1}$ at $(x_{0}, y_{0}, z_{0}),$ by Q[3.2.2]. In particular, any tangent vector, $t,$ to $C$ at $(x_{0}, y_{0}, z_{0})$ must be perpendicular to $\nabla \nabla F (x_{0}, y_{0}, z_{0}),$ the normal vector to $S_{1}$ at $(x_{0}, y_{0}, z_{0}) .$
Since $C$ is contained in $S_{2},$ the tangent line to $C$ at $(x_{0}, y_{0}, z_{0})$ is contained in the tangent plane to $S_{2}$ at $(x_{0}, y_{0}, z_{0}),$ by Q[3.2.2]. In particular, any tangent vector, $t,$ to $C$ at $(x_{0}, y_{0}, z_{0})$ must be perpendicular to $\nabla \nabla G (x_{0}, y_{0}, z_{0}),$ the normal vector to $S_{2}$ at $(x_{0}, y_{0}, z_{0}) .$

So any tangent vector to

C

(x_{0}, y_{0}, z_{0})

must be perpendiular to both

\nabla \nabla F (x_{0}, y_{0}, z_{0})

and

\nabla \nabla G (x_{0}, y_{0}, z_{0}) .

One such tangent vector is

\begin{matrix} t = \nabla \nabla F (x_{0}, y_{0}, z_{0}) \times \nabla \nabla G (x_{0}, y_{0}, z_{0}) \end{matrix}

(Because the vectors

\nabla \nabla F (x_{0}, y_{0}, z_{0})

and

\nabla \nabla G (x_{0}, y_{0}, z_{0})

are nonzero and not parallel,

t

is nonzero.) So the normal plane in question passes through

(x_{0}, y_{0}, z_{0})

and has normal vector

n = t .

Consquently, the normal plane is

\begin{aligned} n \cdot (x - x_{0}, y - y_{0}, z - z_{0}) = 0 \\ where n = t = \nabla \nabla F (x_{0}, y_{0}, z_{0}) \times \nabla \nabla G (x_{0}, y_{0}, z_{0}) \end{aligned}

3.2.5.

Solution.

Use

S_{1}

to denote the surface

z = f (x, y),

S_{2}

to denote the surface

z = g (x, y)

and

C

to denote the curve of intersection of

S_{1}

and

S_{2} .

Since $C$ is contained in $S_{1},$ the tangent line to $C$ at $(x_{0}, y_{0}, z_{0})$ is contained in the tangent plane to $S_{1}$ at $(x_{0}, y_{0}, z_{0}),$ by Q[3.2.2]. In particular, any tangent vector, $t,$ to $C$ at $(x_{0}, y_{0}, z_{0})$ must be perpendicular to $- f_{x} (x_{0}, y_{0}) \hat{ı ı} - f_{y} (x_{0}, y_{0}) \hat{ȷ ȷ} + \hat{k},$ the normal vector to $S_{1}$ at $(x_{0}, y_{0}, z_{0}) .$ (See part (b) of Theorem 3.2.1.)
Since $C$ is contained in $S_{2},$ the tangent line to $C$ at $(x_{0}, y_{0}, z_{0})$ is contained in the tangent plane to $S_{2}$ at $(x_{0}, y_{0}, z_{0}),$ by Q[3.2.2]. In particular, any tangent vector, $t,$ to $C$ at $(x_{0}, y_{0}, z_{0})$ must be perpendicular to $- g_{x} (x_{0}, y_{0}) \hat{ı ı} - g_{y} (x_{0}, y_{0}) \hat{ȷ ȷ} + \hat{k},$ the normal vector to $S_{2}$ at $(x_{0}, y_{0}, z_{0}) .$

So any tangent vector to

C

(x_{0}, y_{0}, z_{0})

must be perpendicular to both of the vectors

- f_{x} (x_{0}, y_{0}) \hat{ı ı} - f_{y} (x_{0}, y_{0}) \hat{ȷ ȷ} + \hat{k}

and

- g_{x} (x_{0}, y_{0}) \hat{ı ı} - g_{y} (x_{0}, y_{0}) \hat{ȷ ȷ} + \hat{k} .

One such tangent vector is

\begin{aligned} t = [- f_{x} (x_{0}, y_{0}) \hat{ı ı} - f_{y} (x_{0}, y_{0}) \hat{ȷ ȷ} + \hat{k}] \times [- g_{x} (x_{0}, y_{0}) \hat{ı ı} - g_{y} (x_{0}, y_{0}) \hat{ȷ ȷ} + \hat{k}] \\ = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - f_{x} (x_{0}, y_{0}) & - f_{y} (x_{0}, y_{0}) & 1 \\ - g_{x} (x_{0}, y_{0}) & - g_{y} (x_{0}, y_{0}) & 1 \end{array}] \\ = (g_{y} (x_{0}, y_{0}) - f_{y} (x_{0}, y_{0}), f_{x} (x_{0}, y_{0}) - g_{x} (x_{0}, y_{0}), \\ f_{x} (x_{0}, y_{0}) g_{y} (x_{0}, y_{0}) - f_{y} (x_{0}, y_{0}) g_{x} (x_{0}, y_{0})) \end{aligned}

So the tangent line in question passes through

(x_{0}, y_{0}, z_{0})

and has direction vector

d = t .

Consquently, the tangent line is

(x - x_{0}, y - y_{0}, z - z_{0}) = t d

\begin{aligned} x & = x_{0} + t [g_{y} (x_{0}, y_{0}) - f_{y} (x_{0}, y_{0})] \\ y & = y_{0} + t [f_{x} (x_{0}, y_{0}) - g_{x} (x_{0}, y_{0})] \\ z & = z_{0} + t [f_{x} (x_{0}, y_{0}) g_{y} (x_{0}, y_{0}) - f_{y} (x_{0}, y_{0}) g_{x} (x_{0}, y_{0})] \end{aligned}

3.2.6. (✳).

Solution.

We are going to use part (b) of Theorem 3.2.1. To do so, we need the first order derivatives of

f (x, y)

(x, y) = (- 1, 1) .

So we find them first.

\begin{aligned} f_{x} (x, y) & = \frac{2 x y}{x^{4} + 2 y^{2}} - \frac{x^{2} y (4 x^{3})}{{(x^{4} + 2 y^{2})}^{2}} & f_{x} (- 1, 1) & = - \frac{2}{3} + \frac{4}{3^{2}} = - \frac{2}{9} \\ f_{y} (x, y) & = \frac{x^{2}}{x^{4} + 2 y^{2}} - \frac{x^{2} y (4 y)}{{(x^{4} + 2 y^{2})}^{2}} & f_{y} (- 1, 1) & = \frac{1}{3} - \frac{4}{3^{2}} = - \frac{1}{9} \end{aligned}

(2 / 9, 1 / 9, 1)

is a normal vector to the surface at

(- 1, 1, 1 / 3)

and the tangent plane is

\begin{aligned} \frac{2}{9} (x + 1) + \frac{1}{9} (y - 1) + (z - \frac{1}{3}) & = 0 \\ \frac{2}{9} x + \frac{1}{9} y + z & = - \frac{2}{9} + \frac{1}{9} + \frac{1}{3} = \frac{2}{9} \end{aligned}

2 x + y + 9 z = 2 .

3.2.7. (✳).

Solution.

The equation of the given surface is of the form

G (x, y, z) = 9

with

G (x, y, z) = \frac{27}{\sqrt{x^{2} + y^{2} + z^{2} + 3}} .

So, by part (c) of Theorem 3.2.1, a normal vector to the surface at

(2, 1, 1)

\begin{aligned} \nabla \nabla G (2, 1, 1) & = - \frac{1}{2} \frac{27}{(x^{2} + y^{2} + z^{2} + 3)^{3 / 2}} (2 x, 2 y, 2 z) |_{(x, y, z) = (2, 1, 1)} \\ = - (2, 1, 1) \end{aligned}

and the equation of the tangent plane is

- (2, 1, 1) \cdot (x - 2, y - 1, z - 1) = 0 or 2 x + y + z = 6

3.2.8. (✳).

Solution.

We may use

G (x, y, z) = x y z^{2} + y^{2} z^{3} - 3 - x^{2} = 0

as an equation for the surface. Note that

(- 1, 1, 2)

really is on the surface since

\begin{matrix} G (- 1, 1, 2) = (- 1) (1) (2)^{2} + (1)^{2} (2)^{3} - 3 - (- 1)^{2} = - 4 + 8 - 3 - 1 = 0 \end{matrix}

By part (c) of Theorem 3.2.1, since

\begin{aligned} G_{x} (x, y, z) & = y z^{2} - 2 x & G_{x} (- 1, 1, 2) & = 6 \\ G_{y} (x, y, z) & = x z^{2} + 2 y z^{3} & G_{y} (- 1, 1, 2) & = 12 \\ G_{z} (x, y, z) & = 2 x y z + 3 y^{2} z^{2} & G_{z} (- 1, 1, 2) & = 8 \end{aligned}

one normal vector to the surface at

(- 1, 1, 2)

\nabla \nabla G (- 1, 1, 2) = (6, 12, 8)

and an equation of the tangent plane to the surface at

(- 1, 1, 2)

\begin{matrix} (6, 12, 8) \cdot (x + 1, y - 1, z - 2) = 0 or 6 x + 12 y + 8 z = 22 \end{matrix}

z = - \frac{3}{4} x - \frac{3}{2} y + \frac{11}{4}

3.2.9. (✳).

Solution.

(a) The surface is

G (x, y, z) = z - x^{2} + 2 x y - y^{2} = 0 .

When

x = a

and

y = 2 a

and

(x, y, z)

is on the surface, we have

z = a^{2} - 2 (a) (2 a) + (2 a)^{2} = a^{2} .

So, by part (c) of Theorem 3.2.1, a normal vector to this surface at

(a, 2 a, a^{2})

\begin{matrix} \nabla \nabla G (a, 2 a, a^{2}) = (- 2 x + 2 y, 2 x - 2 y, 1) |_{(x, y, z) = (a, 2 a, a^{2})} = (2 a, - 2 a, 1) \end{matrix}

and the equation of the tangent plane is

\begin{matrix} (2 a, - 2 a, 1) \cdot (x - a, y - 2 a, z - a^{2}) = 0 or 2 a x - 2 a y + z = - a^{2} \end{matrix}

(b) The two planes are parallel when their two normal vectors, namely

(2 a, - 2 a, 1)

and

(1, - 1, 1),

are parallel. This is the case if and only if

a = \frac{1}{2} .

3.2.10. (✳).

Solution.

A plane is determined by one point on the plane and one vector perpendicular to the plane. We are told that

(8, 1, 5)

is on the plane, so it suffices to find a normal vector. The given surface is parametrized by

\begin{matrix} r (u, v) = 2 u^{2} \hat{ı ı} + v^{2} \hat{ȷ ȷ} + (u^{2} + v^{3}) \hat{k} \end{matrix}

so the vectors

\begin{aligned} \frac{\partial r}{\partial u} (u, v) & = (4 u, 0, 2 u) \\ \frac{\partial r}{\partial v} (u, v) & = (0, 2 v, 3 v^{2}) \end{aligned}

are tangent to

S

r (u, v) .

Note that

r (2, 1) = (8, 1, 5) .

\begin{aligned} \frac{\partial r}{\partial u} (2, 1) & = (8, 0, 4) \\ \frac{\partial r}{\partial v} (2, 1) & = (0, 2, 3) \end{aligned}

are tangent to

S

r (2, 1) = (8, 1, 5)

and

\begin{aligned} \frac{\partial r}{\partial u} (2, 1) \times \frac{\partial r}{\partial v} (2, 1) & = (8, 0, 4) \times (0, 2, 3) \\ = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 8 & 0 & 4 \\ 0 & 2 & 3 \end{array}] \\ = (- 8, - 24, 16) \end{aligned}

\frac{1}{- 8} (- 8, - 24, 16) = (1, 3, - 2)

is normal to

S

(8, 1, 5) .

So the tangent plane is

\begin{matrix} (1, 3, - 2) \cdot {(x, y, z) - (8, 1, 5)} = 0 or x + 3 y - 2 z = 1 \end{matrix}

3.2.11. (✳).

Solution.

To find the tangent plane we have to find a normal vector to the surface at

(2, 2, 0) .

Since

\begin{aligned} \frac{\partial r}{\partial u} & = (1, 2 u, 1) \\ \frac{\partial r}{\partial v} & = (1, 2 v, - 1) \end{aligned}

a normal vector to the surface at

r (u, v)

\begin{aligned} \frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 1 & 2 u & 1 \\ 1 & 2 v & - 1 \end{array}] \\ = (- 2 u - 2 v, 2, 2 v - 2 u) \end{aligned}

r (u, v) = (2, 2, 0)

when (the

x

-coordinate)

u + v = 2

and (the

z

-coordinate)

u - v = 0,

i.e when

u = v = 1,

a normal vector to the surface at

(2, 2, 0) = r (1, 1)

\begin{matrix} (- 4, 2, 0) or (- 2, 1, 0) \end{matrix}

and the equation of the specified tangent plane is

\begin{matrix} - 2 (x - 2) + (y - 2) + 0 z = 0 or y = 2 x - 2 \end{matrix}

3.2.12. (✳).

Solution.

The first order partial derivatives of

f

are

\begin{aligned} f_{x} (x, y) & = - \frac{4 x y}{{(x^{2} + y^{2})}^{2}} & f_{x} (- 1, 2) & = \frac{8}{25} \\ f_{y} (x, y) & = \frac{2}{x^{2} + y^{2}} - \frac{4 y^{2}}{{(x^{2} + y^{2})}^{2}} & f_{y} (- 1, 2) & = \frac{2}{5} - \frac{16}{25} = - \frac{6}{25} \end{aligned}

So, by part (b) of Theorem 3.2.1, a normal vector to the surface at

(x, y) = (- 1, 2)

(\frac{8}{25}, - \frac{6}{25}, - 1) .

f (- 1, 2) = \frac{4}{5},

the tangent plane is

\begin{aligned} (\frac{8}{25}, - \frac{6}{25}, - 1) \cdot (x + 1, y - 2, z - \frac{4}{5}) = 0 \\ or \frac{8}{25} x - \frac{6}{25} y - z = - \frac{8}{5} \end{aligned}

and the normal line is

\begin{matrix} (x, y, z) = (- 1, 2, \frac{4}{5}) + t (\frac{8}{25}, - \frac{6}{25}, - 1) \end{matrix}

3.2.13. (✳).

Solution.

A normal vector to the surface

x^{2} + 9 y^{2} + 4 z^{2} = 17

at the point

(x, y, z)

(2 x, 18 y, 8 z) .

A normal vector to the plane

x - 8 z = 0

(1, 0, - 8) .

So we want

(2 x, 18 y, 8 z)

to be parallel to

(1, 0, - 8),

i.e. to be a nonzero constant times

(1, 0, - 8) .

This is the case whenever

y = 0

and

z = - 2 x

with

x \neq 0 .

In addition, we want

(x, y, z)

to lie on the surface

x^{2} + 9 y^{2} + 4 z^{2} = 17 .

So we want

y = 0,

z = - 2 x

and

\begin{matrix} 17 = x^{2} + 9 y^{2} + 4 z^{2} = x^{2} + 4 (- 2 x)^{2} = 17 x^{2} ⟹ x = \pm 1 \end{matrix}

So the allowed points are

\pm (1, 0, - 2) .

3.2.14. (✳).

Solution.

The equation of

S

is of the form

G (x, y, z) = x^{2} + 2 y^{2} + 2 y - z = 1 .

So one normal vector to

S

at the point

(x_{0}, y_{0}, z_{0})

\nabla \nabla G (x_{0}, y_{0}, z_{0}) = 2 x_{0} \hat{ı ı} + (4 y_{0} + 2) \hat{ȷ ȷ} - \hat{k}

and the normal line to

S

(x_{0}, y_{0}, z_{0})

(x, y, z) = (x_{0}, y_{0}, z_{0}) + t (2 x_{0}, 4 y_{0} + 2, - 1)

For this normal line to pass through the origin, there must be a

t

with

\begin{matrix} (0, 0, 0) = (x_{0}, y_{0}, z_{0}) + t (2 x_{0}, 4 y_{0} + 2, - 1) \end{matrix}

\begin{aligned} (E1) & x_{0} + 2 x_{0} t & = 0 \\ (E2) & y_{0} + (4 y_{0} + 2) t & = 0 \\ (E3) & z_{0} - t & = 0 \end{aligned}

Equation (E3) forces

t = z_{0} .

Substituting this into equations (E1) and (E2) gives

\begin{aligned} (E1) & x_{0} (1 + 2 z_{0}) & = 0 \\ (E2) & y_{0} + (4 y_{0} + 2) z_{0} & = 0 \end{aligned}

The question specifies that

x_{0} \neq 0,

so (E1) forces

z_{0} = - \frac{1}{2} .

Substituting

z_{0} = - \frac{1}{2}

into (E2) gives

- y_{0} - 1 = 0 ⟹ y_{0} = - 1

Finally

x_{0}

is determined by the requirement that

(x_{0}, y_{0}, z_{0})

must lie on

S

and so must obey

\begin{aligned} z_{0} = x_{0}^{2} + 2 y_{0}^{2} + 2 y_{0} - 1 & ⟹ - \frac{1}{2} = x_{0}^{2} + 2 (- 1)^{2} + 2 (- 1) - 1 \\ ⟹ x_{0}^{2} = \frac{1}{2} \end{aligned}

So the allowed points

P

are

(\frac{1}{\sqrt{2}}, - 1, - \frac{1}{2})

and

(- \frac{1}{\sqrt{2}}, - 1, - \frac{1}{2}) .

3.2.15. (✳).

Solution.

Let

(x_{0}, y_{0}, z_{0})

be a point on the hyperboloid

z^{2} = 4 x^{2} + y^{2} - 1

where the tangent plane is parallel to the plane

2 x - y + z = 0 .

A normal vector to the plane

2 x - y + z = 0

(2, - 1, 1) .

Because the hyperboloid is

G (x, y, z) = 4 x^{2} + y^{2} - z^{2} - 1

and

\nabla \nabla G (x, y, z) = (8 x, 2 y, - 2 z),

a normal vector to the hyperboloid at

(x_{0}, y_{0}, z_{0})

\nabla \nabla G (x_{0}, y_{0}, z_{0}) = (8 x_{0}, 2 y_{0}, - 2 z_{0}) .

(x_{0}, y_{0}, z_{0})

satisfies the required conditions if and only if there is a nonzero

t

obeying

\begin{aligned} (8 x_{0}, 2 y_{0}, - 2 z_{0}) = t (2, - 1, 1) and z_{0}^{2} = 4 x_{0}^{2} + y_{0}^{2} - 1 \\ ⟺ x_{0} = \frac{t}{4}, y_{0} = z_{0} = - \frac{t}{2} and z_{0}^{2} = 4 x_{0}^{2} + y_{0}^{2} - 1 \\ ⟺ \frac{t^{2}}{4} = \frac{t^{2}}{4} + \frac{t^{2}}{4} - 1 and x_{0} = \frac{t}{4}, y_{0} = z_{0} = - \frac{t}{2} \\ ⟺ t = \pm 2 (x_{0}, y_{0}, z_{0}) = \pm (\frac{1}{2}, - 1, - 1) \end{aligned}

3.2.16. (✳).

Solution.

(a) A vector perpendicular to

x^{2} + z^{2} = 10

(1, 1, 3)

\nabla \nabla (x^{2} + z^{2}) |_{(1, 1, 3)} = (2 x \hat{ı ı} + 2 z \hat{k}) |_{(1, 1, 3)} = 2 \hat{ı ı} + 6 \hat{k} or \frac{1}{2} (2, 0, 6) = (1, 0, 3)

(b) A vector perpendicular to

y^{2} + z^{2} = 10

(1, 1, 3)

\nabla \nabla (y^{2} + z^{2}) |_{(1, 1, 3)} = (2 y \hat{ȷ ȷ} + 2 z \hat{k}) |_{(1, 1, 3)} = 2 \hat{ȷ ȷ} + 6 \hat{k} or \frac{1}{2} (0, 2, 6) = (0, 1, 3)

A vector is tangent to the specified curve at the specified point if and only if it perpendicular to both

(1, 0, 3)

and

(0, 1, 3) .

One such vector is

(0, 1, 3) \times (1, 0, 3) = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 0 & 1 & 3 \\ 1 & 0 & 3 \end{matrix}] = (3, 3, - 1)

(1, 1, 3)

and has direction vector

(1, 1, 3)

and so has vector parametric equation

r (t) = (1, 1, 3) + t (3, 3, - 1)

3.2.17. (✳).

Solution.

r (t) = (x (t), y (t), z (t))

intersects

z^{3} + x y z - 2 = 0

when

\begin{aligned} z (t)^{3} + x (t) y (t) z (t) - 2 = 0 & ⟺ (t^{2})^{3} + (t^{3}) (t) (t^{2}) - 2 = 0 \\ ⟺ 2 t^{6} = 2 ⟺ t = 1 \end{aligned}

since

t

is required to be positive. The direction vector for the curve at

t = 1

r^{'} (1) = 3 \hat{ı ı} + \hat{ȷ ȷ} + 2 \hat{k}

A normal vector for the surface at

r (1) = (1, 1, 1)

\nabla \nabla (z^{3} + x y z) |_{(1, 1, 1)} = [y z \hat{ı ı} + x z \hat{ȷ ȷ} + (3 z^{2} + x y) \hat{k}]_{(1, 1, 1)} = \hat{ı ı} + \hat{ȷ ȷ} + 4 \hat{k}

The angle

θ

between the curve and the normal vector to the surface is determined by

\begin{aligned} | (3, 1, 2) | | (1, 1, 4) | \cos θ = (3, 1, 2) \cdot (1, 1, 4) & ⟺ \sqrt{14} \sqrt{18} \cos θ = 12 \\ ⟺ \sqrt{7 \times 36} \cos θ = 12 \\ ⟺ \cos θ = \frac{2}{\sqrt{7}} \\ ⟺ θ = {40.89}^{\circ} \end{aligned}

The angle between the curve and the surface is

90 - 40.89 = {49.11}^{\circ}

(to two decimal places).

3.2.18.

Solution.

Let

(x_{0}, y_{0}, z_{0})

be any point on the surface. A vector normal to the surface at

(x_{0}, y_{0}, z_{0})

\begin{aligned} \nabla \nabla (x y e^{- (x^{2} + y^{2}) / 2} - z) |_{(x_{0}, y_{0}, z_{0})} \\ = (y_{0} e^{- (x_{0}^{2} + y_{0}^{2}) / 2} - x_{0}^{2} y_{0} e^{- (x_{0}^{2} + y_{0}^{2}) / 2}, x_{0} e^{- (x_{0}^{2} + y_{0}^{2}) / 2} - x_{0} y_{0}^{2} e^{- (x_{0}^{2} + y_{0}^{2}) / 2}, - 1) \end{aligned}

The tangent plane to the surface at

(x_{0}, y_{0}, z_{0})

is horizontal if and only if this vector is vertical, which is the case if and only if its

x

- and

y

-components are zero, which in turn is the case if and only if

\begin{aligned} y_{0} (1 - x_{0}^{2}) = 0 and x_{0} (1 - y_{0}^{2}) = 0 \\ ⟺ {y_{0} = 0 or x_{0} = 1 or x_{0} = - 1} and {x_{0} = 0 or y_{0} = 1 or y_{0} = - 1} \\ ⟺ (x_{0}, y_{0}) = (0, 0) or (1, 1) or (1, - 1) or (- 1, 1) or (- 1, - 1) \end{aligned}

The values of

z_{0}

at these points are

0,

e^{- 1},

- e^{- 1},

- e^{- 1}

and

e^{- 1},

respectively. So the horizontal tangent planes are

z = 0,

z = e^{- 1}

and

z = - e^{- 1} .

At the highest and lowest points of the surface, the tangent plane is horizontal. So the largest and smallest values of

z

are

e^{- 1}

and

- e^{- 1},

respectively.

3.3 Surface Integrals
3.3.6 Exercises

3.3.6.1.

Solution.

(a)

S

is the part of the plane

z = y \tan θ

that lies above the rectangle in the

x y

-plane with vertices

(0, 0),

(a, 0),

(0, b),

(a, b) .

S

is the rectangle with vertices

(0, 0, 0),

(a, 0, 0),

(0, b, b \tan θ),

(a, b, b \tan θ) .

So it has side lengths

\begin{aligned} | (a, 0, 0) - (0, 0, 0) | & = a \\ | (0, b, b \tan θ) - (0, 0, 0) | & = \sqrt{b^{2} + b^{2} \tan^{2} θ} \end{aligned}

and hence area

a b \sqrt{1 + \tan^{2} θ} = a b \sec θ .

(b)

S

is the part of the surface

z = f (x, y)

with

f (x, y) = y \tan θ

and with

(x, y)

running over

D = {(x, y) | 0 \leq x \leq a, 0 \leq y \leq b}

Hence by (3.3.2)

d S = \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} d x d y = \sqrt{1 + 0^{2} + \tan^{2} θ} d x d y

and

\begin{aligned} Area (S) & = \iint_{S} d S = \iint_{D} \sqrt{1 + \tan^{2} θ} d x d y \\ = \int_{0}^{a} d x \int_{0}^{b} d y \sqrt{1 + \tan^{2} θ} \\ = a b \sqrt{1 + \tan^{2} θ} = a b \sec θ \end{aligned}

3.3.6.2.

Solution.

Note that all three vertices

(a, 0, 0),

(0, b, 0)

and

(0, 0, c)

lie on the plane

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 .

So the triangle is part of that plane.

Method 1.

S

is the part of the surface

z = f (x, y)

with

f (x, y) = c (1 - \frac{x}{a} - \frac{y}{b})

and with

(x, y)

running over the triangle

T_{x y}

in the

x y

-plane with vertices

(0, 0, 0)

(a, 0, 0)

and

(0, b, 0) .

Hence by the first part of (3.3.2),

\begin{aligned} Area (S) & = \iint_{T_{x y}} \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} d x d y \\ = \iint_{T_{x y}} \sqrt{1 + \frac{c^{2}}{a^{2}} + \frac{c^{2}}{b^{2}}} d x d y \\ = \sqrt{1 + \frac{c^{2}}{a^{2}} + \frac{c^{2}}{b^{2}}} A (T_{x y}) \end{aligned}

where

A (T_{x y})

is the area of

T_{x y} .

Since the triangle

T_{x y}

has base

a

and height

b

(see the figure below), it has area

\frac{1}{2} a b .

Area (S) = \frac{1}{2} \sqrt{1 + \frac{c^{2}}{a^{2}} + \frac{c^{2}}{b^{2}}} a b = \frac{1}{2} \sqrt{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}}

Method 2.

S

is the part of the surface

x = g (y, z)

with

g (y, z) = a (1 - \frac{y}{b} - \frac{z}{c})

and with

(y, z)

running over the triangle

T_{y z}

in the

y z

-plane with vertices

(0, 0, 0)

(0, b, 0)

and

(0, 0, c) .

Hence by the second part of (3.3.2),

\begin{aligned} Area (S) & = \iint_{T_{y z}} \sqrt{1 + g_{y} (y, z)^{2} + g_{z} (y, z)^{2}} d y d z \\ = \iint_{T_{y z}} \sqrt{1 + \frac{a^{2}}{b^{2}} + \frac{a^{2}}{c^{2}}} d y d z \\ = \sqrt{1 + \frac{a^{2}}{b^{2}} + \frac{a^{2}}{c^{2}}} A (T_{y z}) \end{aligned}

where

A (T_{y z})

is the area of

T_{y z} .

Since

T_{y z}

has base

b

and height

c,

it has area

\frac{1}{2} b c .

Area (S) = \frac{1}{2} \sqrt{1 + \frac{a^{2}}{b^{2}} + \frac{a^{2}}{c^{2}}} b c = \frac{1}{2} \sqrt{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}}

Method 3.

S

is the part of the surface

y = h (x, z)

with

h (x, z) = b (1 - \frac{x}{a} - \frac{z}{c})

and with

(x, z)

running over the triangle

T_{x z}

in the

x z

-plane with vertices

(0, 0, 0)

(a, 0, 0)

and

(0, 0, c) .

Hence by the third part of (3.3.2),

\begin{aligned} Area (S) & = \iint_{T_{x z}} \sqrt{1 + h_{x} (x, z)^{2} + h_{z} (x, z)^{2}} d x d z \\ = \iint_{T_{x z}} \sqrt{1 + \frac{b^{2}}{a^{2}} + \frac{b^{2}}{c^{2}}} d x d z \\ = \sqrt{1 + \frac{b^{2}}{a^{2}} + \frac{b^{2}}{c^{2}}} A (T_{x z}) \end{aligned}

where

A (T_{x z})

is the area of

T_{x z} .

Since

T_{x z}

has base

a

and height

c,

it has area

\frac{1}{2} a c .

Area (S) = \frac{1}{2} \sqrt{1 + \frac{b^{2}}{a^{2}} + \frac{b^{2}}{c^{2}}} b c = \frac{1}{2} \sqrt{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}}

(b) We have already seen in the solution to part (a) that

Area (T_{x y}) = \frac{a b}{2} Area (T_{x z}) = \frac{a c}{2} Area (T_{y z}) = \frac{b c}{2}

Hence

\begin{aligned} Area (S) & = \sqrt{\frac{a^{2} b^{2}}{4} + \frac{a^{2} c^{2}}{4} + \frac{b^{2} c^{2}}{4}} \\ = \sqrt{Area (T_{x y})^{2} + Area (T_{x z})^{2} + Area (T_{y z})^{2}} \end{aligned}

3.3.6.3.

Solution.

(a) Think of the cylinder as being a piece of paper that has been partially rolled up. If you flatten the piece of paper out, you get a rectangle with the length of one side being

h

and the length of the other side being one quarter of the circumference of a circle of radius

a,

i.e.

\frac{1}{4} (2 π a) = \frac{π a}{2} .

So the area of

S

\frac{π a h}{2} .

(b)

S

is parametrized by

x (θ, y) = a \cos θ y (θ, y) = y z (θ, y) = a \sin θ

with

(θ, y)

running over

0 \leq θ \leq \frac{π}{2}, 0 \leq y \leq h .

Then, by (3.3.1),

\begin{aligned} (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) & = (- a \sin θ, 0, a \cos θ) \\ (\frac{\partial x}{\partial y}, \frac{\partial y}{\partial y}, \frac{\partial z}{\partial y}) & = (0, 1, 0) \\ d S & = | (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) \times (\frac{\partial x}{\partial y}, \frac{\partial y}{\partial y}, \frac{\partial z}{\partial y}) | d θ d y \\ = | (- a \cos θ, 0, - a \sin θ) | d θ d y \\ = a d θ d y \end{aligned}

\begin{aligned} Area (S) & = \iint_{S} d S = \int_{0}^{π / 2} d θ \int_{0}^{h} d y a = a (\frac{π}{2}) h \end{aligned}

3.3.6.4.

Solution.

The surface is

z = f (x, y)

with

f (x, y) = x y .

So, by (3.3.2),

d S = \sqrt{1 + f_{x}^{2} + f_{y}^{2}} d x d y = \sqrt{1 + x^{2} + y^{2}} d x d y

and

\begin{aligned} I & = \iint_{S} (x^{2} + y^{2}) d S = \iint_{x^{2} + y^{2} \leq 3} (x^{2} + y^{2}) \sqrt{1 + x^{2} + y^{2}} d x d y \\ = \int_{0}^{2 π} d θ \int_{0}^{\sqrt{3}} d r r r^{2} \sqrt{1 + r^{2}} \end{aligned}

We switched to polar coordinates in the last step. Making the change of variables

u = 1 + r^{2},

d u = 2 r d r

\begin{aligned} I & = π \int_{1}^{4} d u (u - 1) \sqrt{u} = π {[\frac{2}{5} u^{5 / 2} - \frac{2}{3} u^{3 / 2}]}_{1}^{4} \\ = π [\frac{64}{5} - \frac{16}{3} - \frac{2}{5} + \frac{2}{3}] \\ = \frac{116}{15} π \end{aligned}

3.3.6.5. (✳).

Solution.

First observe that any point

(x, y, z)

on the paraboliod lies above the

x y

-plane if and only if

0 \leq z = a^{2} - x^{2} - y^{2} ⟺ x^{2} + y^{2} \leq a^{2}

That is, if and only if

(x, y)

lies in the circular disk of radius

a

centred on the origin. The equation of the paraboloid is of the form

z = f (x, y)

with

f (x, y) = a^{2} - x^{2} - y^{2} .

So, by (3.3.2),

\begin{aligned} Surface area & = \iint_{x^{2} + y^{2} \leq a^{2}} \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} d x d y \\ = \iint_{x^{2} + y^{2} \leq a^{2}} \sqrt{1 + 4 x^{2} + 4 y^{2}} d x d y \end{aligned}

Switching to polar coordinates,

\begin{aligned} Surface area & = \int_{0}^{a} d r \int_{0}^{2 π} d θ r \sqrt{1 + 4 r^{2}} \\ = 2 π \int_{0}^{a} d r r \sqrt{1 + 4 r^{2}} \\ = 2 π \int_{1}^{1 + 4 a^{2}} \frac{d s}{8} \sqrt{s} with s = 1 + 4 r^{2}, d s = 8 r d r \\ = \frac{π}{4} \frac{2}{3} s^{3 / 2} |_{s = 1}^{s = 1 + 4 a^{2}} \\ = \frac{π}{6} [{(1 + 4 a^{2})}^{3 / 2} - 1] \end{aligned}

3.3.6.6. (✳).

Solution.

First observe that any point

(x, y, z)

on the cone lies between the planes

z = 2

and

z = 3

if and only if

4 \leq x^{2} + y^{2} \leq 9 .

The equation of the cone can be rewritten in the form

z = f (x, y)

with

f (x, y) = \sqrt{x^{2} + y^{2}} .

Note that

\begin{matrix} f_{x} (x, y) = \frac{x}{\sqrt{x^{2} + y^{2}}} f_{y} (x, y) = \frac{y}{\sqrt{x^{2} + y^{2}}} \end{matrix}

So, by (3.3.2),

\begin{aligned} Surface area & = \iint_{4 \leq x^{2} + y^{2} \leq 9} \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} d x d y \\ = \iint_{4 \leq x^{2} + y^{2} \leq 9} \sqrt{1 + \frac{x^{2}}{x^{2} + y^{2}} + \frac{y^{2}}{x^{2} + y^{2}}} d x d y \\ = \sqrt{2} \iint_{4 \leq x^{2} + y^{2} \leq 9} d x d y \end{aligned}

Now the domain of integration is a circular washer with outside radius

3

and inside radius

2

and hence of area

π (3^{2} - 2^{2}) = 5 π .

So the surface area is

5 \sqrt{2} π .

3.3.6.7. (✳).

Solution.

The equation of the surface is of the form

z = f (x, y)

with

f (x, y) = \frac{2}{3} (x^{3 / 2} + y^{3 / 2}) .

Note that

\begin{matrix} f_{x} (x, y) = \sqrt{x} f_{y} (x, y) = \sqrt{y} \end{matrix}

So, by (3.3.2),

\begin{aligned} Surface area & = \int_{0}^{1} d x \int_{0}^{1} d y \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} \\ = \int_{0}^{1} d x \int_{0}^{1} d y \sqrt{1 + x + y} \\ = \int_{0}^{1} d x [\frac{2}{3} (1 + x + y)^{3 / 2}]_{y = 0}^{y = 1} \\ = \frac{2}{3} \int_{0}^{1} d x [(2 + x)^{3 / 2} - (1 + x)^{3 / 2}] \\ = \frac{2}{3} \frac{2}{5} [(2 + x)^{5 / 2} - (1 + x)^{5 / 2}]_{x = 0}^{x = 1} \\ = \frac{4}{15} [3^{5 / 2} - 2^{5 / 2} - 2^{5 / 2} + 1^{5 / 2}] \\ = \frac{4}{15} [9 \sqrt{3} - 8 \sqrt{2} + 1] \end{aligned}

3.3.6.8. (✳).

Solution.

(a) By (3.3.2),

F (x, y) = \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} .

(b) (i) The “dimple” to be painted is part of the upper sphere

x^{2} + y^{2} + (z - 2 \sqrt{3})^{2} = 4 .

It is on the bottom half of the sphere and so has equation

z = f (x, y) = 2 \sqrt{3} - \sqrt{4 - x^{2} - y^{2}} .

Note that

\begin{matrix} f_{x} (x, y) = \frac{x}{\sqrt{4 - x^{2} - y^{2}}} f_{y} (x, y) = \frac{y}{\sqrt{4 - x^{2} - y^{2}}} \end{matrix}

The point on the dimple with the largest value of

x

(1, 0, \sqrt{3}) .

(It is marked by a dot in the figure above.) The dimple is invariant under rotations around the

z

--axis and so has

(x, y)

running over

x^{2} + y^{2} \leq 1 .

So, by (3.3.2),

\begin{aligned} Surface area & = \iint_{x^{2} + y^{2} \leq 1} \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} d x d y \\ = \iint_{x^{2} + y^{2} \leq 1} \sqrt{1 + \frac{x^{2}}{4 - x^{2} - y^{2}} + \frac{y^{2}}{4 - x^{2} - y^{2}}} d x d y \\ = \iint_{x^{2} + y^{2} \leq 1} \frac{2}{\sqrt{4 - x^{2} - y^{2}}} d x d y \end{aligned}

Switching to polar coordinates,

\begin{aligned} Surface area & = \int_{0}^{2 π} d θ \int_{0}^{1} d r \frac{2 r}{\sqrt{4 - r^{2}}} \end{aligned}

(b) (ii) Observe that if we flip the dimple up by reflecting it in the plane

z = \sqrt{3},

as in the figure below, the “Death Star” becomes a perfect ball of radius

2 .

The area of the pink dimple in the figure above is identical to the area of the blue cap in that figure. So the total surface area of the Death Star is exactly the surface area of a sphere of radius

2

and so is

4 π 2^{2} = 16 π .

3.3.6.9.

Solution.

The equation of the half of the cone with

y \geq 0

can be rewritten in the form

y = h (x, z)

with

h (x, z) = \sqrt{x^{2} + z^{2}} .

Note that

\begin{matrix} h_{x} (x, z) = \frac{x}{\sqrt{x^{2} + z^{2}}} h_{z} (x, z) = \frac{z}{\sqrt{x^{2} + z^{2}}} \end{matrix}

The point

(x, y, z)

y = h (x, z)

also obeys

x \geq 0,

0 \leq y \leq 2,

z \geq 0

if and only if

(x, z)

lies in the quarter disk

D = {(x, z) | x^{2} + z^{2} \leq 4, x \geq 0, z \geq 0}

So, by (3.3.2),

\begin{aligned} Surface area & = \iint_{D} \sqrt{1 + h_{x} (x, z)^{2} + h_{z} (x, z)^{2}} d x d z \\ = \iint_{D} \sqrt{1 + \frac{x^{2}}{x^{2} + z^{2}} + \frac{z^{2}}{x^{2} + z^{2}}} d x d z \\ = \sqrt{2} \iint_{D} d x d z \end{aligned}

Now

D

is a one quarter of a circular disk with radius

2 .

Surface area = \sqrt{2} \frac{1}{4} π 2^{2} = \sqrt{2} π

3.3.6.10. (✳).

Solution.

We are to find the surface area of part of a hemisphere. On the hemisphere

\begin{aligned} z = f (x, y) & = \sqrt{a^{2} - x^{2} - y^{2}} \\ f_{x} (x, y) & = - \frac{x}{\sqrt{a^{2} - x^{2} - y^{2}}} \\ f_{y} (x, y) & = - \frac{y}{\sqrt{a^{2} - x^{2} - y^{2}}} \end{aligned}

so that

\begin{aligned} d S & = \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} d x d y \\ = \sqrt{1 + \frac{x^{2}}{a^{2} - x^{2} - y^{2}} + \frac{y^{2}}{a^{2} - x^{2} - y^{2}}} d x d y \\ = \sqrt{\frac{a^{2}}{a^{2} - x^{2} - y^{2}}} d x d y \end{aligned}

In polar coordinates, this is

d S = \frac{a}{\sqrt{a^{2} - r^{2}}} r d r d θ .

We are to find the surface area of the part of the hemisphere that is inside the cylinder,

x^{2} - a x + y^{2} = 0,

which in polar coordinates becomes

r^{2} - a r \cos θ = 0

r = a \cos θ .

The top half of the domain of integration is sketched below.

So the

\begin{aligned} S u r f a c e A r e a & = 2 \int_{0}^{π / 2} d θ \int_{0}^{a \cos θ} d r r \frac{a}{\sqrt{a^{2} - r^{2}}} \\ = 2 a \int_{0}^{π / 2} d θ [- \sqrt{a^{2} - r^{2}}]_{0}^{a \cos θ} \\ = 2 a \int_{0}^{π / 2} d θ [a - a \sin θ] \\ = 2 a^{2} [θ + \cos θ]_{0}^{π / 2} = a^{2} [π - 2] \end{aligned}

3.3.6.11.

Solution.

The upper half cone obeys

f (x, y, z) = x^{2} + y^{2} - z^{2} = 0 .

So, by (3.3.3),

d S = | \frac{\nabla \nabla f}{\nabla \nabla f \cdot \hat{k}} | d x d y = | \frac{2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} - 2 z \hat{k}}{- 2 z} | d x d y = \frac{\sqrt{x^{2} + y^{2} + z^{2}}}{z} d x d y

But on the cone

x^{2} + y^{2} = z^{2},

and

z > 0,

so that

d S = \frac{\sqrt{x^{2} + y^{2} + z^{2}}}{z} d x d y = \frac{\sqrt{2 z^{2}}}{z} d x d y = \sqrt{2} d x d y

and

\begin{aligned} x^{4} - y^{4} + y^{2} z^{2} - z^{2} x^{2} + 1 & = x^{4} - y^{4} + y^{2} (x^{2} + y^{2}) - (x^{2} + y^{2}) x^{2} + 1 = 1 \end{aligned}

We have to integrate

(x, y)

over the interior of

x^{2} + y^{2} = 2 x,

or equivalently, the interior of

(x - 1)^{2} + y^{2} = 1,

which is the disk

D = {(x, y) | (x - 1)^{2} + y^{2} \leq 1}

\begin{aligned} \iint_{S} \overset{= 1}{\overset{⏞}{(x^{4} - y^{4} + y^{2} z^{2} - z^{2} x^{2} + 1)}} d S & = \sqrt{2} \iint_{D} d x d y = \sqrt{2} Area (D) = \sqrt{2} π \end{aligned}

3.3.6.12.

Solution.

As we saw in Example 3.1.5, the torus may be parametrized by

\begin{aligned} r (θ, ψ) = (R + r \cos θ) \cos ψ \hat{ı ı} + (R + r \cos θ) \sin ψ \hat{ȷ ȷ} + r \sin θ \hat{k} \\ 0 \leq θ, ψ \leq 2 π \end{aligned}

Then

\begin{aligned} \frac{\partial r}{\partial ψ} & = (R + r \cos θ) [- \sin ψ \hat{ı ı} + \cos ψ \hat{ȷ ȷ}] \\ \frac{\partial r}{\partial θ} & = r [- \sin θ \cos ψ \hat{ı ı} - \sin θ \sin ψ \hat{ȷ ȷ} + \cos θ \hat{k}] \end{aligned}

and

\begin{aligned} \frac{\partial r}{\partial ψ} \times \frac{\partial r}{\partial θ} & = r (R + r \cos θ) [- \sin ψ \hat{ı ı} + \cos ψ \hat{ȷ ȷ}] \times \\ [- \sin θ \cos ψ \hat{ı ı} - \sin θ \sin ψ \hat{ȷ ȷ} + \cos θ \hat{k}] \\ = r (R + r \cos θ) det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - \sin ψ & \cos ψ & 0 \\ - \sin θ \cos ψ & - \sin θ \sin ψ & \cos θ \end{array}] \\ = r (R + r \cos θ) [\cos ψ \cos θ \hat{ı ı} + \sin ψ \cos θ \hat{ȷ ȷ} + \sin θ \hat{k}] \end{aligned}

[\cos ψ \cos θ \hat{ı ı} + \sin ψ \cos θ \hat{ȷ ȷ} + \sin θ \hat{k}]

is a unit vector, (we could have shortened this computation by observing that

- \sin ψ \hat{ı ı} + \cos ψ \hat{ȷ ȷ}

and

- \sin θ \cos ψ \hat{ı ı} - \sin θ \sin ψ \hat{ȷ ȷ} + \cos θ \hat{k}

are mutually perpendicular unit vectors, so that their cross product is automatically a unit vector) and

| \frac{\partial r}{\partial ψ} \times \frac{\partial r}{\partial θ} | = r (R + r \cos θ) ⟹ d S = r (R + r \cos θ) d ψ d θ

The total surface area of the torus is

r \int_{0}^{2 π} d θ \int_{0}^{2 π} d ψ (R + r \cos θ) = 2 π r \int_{0}^{2 π} d θ (R + r \cos θ) = (2 π)^{2} R r

3.3.6.13.

Solution.

By symmetry, the centroid

(\bar{x}, \bar{y}, \bar{z})

obeys

\bar{x} = \bar{y} = \bar{z} .

Parametrize the sphere using spherical coordinates.

r (θ, φ) = a \sin φ \cos θ \hat{ı ı} + a \sin φ \sin θ \hat{ȷ ȷ} + a \cos φ \hat{k}

Then

\begin{aligned} \frac{\partial r}{\partial θ} & = - a \sin φ \sin θ \hat{ı ı} + a \sin φ \cos θ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial φ} & = a \cos φ \cos θ \hat{ı ı} + a \cos φ \sin θ \hat{ȷ ȷ} - a \sin φ \hat{k} \end{aligned}

so that

\begin{aligned} \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - a \sin φ \sin θ & a \sin φ \cos θ & 0 \\ a \cos φ \cos θ & a \cos φ \sin θ & - a \sin φ \end{array}] \\ = - a^{2} \sin^{2} φ \cos θ \hat{ı ı} - a^{2} \sin^{2} φ \sin θ \hat{ȷ ȷ} - a^{2} \sin φ \cos φ \hat{k} \\ ⟹ d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} | d θ d φ = a^{2} \sin φ d θ d φ \end{aligned}

As the surface area of the part of the sphere in the first octant is

\frac{1}{8} 4 π a^{2} = \frac{π a^{2}}{2}

\begin{aligned} \bar{x} = \bar{y} = \bar{z} & = \frac{\iint_{S} z d S}{\iint_{S} d S} = \frac{1}{π a^{2} / 2} \int_{0}^{π / 2} d θ \int_{0}^{π / 2} d φ (a^{2} \sin φ) (\overset{z}{\overset{⏞}{a \cos φ}}) \\ = \frac{2 a}{π} \frac{π}{2} \int_{0}^{π / 2} d φ \sin φ \cos φ = a [\frac{1}{2} \sin^{2} φ]_{0}^{π / 2} = \frac{a}{2} \end{aligned}

3.3.6.14.

Solution.

In cylindrical coordinates

x = r \cos θ y = r \sin θ z = z

In these coordinates the equation,

x^{2} + y^{2} = 2 a y,

of the cylinder becomes

r^{2} = 2 a r \sin θ or r = 2 a \sin θ

That is,

r = f (θ)

with

f (θ) = 2 a \sin θ .

Parametrize the cylinder by

r (θ, z) = x (θ, z) \hat{ı ı} + y (θ, z) \hat{ȷ ȷ} + z (θ, z) \hat{k}

with

\begin{aligned} x (θ, z) & = f (θ) \cos θ = 2 a \sin θ \cos θ = a \sin 2 θ \\ y (θ, z) & = f (θ) \sin θ = 2 a \sin θ \sin θ = a (1 - \cos 2 θ) \\ z (θ, z) & = z \end{aligned}

Under this parametrization,

\begin{aligned} \frac{\partial r}{\partial θ} = 2 a \cos 2 θ \hat{ı ı} + 2 a \sin 2 θ \hat{ȷ ȷ} \frac{\partial r}{\partial z} = \hat{k} \\ ⟹ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} = - 2 a \cos 2 θ \hat{ȷ ȷ} + 2 a \sin 2 θ \hat{ı ı} \\ ⟹ d S = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} | d θ d z = 2 a d θ d z \end{aligned}

We still have to determine the limits of integration. The figure on the left below provides a top view of the cylinder.

From it we see that

0 \leq θ \leq π .

The cone

z^{2} = x^{2} + y^{2} = r^{2}

(i.e.

z = \pm r

) and the cylinder

r = 2 a \sin θ

intersect at

z^{2} = r^{2} = 4 a^{2} \sin^{2} θ .

So, for each fixed

θ,

z

runs from

- 2 a \sin θ

z = + 2 a \sin θ .

(See the figure on the right above. It shows a constant

θ

cross-section.) Finally,

\begin{aligned} A r e a & = \int_{| z | \leq 2 a \sin θ} 2 a d θ d z = 2 a \int_{0}^{π} d θ \int_{- 2 a \sin θ}^{2 a \sin θ} d z = 8 a^{2} \int_{0}^{π} d θ \sin θ \\ = 8 a^{2} [- \cos θ]_{0}^{π} \\ = 16 a^{2} \end{aligned}

3.3.6.15. (✳).

Solution.

(a) This right circular cone symmetric about the

z

-axis projects down onto a disk

D

in the plane

z = 0 .

Setting

z = b

gives

D = {(x, y, z) | x^{2} + y^{2} \leq a^{2}, z = 0}

Since

G (x, y, z) = b^{2} (x^{2} + y^{2}) - a^{2} z^{2}

is constant on

S,

the area elements

d S

S

are related to area elements

d x d y

D

as follows:

\begin{aligned} d S & = \frac{| \nabla G (x, y, z) |}{| \nabla G (x, y, z) \cdot \hat{k} |} d x d y = \frac{2 | (b^{2} x, b^{2} y, - a^{2} z) |}{2 | a^{2} z |} d x d y \\ = \frac{\sqrt{b^{4} (x^{2} + y^{2}) + a^{4} z^{2}}}{a^{2} z} d x d y \end{aligned}

by (3.3.3). The defining equation for

S

gives

z = \frac{b}{a} \sqrt{x^{2} + y^{2}},

d S = \frac{\sqrt{b^{4} (x^{2} + y^{2}) + a^{2} b^{2} (x^{2} + y^{2})}}{a \sqrt{b^{2} (x^{2} + y^{2})}} d x d y = \frac{1}{a} \sqrt{a^{2} + b^{2}} d x d y .

Hence

I = \frac{\sqrt{a^{2} + b^{2}}}{a} \iint_{D} (x^{2} + y^{2}) d x d y .

(b) Or, parametrize the surface

S

using

θ

and

t

as follows:

\begin{aligned} (*) & x = t \cos θ, y = t \sin θ, z = \frac{b}{a} \sqrt{x^{2} + y^{2}} = \frac{b}{a} t, \\ 0 \leq θ \leq 2 π, 0 \leq t \leq a . \end{aligned}

Then we have, by (3.3.1),

\begin{aligned} \frac{\partial r}{\partial t} \times \frac{\partial r}{\partial θ} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \cos θ & \sin θ & b / a \\ - t \sin θ & t \cos θ & 0 \end{array}] \\ = (- \frac{b}{a} t \cos θ, - \frac{b}{a} t \sin θ, t), \end{aligned}

d S = | \frac{\partial r}{\partial t} \times \frac{\partial r}{\partial θ} | d t d θ = t \sqrt{1 + b^{2} / a^{2}} d t d θ .

It follows that for the rectangular region

R

of the

t θ

-plane described in

(*),

I = \iint_{R} (t^{2}) t \sqrt{1 + b^{2} / a^{2}} d t d θ .

I = \frac{\sqrt{a^{2} + b^{2}}}{a} \int_{θ = 0}^{2 π} \int_{r = 0}^{a} r^{2} r d r d θ = \frac{π}{2} a^{3} \sqrt{a^{2} + b^{2}} .

Direct integration in (b) gives the same thing, because

I = \iint_{D} (t^{2}) t \sqrt{1 + b^{2} / a^{2}} d t d θ = \frac{\sqrt{a^{2} + b^{2}}}{a} \int_{θ = 0}^{2 π} \int_{t = 0}^{a} t^{3} d t d θ .

3.3.6.16.

Solution.

(a) The surface is

g (x, y, z) = x^{2} + y^{2} + z^{2} - a^{2} = 0 .

So, on the surface of the sphere,

\begin{aligned} \hat{n} = \frac{\nabla \nabla g}{| \nabla \nabla g |} = \frac{x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ ⟹ F \cdot \hat{n} = (x^{2} + y^{2} + z^{2})^{n + + 1 - 1 / 2} = {(a^{2})}^{n + 1 / 2} = a^{2 n + 1} \\ ⟹ \iint_{S} F \cdot \hat{n} d S = a^{2 n + 1} \iint_{S} d S = a^{2 n + 1} Area (S) = 4 π a^{2 n + 3} \end{aligned}

since the surface area of a sphere of radius

a

4 π a^{2} .

(b) The box has six faces.

\begin{aligned} S_{1} & = {(x, y, z) | 0 \leq x \leq a, 0 \leq y \leq b, z = c} with outward normal \hat{n} = \hat{k} \\ S_{2} & = {(x, y, z) | 0 \leq x \leq a, 0 \leq y \leq b, z = 0} with outward normal \hat{n} = - \hat{k} \\ S_{3} & = {(x, y, z) | 0 \leq x \leq a, 0 \leq z \leq c, y = b} with outward normal \hat{n} = \hat{ȷ ȷ} \\ S_{4} & = {(x, y, z) | 0 \leq x \leq a, 0 \leq z \leq c, y = 0} with outward normal \hat{n} = - \hat{ȷ ȷ} \\ S_{5} & = {(x, y, z) | 0 \leq y \leq b, 0 \leq z \leq c, x = a} with outward normal \hat{n} = \hat{ı ı} \\ S_{6} & = {(x, y, z) | 0 \leq y \leq b, 0 \leq z \leq c, x = 0} with outward normal \hat{n} = - \hat{ı ı} \end{aligned}

For

S_{1},

i.e. the

z = c

face, and

S_{2},

i.e. the

z = 0

face,

\begin{aligned} \int_{\binom{z = c}{f a c e}} F \cdot \hat{n} d S & = \int_{\binom{z = c}{f a c e}} (x \hat{ı ı} + y \hat{ȷ ȷ} + c \hat{k}) \cdot \hat{k} d x d y & = c \int_{\binom{z = c}{f a c e}} d x d y = a b c \\ \int_{\binom{z = 0}{f a c e}} F \cdot \hat{n} d S & = \int_{\binom{z = 0}{f a c e}} (x \hat{ı ı} + y \hat{ȷ ȷ} + 0 \hat{k}) \cdot (- \hat{k}) d x d y & = 0 \end{aligned}

because the

z = c

face has area

a b .

Similarly,

\int_{\binom{x = 0}{f a c e}} F \cdot \hat{n} d S = \int_{\binom{y = 0}{f a c e}} F \cdot \hat{n} d S = 0 \int_{\binom{x = a}{f a c e}} F \cdot \hat{n} d S = \int_{\binom{y = b}{f a c e}} F \cdot \hat{n} d S = a b c

The total flux is

3 a b c .

{(x, y, z) | x^{2} + y^{2} \leq 1, z = 0}

and has (outward) normal

\hat{n} = - \hat{k} .

So The flux through the base is

\int_{b a s e} F \cdot \hat{n} d S = \iint_{x^{2} + y^{2} \leq 1} (y \hat{ı ı}) \cdot (- \hat{k}) d x d y = 0

In cylindrical coordinates

x = r \cos θ,

y = r \sin θ,

z = z

and the equation

z = 1 - \sqrt{x^{2} + y^{2}}

of the top part of the cone becomes

z = 1 - r .

So we may parametrize the top part of the cone by

r (r, θ) = r \cos θ \hat{ı ı} + r \sin θ \hat{ȷ ȷ} + (1 - r) \hat{k} with 0 \leq θ \leq 2 π, 0 \leq r \leq 1

Then

\begin{aligned} \frac{\partial r}{\partial r} & = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} - \hat{k} \\ \frac{\partial r}{\partial θ} & = - r \sin θ \hat{ı ı} + r \cos θ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial r} \times \frac{\partial r}{\partial θ} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \cos θ & \sin θ & - 1 \\ - r \sin θ & r \cos θ & 0 \end{array}] \\ = - r \cos θ \hat{ı ı} - r \sin θ \hat{ȷ ȷ} + r \hat{k} \\ ⟹ \hat{n} d S & = \frac{\partial r}{\partial r} \times \frac{\partial r}{\partial θ} d r d θ \\ = (- r \cos θ \hat{ı ı} - r \sin θ \hat{ȷ ȷ} + r \hat{k}) d r d θ \end{aligned}

by (3.1.1). We have the orientation correct because the

\hat{k}

component of

\hat{n}

is positive. The flux through the top, as well as the total flux, is

\begin{aligned} \int_{t o p} F \cdot \hat{n} d S \\ = \int_{0}^{1} d r \int_{0}^{2 π} d θ (\overset{y}{\overset{⏞}{r \sin θ}} \hat{ı ı} + \overset{z}{\overset{⏞}{(1 - r)}} \hat{k}) \cdot (- r \cos θ \hat{ı ı} - r \sin θ \hat{ȷ ȷ} + r \hat{k}) \\ = \int_{0}^{1} d r \int_{0}^{2 π} d θ (- r^{2} \sin θ \cos θ + r (1 - r)) \\ = - [\int_{0}^{1} d r r^{2}] [\int_{0}^{2 π} d θ \frac{1}{2} \sin (2 θ)] + 2 π \int_{0}^{1} d r [r - r^{2}] \\ = - \frac{1}{3} \times 0 + 2 π [\frac{1}{2} - \frac{1}{3}] = \frac{π}{3} \end{aligned}

3.3.6.17. (✳).

Solution.

Let

G (x, y, z) = x^{2} + y^{2} + 2 z .

Then, by (3.3.3),

\begin{aligned} \hat{n} d S & = \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} d x d y = \frac{2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + 2 \hat{k}}{2} d x d y = (x \hat{ı ı} + y \hat{ȷ ȷ} + \hat{k}) d x d y \end{aligned}

so that

\begin{aligned} \frac{x^{2} + y^{2}}{\sqrt{1 + x^{2} + y^{2}}} d S & = \frac{x^{2} + y^{2}}{\sqrt{1 + x^{2} + y^{2}}} \sqrt{1 + x^{2} + y^{2}} d x d y = (x^{2} + y^{2}) d x d y \end{aligned}

and

\begin{aligned} F \cdot \hat{n} d S & = [x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}] \cdot [x \hat{ı ı} + y \hat{ȷ ȷ} + \hat{k}] d x d y = [x^{2} + y^{2} + z] d x d y \\ = [1 + \frac{1}{2} (x^{2} + y^{2})] d x d y \end{aligned}

since

z = 1 - \frac{1}{2} (x^{2} + y^{2})

S .

(a)

\begin{aligned} \iint_{S} \frac{x^{2} + y^{2}}{\sqrt{1 + x^{2} + y^{2}}} d S & = \iint_{S} (x^{2} + y^{2}) d x d y = 4 \int_{0}^{1} d x \int_{0}^{1} d y (x^{2} + y^{2}) \\ = 4 \int_{0}^{1} d x (x^{2} + \frac{1}{3}) = 4 (\frac{1}{3} + \frac{1}{3}) = \frac{8}{3} \end{aligned}

(b)

\iint_{S} F \cdot \hat{n} d S = \int_{- 1}^{1} d x \int_{- 1}^{1} d y [1 + \frac{1}{2} (x^{2} + y^{2})] = 2 \times 2 + \frac{1}{2} \times \frac{8}{3} = \frac{16}{3}

3.3.6.18. (✳).

Solution.

Let

G (x, y, z) = z - x y .

Then, using (3.3.3),

\begin{aligned} \hat{n} d S & = \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} d x d y = \frac{- y \hat{ı ı} - x \hat{ȷ ȷ} + \hat{k}}{1} d x d y = (- y \hat{ı ı} - x \hat{ȷ ȷ} + \hat{k}) d x d y \end{aligned}

so that

\begin{aligned} \frac{x^{2} y}{\sqrt{1 + x^{2} + y^{2}}} d S & = \frac{x^{2} y}{\sqrt{1 + x^{2} + y^{2}}} \sqrt{y^{2} + x^{2} + 1} d x d y = x^{2} y d x d y \end{aligned}

and

\begin{aligned} F \cdot \hat{n} d S & = [x \hat{ı ı} + y \hat{ȷ ȷ} + \hat{k}] \cdot [- y \hat{ı ı} - x \hat{ȷ ȷ} + \hat{k}] d x d y = [1 - 2 x y] d x d y \end{aligned}

(a)

\begin{aligned} \iint_{S} \frac{x^{2} y}{\sqrt{1 + x^{2} + y^{2}}} d S & = \iint_{S} x^{2} y d x d y = \int_{0}^{1} d x \int_{0}^{1} d y x^{2} y \\ = \int_{0}^{1} d x \frac{1}{2} x^{2} = \frac{1}{6} \end{aligned}

(b)

\iint_{S} F \cdot \hat{n} d S = \int_{0}^{1} d x \int_{0}^{1} d y [1 - 2 x y] = \int_{0}^{1} d x [1 - x] = 1 - \frac{1}{2} = \frac{1}{2}

3.3.6.19. (✳).

Solution.

For the surface

z = f (x, y) = y^{3 / 2},

d S = \sqrt{1 + f_{x}^{2} + f_{y}^{2}} d x d y = \sqrt{1 + (\frac{3}{2} \sqrt{y})^{2}} d x d y = \sqrt{1 + \frac{9}{4} y} d x d y

by (3.3.2). So the area is

\begin{aligned} \int_{0}^{1} d x \int_{0}^{1} d y \sqrt{1 + \frac{9}{4} y} & = \int_{0}^{1} d x \frac{8}{27} [(1 + \frac{9}{4} y)^{3 / 2}]_{0}^{1} \\ = \int_{0}^{1} d x \frac{8}{27} [(\frac{13}{4})^{3 / 2} - 1] \\ = \frac{8}{27} [{(\frac{13}{4})}^{3 / 2} - 1] \end{aligned}

3.3.6.20. (✳).

Solution.

The surface is a sphere of radius

2

centered on

(0, 0, 2),

The plane

z = 1

intersects the sphere on the circle

x^{2} + y^{2} = 3 .

Let

F (x, y, z) = x^{2} + y^{2} + (z - 2)^{2} .

Then, by (3.3.3),

\begin{aligned} d S & = | \frac{\nabla \nabla F}{\nabla \nabla F \cdot \hat{k}} | d x d y = | \frac{2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + 2 (z - 2) \hat{k}}{2 (z - 2)} | d x d y \\ = | \frac{x \hat{ı ı} + y \hat{ȷ ȷ} + (z - 2) \hat{k}}{(z - 2)} | d x d y \\ = \frac{\sqrt{x^{2} + y^{2} + (z - 2)^{2}}}{| z - 2 |} d x d y \\ = \frac{2}{| z - 2 |} d x d y \end{aligned}

since

x^{2} + y^{2} + (z - 2)^{2} = 4

S .

S,

z \leq 2,

| z - 2 | = 2 - z

and

\begin{aligned} \iint_{S} f (x, y, z) d S & = \iint_{x^{2} + y^{2} \leq 3} (2 - z) (x^{2} + y^{2}) \frac{2}{| z - 2 |} d x d y \\ = 2 \iint_{x^{2} + y^{2} \leq 3} (x^{2} + y^{2}) d x d y \end{aligned}

Switching to polar coordinates

\iint_{S} f (x, y, z) d S = 2 \int_{0}^{\sqrt{3}} d r r \int_{0}^{2 π} d θ r^{2} = 2 (2 π) \frac{r^{4}}{4} |_{0}^{\sqrt{3}} = 9 π

3.3.6.21. (✳).

Solution.

(a) Each (horizontal) constant

z

cross-section is a circle centred on the

z

-axis. The radius varies linearly from

2,

when

z = 0

0,

when

z = 3 .

So the radius at height

z

\frac{2}{3} (3 - z)

and we can use

\begin{aligned} r (θ, z) = \frac{2}{3} (3 - z) \cos θ \hat{ı ı} + \frac{2}{3} (3 - z) \sin θ \hat{ȷ ȷ} + z \hat{k} \\ 0 \leq θ < 2 π, 0 \leq z \leq 3 \end{aligned}

as the parametrization.

(b) By symmetry the centre of mass will lie on the

z

-axis. We are only asked for the

z

-coordinate anyway. The

z

-coordinate of the centre of mass is the weighted average of

z

over the cone. Since a density has not been specified, we assume that it is a constant. We may take the density to be

1,

so the

z

-coordinate of the centre of mass is

\iint_{S} z d S / \iint_{S} d S .

Since

\begin{aligned} \frac{\partial r}{\partial θ} & = (- \frac{2}{3} (3 - z) \sin θ, \frac{2}{3} (3 - z) \cos θ, 0) \\ \frac{\partial r}{\partial z} & = (- \frac{2}{3} \cos θ, - \frac{2}{3} \sin θ, 1) \\ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} & = (\frac{2}{3} (3 - z) \cos θ, \frac{2}{3} (3 - z) \sin θ, \frac{4}{9} (3 - z)) \end{aligned}

the element of surface area for this parametrization is

\begin{aligned} d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} | d θ d z = \frac{2}{3} (3 - z) | (\cos θ, \sin θ, \frac{2}{3}) | d θ d z \\ = \frac{2 \sqrt{13}}{9} (3 - z) d θ d z \end{aligned}

So the surface area,

\iint_{S} d S,

of the cone is

\begin{aligned} \int_{0}^{3} d z \int_{0}^{2 π} d θ \frac{2 \sqrt{13}}{9} (3 - z) & = \frac{4 \sqrt{13}}{9} π \int_{0}^{3} d z (3 - z) \\ = - \frac{2 \sqrt{13}}{9} π (3 - z)^{2} |_{0}^{3} \\ = 2 \sqrt{13} π \end{aligned}

and the

z

-coordinate of the centre of mass is

\begin{aligned} \bar{z} & = \frac{1}{2 \sqrt{13} π} \int_{0}^{3} d z \int_{0}^{2 π} d θ \frac{2 \sqrt{13}}{9} (3 - z) z \\ = \frac{2}{9} \int_{0}^{3} d z (3 z - z^{2}) \\ = \frac{2}{9} {[\frac{3 z^{2}}{2} - \frac{z^{3}}{3}]}_{0}^{3} \\ = \frac{2}{9} \frac{27}{6} = 1 \end{aligned}

This is a little less than half way up the cone, which is reasonable since the cone is “bottom heavy”.

3.3.6.22. (✳).

Solution.

Each constant

z

cross-section of the cone is a circle. When

z = 0,

that circle has radius

a .

When

z = a

that circle has radius

0 .

Thus the radius decreases linearly from

a

0

z

increases from

0

a .

So the radius at height

z

a - z

and we can parametrize the cone by

r (θ, z) = (a - z) \cos θ \hat{ı ı} + (a - z) \sin θ \hat{ȷ ȷ} + z \hat{k} 0 \leq θ < 2 π, 0 \leq z \leq a

Since

\begin{aligned} \frac{\partial r}{\partial θ} & = (- (a - z) \sin θ, (a - z) \cos θ, 0) \\ \frac{\partial r}{\partial z} & = (- \cos θ, - \sin θ, 1) \\ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} & = ((a - z) \cos θ, (a - z) \sin θ, a - z) \end{aligned}

the element of surface area for this parametrization is

\begin{aligned} d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} | d θ d z = (a - z) | (\cos θ, \sin θ, 1) | d θ d z \\ = \sqrt{2} (a - z) d θ d z \end{aligned}

by (3.3.1). So the surface area of the cone is

\begin{aligned} \iint_{S} d S & = \int_{0}^{a} d z \int_{0}^{2 π} d θ \sqrt{2} (a - z) \\ = 2 \sqrt{2} π \int_{0}^{a} d z (a - z) = - \sqrt{2} π (a - z)^{2} |_{0}^{a} \\ = \sqrt{2} π a^{2} \end{aligned}

and the

z

-coordinate of the centre of mass is

\begin{aligned} \bar{z} & = \frac{\iint_{S} z d S}{\iint_{S} d S} = \frac{1}{\sqrt{2} π a^{2}} \int_{0}^{a} d z \int_{0}^{2 π} d θ \sqrt{2} (a - z) z = \frac{2}{a^{2}} \int_{0}^{a} d z (a z - z^{2}) \\ = \frac{2}{a^{2}} {[\frac{a z^{2}}{2} - \frac{z^{3}}{3}]}_{0}^{a} = \frac{2}{a^{2}} \frac{a^{3}}{6} = \frac{a}{3} \end{aligned}

This is a little less than half way up the cone, which is reasonable since the cone is “bottom heavy”.

3.3.6.23. (✳).

Solution.

Parametrize the surface by

x (θ, z) = \cos θ y (θ, z) = 2 \sin θ z (θ, z) = z

with

(θ, z)

running over

0 \leq θ \leq 2 π, 0 \leq z \leq 1 .

Then, by (3.3.1),

\begin{aligned} (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) & = (- \sin θ, 2 \cos θ, 0) \\ (\frac{\partial x}{\partial z}, \frac{\partial y}{\partial z}, \frac{\partial z}{\partial z}) & = (0, 0, 1) \\ \hat{n} d S & = \pm (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) \times (\frac{\partial x}{\partial z}, \frac{\partial y}{\partial z}, \frac{\partial z}{\partial z}) d θ d z \\ = (2 \cos θ, \sin θ, 0) d θ d z (+ for outward normal) \\ F (x (θ, z), y (θ, z), z (θ, z)) & = \cos θ \hat{ı ı} + 2 z \sin θ \cos θ \hat{ȷ ȷ} + 16 z \sin^{4} θ \hat{k} \end{aligned}

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{0}^{1} d z \int_{0}^{2 π} d θ [2 \cos^{2} θ + 2 z \sin^{2} θ \cos θ] \\ = \int_{0}^{1} d z \int_{0}^{2 π} d θ [1 + \cos (2 θ) + 2 z \sin^{2} θ \cos θ] \\ = \int_{0}^{1} d z [θ + \frac{1}{2} \sin (2 θ) + \frac{2}{3} z \sin^{3} θ]_{0}^{2 π} = 2 π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} θ d θ,

see Example 2.4.4.

3.3.6.24. (✳).

Solution.

By (3.3.2), with

f (x, y) = 4 - x^{2} - y^{2},

\begin{aligned} \hat{n} d S & = \pm (- f_{x}, - f_{y}, 1) d x d y \\ = \pm (2 x, 2 y, 1) d x d y \end{aligned}

To get the downward pointing normal, we want the minus sign. Set

T = {(x, y) | 0 \leq x \leq 1, 0 \leq y \leq 1 - x}

Then

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = - \iint_{T} (x + 1, y + 1, 2 \overset{z}{\overset{⏞}{(4 - x^{2} - y^{2})}}) \cdot (2 x, 2 y, 1) d x d y \\ = - \iint_{T} (8 + 2 x + 2 y) d x d y \\ = - \int_{0}^{1} d x \int_{0}^{1 - x} d y (8 + 2 x + 2 y) \\ = - \int_{0}^{1} d x (8 (1 - x) + 2 x (1 - x) + (1 - x)^{2}) \\ = - \int_{0}^{1} d x (9 - 8 x - x^{2}) \\ = - (9 - 4 - \frac{1}{3}) = - \frac{14}{3} \end{aligned}

3.3.6.25. (✳).

Solution.

First we have to parametrize

S .

It is natural to use spherical coordinates with

ρ = \sqrt{2} .

However if we use the standard spherical coordinates

\begin{matrix} x = \sqrt{2} \sin φ \cos θ y = \sqrt{2} \sin φ \sin θ z = \sqrt{2} \cos φ \end{matrix}

the condition

x \geq \sqrt{y^{2} + z^{2}},

i.e.

\frac{\sqrt{y^{2} + z^{2}}}{x} \leq 1,

becomes

\frac{\sqrt{\sin^{2} φ \sin^{2} θ + \cos^{2} φ}}{\sin φ \cos θ} \leq 1,

which is very complicated. So let’s back up and think a bit before we compute. From the sketch below

we see that

\frac{\sqrt{y^{2} + z^{2}}}{x}

is the tangent of the angle between the radius vector

(x, y, z)

and the

x

-axis. The angle between the radius vector

(x, y, z)

and the

z

-axis (not the

x

-axis) is exactly spherical coordinate

φ .

So let’s modify spherical coordinates to make the

x

-axis play the role of the

z

-axis. The easy way to do is to just rename

x = Z,

y = X,

z = Y .

Then the integral we are to compute becomes

\iint_{S} Z X^{2} d S,

and the condition

x \geq \sqrt{y^{2} + z^{2}}

becomes

Z \geq \sqrt{X^{2} + Y^{2}} .

Under the parametrization

\begin{matrix} X = \sqrt{2} \sin φ \cos θ Y = \sqrt{2} \sin φ \sin θ Z = \sqrt{2} \cos φ \end{matrix}

the condition

Z \geq \sqrt{X^{2} + Y^{2}}

\frac{\sqrt{X^{2} + Y^{2}}}{Z} = \frac{\sin φ}{\cos φ} \leq 1,

which is turn is

0 \leq φ \leq \frac{π}{4} .

d S = 2 \sin φ d θ d φ

(see Appendix A.6.3 and recall that

ρ = \sqrt{2}

) the specified integral is

\begin{aligned} \iint_{S} x y^{2} d S & = \iint_{S} Z X^{2} d S \\ = 2 \int_{0}^{π / 4} d φ \sin φ \int_{0}^{2 π} d θ (\sqrt{2} \cos φ) (\sqrt{2} \sin φ \cos θ)^{2} \\ = 4 \sqrt{2} {\int_{0}^{π / 4} d φ \cos φ \sin^{3} φ} {\int_{0}^{2 π} d θ \cos^{2} θ} \\ = 4 \sqrt{2} {\int_{0}^{π / 4} d φ \cos φ \sin^{3} φ} {\int_{0}^{2 π} d θ \frac{\cos (2 θ) + 1}{2}} \\ = 4 \sqrt{2} {[\frac{\sin^{4} φ}{4}]}_{0}^{π / 4} {[\frac{\sin (2 θ)}{4} + \frac{θ}{2}]}_{0}^{2 π} \\ = \frac{\sqrt{2} π}{4} \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} θ d θ,

see Example 2.4.4.

3.3.6.26. (✳).

Solution.

Here is a sketch of the part of

S

that is in the first octant.

For each fixed

y,

x^{2} + z^{2} = \sin^{2} y

is a circle of radius

\sin y .

(It’s the blue circle in the sketch above.) So we may parametrize the surface by

\begin{matrix} r (θ, y) = (\sin y \cos θ, y, \sin y \sin θ) 0 \leq θ < 2 π, 0 \leq y \leq π \end{matrix}

Then, by (3.3.1),

\begin{aligned} \frac{\partial r}{\partial θ} & = (- \sin y \sin θ, 0, \sin y \cos θ) \\ \frac{\partial r}{\partial y} & = (\cos y \cos θ, 1, \cos y \sin θ) \\ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial y} & = (- \sin y \cos θ, \sin y \cos y, - \sin y \sin θ) \\ d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial y} | d θ d y = \sin y \sqrt{1 + \cos^{2} y} d θ d y \end{aligned}

So the specified integral is

\begin{aligned} \iint_{S} \sqrt{1 + \cos^{2} y} d S & = \int_{0}^{π} d y \int_{0}^{2 π} d θ \sin y {1 + \cos^{2} y} \\ = 2 π \int_{0}^{π} d y \sin y {1 + \cos^{2} y} \\ = - 2 π \int_{1}^{- 1} d u {1 + u^{2}} \\ with u = \cos y, d u = - \sin y d y \\ = 4 π \int_{0}^{1} d u {1 + u^{2}} \\ = 4 π {[u + \frac{u^{3}}{3}]}_{0}^{1} \\ = \frac{16}{3} π \end{aligned}

3.3.6.27. (✳).

Solution.

The paraboloid is

\begin{aligned} S & = {(x, y, z) | z = 1 - x^{2} - y^{2}, z \geq 0} \\ = {(x, y, z) | z = 1 - x^{2} - y^{2}, x^{2} + y^{2} \leq 1} \end{aligned}

By (3.3.2), the paraboloid has

\begin{aligned} d S & = \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}} d x d y with z = f (x, y) = 1 - x^{2} - y^{2} \\ = \sqrt{1 + 4 x^{2} + 4 y^{2}} d x d y \end{aligned}

By symmetry, the centre of mass will lie on the

z

-axis. By definition, the

z

-coordinate of the centre of mass is the weighted average of

z

over

S,

which is

\bar{z} = \frac{\iint_{S} z ρ (x, y, z) d S}{\iint_{S} ρ (x, y, z) d S}

S,

\begin{matrix} ρ (x, y, z) = \frac{z}{\sqrt{5 - 4 z}} = \frac{1 - x^{2} - y^{2}}{\sqrt{1 + 4 x^{2} + 4 y^{2}}} \end{matrix}

so that

\begin{matrix} ρ (x, y, z) d S = (1 - x^{2} - y^{2}) d x d y \end{matrix}

So, using polar coordinates, the denominator of

\bar{z}

\begin{aligned} \iint_{S} ρ (x, y, z) d S & = \iint_{x^{2} + y^{2} \leq 1} (1 - x^{2} - y^{2}) d x d y = \int_{0}^{1} d r r \int_{0}^{2 π} d θ (1 - r^{2}) \\ = 2 π \int_{0}^{1} r (1 - r^{2}) d r \\ = 2 π [\frac{r^{2}}{2} - \frac{r^{4}}{4}]_{0}^{1} \\ = \frac{π}{2} \end{aligned}

and the numerator of

\bar{z}

\begin{aligned} \iint_{S} z ρ (x, y, z) d S & = \iint_{x^{2} + y^{2} \leq 1} {(1 - x^{2} - y^{2})}^{2} d x d y \\ = \int_{0}^{1} d r r \int_{0}^{2 π} d θ {(1 - r^{2})}^{2} \\ = 2 π \int_{0}^{1} r {(1 - r^{2})}^{2} d r \\ = 2 π [\frac{r^{2}}{2} - 2 \frac{r^{4}}{4} + \frac{r^{6}}{6}]_{0}^{1} \\ = \frac{π}{3} \end{aligned}

and

\bar{z} = \frac{π / 3}{π / 2} = \frac{2}{3}

3.3.6.28. (✳).

Solution.

The equation of the plane is

z = f (x, y) = 2 - x - y .

So by (3.3.2),

\hat{n} d S = [- f_{x} (x, y) \hat{ı ı} - f_{y} (x, y) \hat{ȷ ȷ} + \hat{k}] d x d y = [\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}] d x d y

A point

(x, y, z)

on the plane lies in the first octant if and only if

x \geq 0 and y \geq 0 and z = 2 - x - y \geq 0

So the domain of integration is the triangle

T = {(x, y) | x \geq 0, y \geq 0, x + y \leq 2}

and

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \iint_{T} [x \hat{ı ı} + y \hat{ȷ ȷ} + (\overset{z}{\overset{⏞}{2 - x - y}}) \hat{k}] \cdot [\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}] d x d y \\ = 2 \iint_{T} d x d y \\ = 2 \frac{1}{2} (2) (2) = 4 \end{aligned}

3.3.6.29. (✳).

Solution.

Since

\begin{aligned} \frac{\partial r}{\partial u} & = (v^{2}, 2 u v, v) \\ \frac{\partial r}{\partial v} & = (2 u v, u^{2}, u) \\ \frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v} & = (u^{2} v, u v^{2}, - 3 u^{2} v^{2}) \end{aligned}

(3.3.1) gives

\hat{n} d S = \pm (u^{2} v, u v^{2}, - 3 u^{2} v^{2}) d u d v

We are told that

\hat{n}

should have a positive

z

-component, so

\hat{n} d S = - (u^{2} v, u v^{2}, - 3 u^{2} v^{2}) d u d v = (- u^{2} v, - u v^{2}, 3 u^{2} v^{2}) d u d v

and

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \iint_{S} \overset{F}{\overset{⏞}{(u v^{2}, u^{2} v, u v)}} \cdot (- u^{2} v, - u v^{2}, 3 u^{2} v^{2}) d u d v \\ = \int_{0}^{1} d u \int_{0}^{3} d v u^{3} v^{3} = [\int_{0}^{1} d u u^{3}] [\int_{0}^{3} d v v^{3}] \\ = \frac{1}{4} \frac{3^{4}}{4} = \frac{81}{16} \end{aligned}

3.3.6.30. (✳).

Solution.

(a) We start by just sketching the curve

z = e^{y},

considering the

y z

-plane as the plane

x = 0

R^{3} .

This curve is the red curve in the figure below. Concentrate on any one point on that curve. It is the blue dot at

(0, Y, e^{Y})

in the figure. When our curve is rotated about the

y

-axis, the blue dot sweeps out a circle. The circle that the blue dot sweeps out

lies in the vertical plane $y = Y$ and
is centred on the $y$ -axis and
has radius $e^{Y} .$

We can parametrize the circle swept out in the usual way. Here is an end view of the circle (looking down the

y

-axis), with the parameter, named

θ,

indicated.

The coordinates of the red dot are

(e^{Y} \sin θ, Y, e^{Y} \cos θ) .

This also gives a parametrization of the surface of revolution

\begin{aligned} x (Y, θ) & = e^{Y} \sin θ \\ y (Y, θ) & = Y \\ z (Y, θ) & = e^{Y} \cos θ \\ 0 \leq Y \leq 1, 0 \leq θ < 2 π \end{aligned}

Finally here is a sketch of the part of the surface in the first octant,

x, y, z \geq 0 .

(b) We are using the parametrization

r (Y, θ) = e^{Y} \sin θ \hat{ı ı} + Y \hat{ȷ ȷ} + e^{Y} \cos θ \hat{k} 0 \leq Y \leq 1, 0 \leq θ \leq 2 π

so that

\frac{\partial r}{\partial Y} \times \frac{\partial r}{\partial θ} = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ e^{Y} \sin θ & 1 & e^{Y} \cos θ \\ e^{Y} \cos θ & 0 & - e^{Y} \sin θ \end{matrix}] = (- e^{Y} \sin θ, e^{2 Y}, - e^{Y} \cos θ),

and, by (3.3.1),

\begin{matrix} d S = | \frac{\partial r}{\partial Y} \times \frac{\partial r}{\partial θ} | d Y d θ = \sqrt{e^{2 Y} + e^{4 Y}} d Y d θ = e^{Y} \sqrt{1 + e^{2 Y}} d Y d θ \end{matrix}

So the integral is

\begin{aligned} \iint_{S} e^{y} d S & = \int_{0}^{1} d Y \int_{0}^{2 π} d θ e^{2 Y} \sqrt{1 + e^{2 Y}} = 2 π \int_{0}^{1} d Y e^{2 Y} \sqrt{1 + e^{2 Y}} \\ = \frac{2 π}{3} [1 + e^{2 Y}]^{3 / 2} |_{0}^{1} \\ = \frac{2 π}{3} [(1 + e^{2})^{3 / 2} - 2^{3 / 2}] \end{aligned}

r (Y, θ) = e^{Y} \sin θ \hat{ı ı} + Y \hat{ȷ ȷ} + e^{Y} \cos θ \hat{k} 0 \leq Y \leq 1, 0 \leq θ \leq 2 π

so that

\frac{\partial r}{\partial Y} \times \frac{\partial r}{\partial θ} = (- e^{Y} \sin θ, e^{2 Y}, - e^{Y} \cos θ),

and, by (3.3.1),

\begin{matrix} \hat{n} d S = \pm \frac{\partial r}{\partial Y} \times \frac{\partial r}{\partial θ} d Y d θ = \pm (- e^{Y} \sin θ, e^{2 Y}, - e^{Y} \cos θ) d Y d θ \end{matrix}

We choose the “

+

” sign so that

\hat{n}

points towards the

y

-axis. As an example, when

0 \leq θ \leq \frac{π}{2},

then

z = e^{Y} \cos θ > 0

while the

z

-coordinate of

\hat{n}

- e^{y} \cos θ < 0 .

So the integral is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S \\ = \int_{0}^{1} d Y \int_{0}^{2 π} d θ (\overset{x}{\overset{⏞}{e^{Y} \sin θ}}, 0, \overset{z}{\overset{⏞}{e^{Y} \cos θ}}) \cdot (- e^{Y} \sin θ, e^{2 Y}, - e^{Y} \cos θ) \\ = - \int_{0}^{1} d Y \int_{0}^{2 π} d θ e^{2 Y} = - 2 π \int_{0}^{1} d Y e^{2 Y} \\ = - π (e^{2} - 1) = π (1 - e^{2}) \end{aligned}

3.3.6.31. (✳).

Solution.

Write

V = {(x, y, z) | 1 \leq x^{2} + y^{2} + z^{2} \leq 4}

The boundary of

V

consists of two parts — the sphere,

S_{2},

of radius

2,

centred on the origin, with (outward) normal

\hat{n} = \frac{r}{| r |} = \frac{r}{2},

and the sphere

S_{1}

of radius

1,

centred on the origin, with (inward) normal

\hat{n} = - r,

So,

\begin{aligned} \iint_{\partial V} F \cdot \hat{n} d S & = \iint_{S_{2}} \frac{r}{| r |} \cdot \frac{r}{2} d S - \iint_{S_{1}} \frac{r}{| r |} \cdot r d S \\ = \iint_{S_{2}} d S - \iint_{S_{1}} d S \\ = 4 π (2)^{2} - 4 π (1)^{2} \\ = 12 π \end{aligned}

3.3.6.32. (✳).

Solution.

The part of the cone that has some fixed value,

Z,

z

with

0 \leq Z \leq 1

is the part of the circle

{(x, y, z) | x^{2} + y^{2} = 4 Z^{2}, z = Z}

of radius

2 Z

that has

0 \leq x \leq y .

Here is a sketch of the top view of that part of that circle.

So we can parametrize

S

r (θ, Z) = 2 Z \sin θ \hat{ı ı} + 2 Z \cos θ \hat{ȷ ȷ} + Z \hat{k} 0 \leq θ \leq \frac{π}{4}, 0 \leq Z \leq 1

\begin{aligned} \frac{\partial r}{\partial θ} & = 2 Z \cos θ \hat{ı ı} - 2 Z \sin θ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial Z} & = 2 \sin θ \hat{ı ı} + 2 \cos θ \hat{ȷ ȷ} + \hat{k} \end{aligned}

so that

\frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial Z} = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 2 Z \cos θ & - 2 Z \sin θ & 0 \\ 2 \sin θ & 2 \cos θ & 1 \end{matrix}] = (- 2 Z \sin θ, - 2 Z \cos θ, 4 Z),

and, by (3.3.1),

\begin{matrix} d S = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial Z} | d θ d Z = \sqrt{20} Z d θ d Z \end{matrix}

and

\begin{aligned} \iint_{S} z^{2} d S & = \sqrt{20} \int_{0}^{1} d Z \int_{0}^{π / 4} d θ Z^{3} = \sqrt{20} \frac{π}{4} \frac{1}{4} = \frac{\sqrt{5} π}{8} \end{aligned}

3.3.6.33. (✳).

Solution.

We’ll start by parametrizing

S .

Note that as

x^{2} + y^{2}

runs from

0

4,

z

runs from

5

1,

and that, for each fixed

1 \leq Z \leq 5,

the cross-section of

S

with

z = Z

is the circle

x^{2} + y^{2} = 5 - Z,

z = Z .

So we may parametrize

S

\begin{aligned} r (θ, Z) = \sqrt{5 - Z} \cos θ \hat{ı ı} + \sqrt{5 - Z} \sin θ \hat{ȷ ȷ} + Z \hat{k} \\ 0 \leq θ \leq 2 π, 1 \leq Z \leq 5 \end{aligned}

Since

\begin{aligned} \frac{\partial r}{\partial θ} & = - \sqrt{5 - Z} \sin θ \hat{ı ı} + \sqrt{5 - Z} \cos θ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial Z} & = - \frac{1}{2 \sqrt{5 - Z}} \cos θ \hat{ı ı} - \frac{1}{2 \sqrt{5 - Z}} \sin θ \hat{ȷ ȷ} + \hat{k} \end{aligned}

so that

\begin{aligned} \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial Z} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - \sqrt{5 - Z} \sin θ & \sqrt{5 - Z} \cos θ & 0 \\ - \frac{1}{2 \sqrt{5 - Z}} \cos θ & - \frac{1}{2 \sqrt{5 - Z}} \sin θ & 1 \end{array}] \\ = (\sqrt{5 - Z} \cos θ, \sqrt{5 - Z} \sin θ, 1 / 2), \end{aligned}

(3.3.1) gives

\begin{matrix} \hat{n} d S = \pm \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial Z} d θ d Z = \pm (\sqrt{5 - Z} \cos θ, \sqrt{5 - Z} \sin θ, 1 / 2) d θ d Z \end{matrix}

Choosing the minus sign to give the downward pointing normal

\begin{aligned} \iint_{S} F \cdot \hat{n} d S \\ = - \int_{1}^{5} d Z \int_{0}^{2 π} d θ (- \frac{1}{2} \overset{x^{3}}{\overset{⏞}{[5 - Z]^{3 / 2} \cos^{3} θ}} - \overset{x y^{2}}{\overset{⏞}{[5 - Z]^{3 / 2} \cos θ \sin^{2} θ}}, \\ - \frac{1}{2} \overset{y^{3}}{\overset{⏞}{[5 - Z]^{3 / 2} \sin^{3} θ}}, \overset{z^{2}}{\overset{⏞}{Z^{2}}}) \\ \cdot (\sqrt{5 - Z} \cos θ, \sqrt{5 - Z} \sin θ, 1 / 2) \\ = - \int_{1}^{5} d Z \int_{0}^{2 π} d θ (- \frac{1}{2} [5 - Z]^{2} \cos^{4} θ - [5 - Z]^{2} \cos^{2} θ \sin^{2} θ \\ - \frac{1}{2} [5 - Z]^{2} \sin^{4} θ + \frac{1}{2} Z^{2}) \end{aligned}

Since

\frac{1}{2} \cos^{4} θ + \cos^{2} θ \sin^{2} θ + \frac{1}{2} \sin^{4} θ = \frac{1}{2} (\cos^{2} θ + \sin^{2} θ)^{2} = \frac{1}{2}

the flux

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{1}^{5} d Z \int_{0}^{2 π} d θ (\frac{1}{2} [5 - Z]^{2} - \frac{1}{2} Z^{2}) \\ = π \int_{1}^{5} d Z ([5 - Z]^{2} - Z^{2}) \\ = π {[- \frac{1}{3} [5 - Z]^{3} - \frac{Z^{3}}{3}]}_{1}^{5} = π [\frac{4^{3}}{3} - \frac{5^{3}}{3} + \frac{1}{3}] \\ = - 20 π \end{aligned}

3.3.6.34. (✳).

Solution.

The surface is

z = f (x, y)

with

f (x, y) = \sqrt{2 x y} .

Since

f_{x} = \sqrt{\frac{y}{2 x}}

and

f_{y} = \sqrt{\frac{x}{2 y}},

(3.3.2) gives

\begin{aligned} d S & = \sqrt{1 + f_{x}^{2} + f_{y}^{2}} d x d y = \sqrt{1 + \frac{y}{2 x} + \frac{x}{2 y}} d x d y \\ = \sqrt{\frac{2 x y + y^{2} + x^{2}}{2 x y}} d x d y \\ = \frac{x + y}{\sqrt{2 x y}} d x d y \end{aligned}

On the shell,

z^{2} = 2 x y \leq 4 .

So the

x

and

y

components of points

(x, y, z)

on the shell run over the region

x \geq 1,

y \geq 1,

x y \leq 2,

which is sketched below

So the mass is

\begin{aligned} \iint_{S} ρ (x, y, z) d S & = \int_{1}^{2} d x \int_{1}^{2 / x} d y 3 f (x, y) \frac{x + y}{\sqrt{2 x y}} \\ = \int_{1}^{2} d x \int_{1}^{2 / x} d y 3 (x + y) \\ = 3 \int_{1}^{2} d x {[x y + \frac{1}{2} y^{2}]}_{1}^{2 / x} = 3 \int_{1}^{2} d x [2 + \frac{2}{x^{2}} - x - \frac{1}{2}] \\ = 3 {\frac{3}{2} - \frac{2}{x} |_{1}^{2} - \frac{x^{2}}{2} |_{1}^{2}} = 3 {\frac{3}{2} - 1 + 2 - 2 + \frac{1}{2}} \\ = 3 \end{aligned}

3.3.6.35. (✳).

Solution.

Since

x = g (y, z)

with

g (x, y) = y^{2} + z^{2},

(3.3.2) gives

\hat{n} d S = \pm (1, - g_{y}, - g_{z}) d y d z = \pm (1, - 2 y, - 2 z) d y d z

We choose the

+

sign so that

\hat{n} \cdot \hat{ı ı} > 0 .

Furthermore

\begin{aligned} S & = {(x, y, z) | x = y^{2} + z^{2}, x \leq 2 y} \\ = {(x, y, z) | x = y^{2} + z^{2}, y^{2} + z^{2} \leq 2 y} \\ = {(x, y, z) | x = y^{2} + z^{2}, (y - 1)^{2} + z^{2} \leq 1} \\ = {(x, y, z) | x = y^{2} + z^{2}, (y, z) in D} \end{aligned}

where

D = {(x, y) | (y - 1)^{2} + z^{2} \leq 1}

is a disk with radius

1 .

Hence

\iint_{S} F \cdot \hat{n} d S = \iint_{D} (2, z, y) \cdot (1, - 2 y, - 2 z) d y d z = \iint_{D} (2 - 4 y z) d y d z

Since

- 4 y z

is odd under

z \to - z

the integral of

- 4 y z

is zero and

\iint_{S} F \cdot \hat{n} d S = 2 Area (D) = 2 π

3.3.6.36. (✳).

Solution.

For the specified

F

and the surface

x = f (x, y) = 1 - \frac{1}{4} x^{2} - y^{2},

by (3.3.2),

\begin{aligned} \hat{n} d S & = (- f_{x} \hat{ı ı} - f_{y} \hat{ȷ ȷ} + \hat{k}) d x d y = (\frac{x}{2} \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}) d x d y \\ \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3 y^{2} + z & x - x^{2} & 1 \end{array}] \\ = \hat{ȷ ȷ} + (1 - 2 x - 6 y) \hat{k} \\ \nabla \times F \cdot \hat{n} d S & = (2 y + 1 - 2 x - 6 y) d x d y = (1 - 2 x - 4 y) d x d y \end{aligned}

The domain of integration is

1 - \frac{1}{4} x^{2} - y^{2} \geq 0

\frac{1}{4} x^{2} + y^{2} \leq 1 .

This is an ellipse. Call it

D .

\iint_{S} \nabla \times F \cdot \hat{n} d S = \iint_{D} (1 - 2 x - 4 y) d x d y

The integrals over

D

x,

which is odd under

x \to - x,

and of

y,

which is odd under

y \to - y,

are both zero. As the ellipse

D

has area

A = π \times 2 \times 1 = 2 π

\iint_{S} \nabla \times F \cdot \hat{n} d S = \iint_{D} (1 - 2 x - 4 y) d x d y = A = 2 π

3.3.6.37.

Solution.

Due to the symmetry of the surface and the vector field under reflection in the

x y

-plane, i.e. under

z \to - z,

it is sufficient to compute the integral over the upper half of the surface, where

z \geq 0,

and then multiply the result by 2. The upper half of the surface consists of two pieces,

S_{1}

and

S_{2},

where

S_{1}

is the part on the sphere and

S_{2}

is the part on the hyperboloid.

S_{1}

and

S_{2}

intersect on a circle. The circle is obtained by imposing the two equations

x^{2} + y^{2} + z^{2} = 16

and

x^{2} + y^{2} - z^{2} = 8

simultaneously. Thus we have

x^{2} + y^{2} = 12

and

z = 2,

or in cylindrical coordinates

r = \sqrt{12},

z = 1,

on the circle. Here is a sketch of a cross-section of the apple core.

Let

ϕ_{1}

be the angle between

z

-axis and the cone formed by connecting the circle to the origin. We have

\tan ϕ_{1} = \sqrt{12} / 2 = \sqrt{3} .

Thus

ϕ_{1} = π / 3 .

We’ll use spherical coordinates to compute the flux integral

\iint_{S_{1}} F \cdot \hat{n} d S .

As the spherical coordinate

ρ = 4

on all of

S_{1},

we can paramerize

S_{1}

\begin{aligned} r (θ, φ) = 4 \cos θ \sin φ \hat{ı ı} + 4 \sin θ \sin φ \hat{ȷ ȷ} + 4 \cos φ \hat{k} \\ 0 \leq θ \leq 2 π, 0 \leq ϕ \leq π / 3 \end{aligned}

\begin{aligned} \frac{\partial r}{\partial θ} & = - 4 \sin θ \sin φ \hat{ı ı} + 4 \cos θ \sin φ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial φ} & = 4 \cos θ \cos φ \hat{ı ı} + 4 \sin θ \cos φ \hat{ȷ ȷ} - 4 \sin φ \hat{k} \\ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - 4 \sin θ \sin φ & 4 \cos θ \sin φ & 0 \\ 4 \cos θ \cos φ & 4 \sin θ \cos φ & - 4 \sin φ \end{array}] \\ = - 16 (\cos θ \sin^{2} φ, \sin θ \sin^{2} φ, \sin φ \cos φ) \\ = - 4 (\sin φ) r (θ, φ) \end{aligned}

and, by (3.3.1),

\begin{aligned} \hat{n} d S & = \pm \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} d θ d φ = \mp 4 (\sin φ) r (θ, φ) d θ d φ \end{aligned}

To get the outward pointing normal, i.e. the normal point in the same direction as

r (θ, φ),

we take the plus sign. As

F = r (θ, φ),

F \cdot \hat{n} d S = 4 \overset{4^{2}}{\overset{⏞}{| r (θ, φ) |^{2}}} \sin φ d θ d φ = 64 \sin φ d θ d φ

and

\begin{matrix} \iint_{S_{1}} F \cdot \hat{n} d S = 64 \int_{0}^{2 π} d θ \int_{0}^{π / 3} d φ \sin φ = 64 \cdot 2 π [- \cos φ]_{0}^{π / 3} = 64 π \end{matrix}

The surface

S_{2}

can be parametrized using the cylindrical coordinates

θ

and

z .

Indeed, we have

r = \sqrt{x^{2} + y^{2}} = (8 + z^{2})^{1 / 2}

for the hyperboloid and we always have

x = r \cos θ

and

y = r \sin θ .

Thus the hyperboloid has the following parametrization:

\begin{matrix} R (θ, z) = (8 + z^{2})^{1 / 2} \cos θ \hat{ı ı} + (8 + z^{2})^{1 / 2} \sin θ \hat{ȷ ȷ} + z \hat{k} \end{matrix}

The range for the parameters of

S_{2}

0 \leq θ \leq 2 π

and

0 \leq z \leq 2 .

We have

\begin{aligned} \frac{\partial R}{\partial θ} & = - (8 + z^{2})^{1 / 2} \sin θ \hat{ı ı} + (8 + z^{2})^{1 / 2} \cos θ \hat{ȷ ȷ} + 0 \hat{k} \\ \frac{\partial R}{\partial z} & = z (8 + z^{2})^{- 1 / 2} \cos θ \hat{ı ı} + z (8 + z^{2})^{- 1 / 2} \sin θ \hat{ȷ ȷ} + \hat{k} \end{aligned}

and

\begin{aligned} \frac{\partial R}{\partial θ} \times \frac{\partial R}{\partial z} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - (8 + z^{2})^{1 / 2} \sin θ & (8 + z^{2})^{1 / 2} \cos θ & 0 \\ z (8 + z^{2})^{- 1 / 2} \cos θ & z (8 + z^{2})^{- 1 / 2} \sin θ & 1 \end{array}] \\ = (8 + z^{2})^{1 / 2} \cos θ \hat{ı ı} + (8 + z^{2})^{1 / 2} \sin θ \hat{ȷ ȷ} - z \hat{k} \\ = x \hat{ı ı} + y \hat{ȷ ȷ} - z \hat{k} \end{aligned}

Note that

\frac{\partial R}{\partial θ} \times \frac{\partial R}{\partial z}

is pointing downward (since

z > 0

) and hence outward. Since

F \cdot (\frac{\partial R}{\partial θ} \times \frac{\partial R}{\partial z}) = (x, y, z) \cdot (x, y, - z) = x^{2} + y^{2} - z^{2} = 8

S_{2},

we have

\iint_{S_{2}} F \cdot \hat{n} d S = \iint_{S_{2}} F \cdot (R_{θ} \times R_{z}) d θ d z = \int_{0}^{2} d z \int_{0}^{2 π} d θ 8 = 32 π

Finally, the flux integral over the whole apple core surface is

2 (\iint_{S_{1}} F \cdot \hat{n} d S + \iint_{S_{2}} F \cdot \hat{n} d S) = 2 (64 π + 32 π) = 192 π

3.3.6.38. (✳).

Solution.

(a) The specified surface is of the form

G (x, y, z) = x^{2} + z^{2} - \cos^{2} y = 0

So one normal vector at the point

(\frac{1}{2}, \frac{π}{4}, \frac{1}{2})

\nabla \nabla G (\frac{1}{2}, \frac{π}{4}, \frac{1}{2}) = (2 x, 2 \sin y \cos y, 2 z) |_{(\frac{1}{2}, \frac{π}{4}, \frac{1}{2})} = (1, 1, 1)

and an equation for the tangent plane at

(\frac{1}{2}, \frac{π}{4}, \frac{1}{2})

\begin{matrix} (1, 1, 1) \cdot (x - 1 / 2, y - π / 4, z - 1 / 2) = 0 or x + y + z = 1 + π / 4 \end{matrix}

(b) For each fixed

y,

x^{2} + z^{2} = \cos^{2} y

is a circle of radius

| \cos y | .

So we may parametrize the surface by

\begin{matrix} r (θ, y) = (\cos y \cos θ, y, \cos y \sin θ) 0 \leq θ < 2 π, 0 \leq y \leq \frac{π}{2} \end{matrix}

Then

\begin{aligned} \frac{\partial r}{\partial θ} & = (- \cos y \sin θ, 0, \cos y \cos θ) \\ \frac{\partial r}{\partial y} & = (- \sin y \cos θ, 1, - \sin y \sin θ) \\ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial y} & = (- \cos y \cos θ, - \sin y \cos y, - \cos y \sin θ) \\ d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial y} | d θ d y = \cos y \sqrt{1 + \sin^{2} y} d θ d y \end{aligned}

So the specified integral is

\begin{aligned} \iint_{S} \sin y d S & = \int_{0}^{\frac{π}{2}} d y \int_{0}^{2 π} d θ \cos y \sqrt{1 + \sin^{2} y} \sin y \\ = 2 π \int_{0}^{\frac{π}{2}} d y \sqrt{1 + \sin^{2} y} \sin y \cos y \\ = π \int_{1}^{2} d u \sqrt{u} with u = 1 + \sin^{2} y, d u = 2 \sin y \cos y d y \\ = π {[\frac{u^{3 / 2}}{3 / 2}]}_{1}^{2} \\ = \frac{2 π}{3} [2 \sqrt{2} - 1] \end{aligned}

3.3.6.39. (✳).

Solution.

(a) By definition

F

is a conservative vector field with potential

f .

Suppose that the curve

C

starts at

P_{1},

S,

and ends at

P_{2},

S .

Then

f (P_{1}) = f (P_{2}) = c

and, by Theorem 2.4.2,

\int_{C} F \cdot d r = f (P_{2}) - f (P_{1}) = c - c = 0

(b) Since

F = \nabla \nabla f,

F

is normal to the level surfaces of

f

by Lemma 2.3.6. So, at any point of

S,

F

is a scalar multiple of

\hat{n}

and

F \times G

is perpendicular to

\hat{n} .

Thus

(F \times G) \cdot \hat{n} = 0

and

\iint_{S} (F \times G) \cdot \hat{n} d S = 0.

3.3.6.40. (✳).

Solution.

(a) (i) Here is a sketch of the part of the plane in question.

We can use

x

and

y

as parameters. As we can rewrite the equation of the plane as

z = \frac{1}{3} (16 - 2 x - 4 y),

we have the parametrization

r (x, y) = x \hat{ı ı} + y \hat{ȷ ȷ} + \frac{1}{3} (16 - 2 x - 4 y) \hat{k}

In terms of

x

and

y,

the condition

z = \frac{1}{3} (16 - 2 x - 4 y) \geq 0

16 - 2 x - 4 y \geq 0

x + 2 y \leq 8 .

So the domain is

{(x, y) | x \geq 0, y \geq 0, x + 2 y \leq 8}

Renaming

x

u

and

y

v,

the parametrization is also

r (u, v) = (u, v, \frac{1}{3} (16 - 2 u - 4 v)) \hat{k} u \geq 0, v \geq 0, u + 2 v \leq 8

(a) (ii) Here is a sketch of the part of the cap in the first octant.

The full sphere can be parametrized, using spherical coordinates with

ρ = 4,

\begin{aligned} r (θ, φ) = 4 \cos θ \sin φ \hat{ı ı} + 4 \sin θ \sin φ \hat{ȷ ȷ} + 4 \cos φ \hat{k} \\ 0 \leq θ \leq 2 π, 0 \leq φ \leq π \end{aligned}

In these coordinates, the condition

4 / \sqrt{2} \leq z \leq 4

\begin{aligned} \frac{4}{\sqrt{2}} \leq 4 \cos φ \leq 4 & ⟺ \frac{1}{\sqrt{2}} \leq \cos φ \leq 1 \\ ⟺ 0 \leq φ \leq \frac{π}{4} \end{aligned}

So our parametrization is

\begin{aligned} r (θ, φ) = 4 \cos θ \sin φ \hat{ı ı} + 4 \sin θ \sin φ \hat{ȷ ȷ} + 4 \cos φ \hat{k} \\ 0 \leq θ \leq 2 π, 0 \leq φ \leq \frac{π}{4} \end{aligned}

Renaming

θ

u

and

φ

v,

the parametrization is also

r (u, v) = (4 \cos u \sin v, 4 \sin u \sin v, 4 \cos v) 0 \leq u \leq 2 π, 0 \leq v \leq \frac{π}{4}

(a) (iii) Here is a sketch of the hyperboloid.

If we use

x

and

y

as parameters, then, since

z = \sqrt{1 + x^{2} + y^{2}},

we have the parametrization

r (x, y) = x \hat{ı ı} + y \hat{ȷ ȷ} + \sqrt{1 + x^{2} + y^{2}} \hat{k}

In terms of

x

and

y,

the condition

1 \leq z \leq 10

1 \leq \sqrt{1 + x^{2} + y^{2}} \leq 10 or 0 \leq x^{2} + y^{2} \leq 99

So the domain is

{(x, y) | x^{2} + y^{2} \leq 99}

Renaming

x

u

and

y

v,

the parametrization is also

r (u, v) = (u, v, \sqrt{1 + u^{2} + v^{2}}) u^{2} + v^{2} \leq 99

Alternatively, if we replace

x

and

y

with the polar coordinates

r

and

θ,

we get the parametrization

r (r, θ) = r \cos θ \hat{ı ı} + r \sin θ \hat{ȷ ȷ} + \sqrt{1 + r^{2}} \hat{k} 0 \leq θ \leq 2 π, 0 \leq r \leq \sqrt{99}

Renaming

r

u

and

θ

v,

the parametrization is also

r (u, v) = (u \cos v, u \sin v, \sqrt{1 + u^{2}}) 0 \leq v \leq 2 π, 0 \leq u \leq \sqrt{99}

(b) Let’s use the parametrization

\begin{aligned} r (θ, φ) = 4 \cos θ \sin φ \hat{ı ı} + 4 \sin θ \sin φ \hat{ȷ ȷ} + 4 \cos φ \hat{k} \\ 0 \leq θ \leq 2 π, 0 \leq φ \leq \frac{π}{4} \end{aligned}

from part (a) (ii), so that

\begin{aligned} \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - 4 \sin θ \sin φ & 4 \cos θ \sin φ & 0 \\ 4 \cos θ \cos φ & 4 \sin θ \cos φ & - 4 \sin φ \end{array}] \\ = - 16 (\cos θ \sin^{2} φ, \sin θ \sin^{2} φ, \sin φ \cos φ) \end{aligned}

and, by (3.3.1),

\begin{aligned} d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} | d θ d φ \\ = 16 \sin φ \sqrt{\cos^{2} θ \sin^{2} φ + \sin^{2} θ \sin^{2} φ + \cos^{2} φ} d θ d φ \\ = 16 \sin φ d θ d φ \end{aligned}

So the area is

\begin{aligned} Area = \iint_{S} d S & = \int_{0}^{π / 4} d φ \int_{0}^{2 π} d θ 16 \sin φ = 32 π \int_{0}^{π / 4} d φ \sin φ \\ = 32 π [- \cos φ]_{0}^{π / 4} \\ = 32 π [1 - \frac{1}{\sqrt{2}}] \end{aligned}

3.3.6.41. (✳).

Solution.

Solution 1 — using tweaked spherical coordinates.

First we have to parametrize

S .

It is natural to use spherical coordinates with

ρ = \sqrt{2} .

However if we use the standard spherical coordinates

\begin{matrix} x = \sqrt{2} \sin φ \cos θ y = \sqrt{2} \sin φ \sin θ z = \sqrt{2} \cos φ \end{matrix}

the condition

y \geq 1

becomes

\sin φ \sin θ \geq \frac{1}{\sqrt{2}},

which is very complicated. So let’s back up and think a bit before we compute. The condition

z \geq 1,

as opposed to

y \geq 1,

is easy to implement in spherical coordinates. It is

\cos φ \geq \frac{1}{\sqrt{2}}

0 \leq φ \leq \frac{π}{4} .

So let’s modify spherical coordinates to make the

y

-axis play the role of the

z

-axis, by just exchanging

y

and

z

in the parametrization.

\begin{matrix} x = \sqrt{2} \sin φ \cos θ y = \sqrt{2} \cos φ z = \sqrt{2} \sin φ \sin θ \end{matrix}

The condition

y \geq 1

is then

\sqrt{2} \cos φ \geq 1,

which is turn is

0 \leq φ \leq \frac{π}{4} .

Since we have just exchanged

y

and

z

we could probably just guess

\hat{n} d S

and

d S

from standard spherical coordinates. (See Appendix A.6.3 and recall that

ρ = \sqrt{2} .

) But to be on the safe side, let’s derive them. We are using the parametrization

\begin{aligned} r (θ, φ) = \sqrt{2} \sin φ \cos θ \hat{ı ı} + \sqrt{2} \cos φ \hat{ȷ ȷ} + \sqrt{2} \sin φ \sin θ \hat{k} \\ 0 \leq θ \leq 2 π, 0 \leq φ \leq \frac{π}{4} \end{aligned}

Since

\begin{aligned} \frac{\partial r}{\partial θ} & = - \sqrt{2} \sin φ \sin θ \hat{ı ı} + \sqrt{2} \sin φ \cos θ \hat{k} \\ \frac{\partial r}{\partial φ} & = \sqrt{2} \cos φ \cos θ \hat{ı ı} - \sqrt{2} \sin φ \hat{ȷ ȷ} + \sqrt{2} \cos φ \sin θ \hat{k} \end{aligned}

so that

\begin{aligned} \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - \sqrt{2} \sin φ \sin θ & 0 & \sqrt{2} \sin φ \cos θ \\ \sqrt{2} \cos φ \cos θ & - \sqrt{2} \sin φ & \sqrt{2} \cos φ \sin θ \end{array}] \\ = 2 \sin^{2} φ \cos θ \hat{ı ı} + 2 \sin φ \cos φ \hat{ȷ ȷ} + 2 \sin^{2} φ \sin θ \hat{k} \end{aligned}

(3.3.1) gives

\begin{aligned} \hat{n} d S & = \pm \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} d θ d φ & = \pm 2 \sin φ (\sin φ \cos θ, \cos φ, \sin φ \sin θ) d θ d φ \\ d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} | d θ d φ & = 2 \sin φ d θ d φ \end{aligned}

Choose the plus sign to give the outward pointing normal.

(a) The specified integral is

\begin{aligned} \iint_{S} y^{3} d S & = 2 \int_{0}^{π / 4} d φ \int_{0}^{2 π} d θ \sin φ \overset{y^{3}}{\overset{⏞}{(\sqrt{2} \cos φ)^{3}}} \\ = 8 \sqrt{2} π \int_{0}^{π / 4} d φ \sin φ \cos^{3} φ \\ = 8 \sqrt{2} π {[- \frac{\cos^{4} φ}{4}]}_{0}^{π / 4} \\ = 2 \sqrt{2} π [1 - \frac{1}{4}] = \frac{3}{\sqrt{2}} π \end{aligned}

(b) The specified integral is

\begin{aligned} \iint_{S} (x y \hat{ı ı} + x z \hat{ȷ ȷ} + z y \hat{k}) \cdot \hat{n} d S \\ = 2 \int_{0}^{π / 4} d φ \int_{0}^{2 π} d θ \sin φ (2 \sin φ \cos φ \cos θ, 2 \sin^{2} φ \sin θ \cos θ, \\ 2 \sin φ \cos φ \sin θ) \cdot \\ \cdot (\sin φ \cos θ, \cos φ, \sin φ \sin θ) \\ = 4 \int_{0}^{π / 4} d φ \int_{0}^{2 π} d θ {\sin^{3} φ \cos φ \cos^{2} θ + \sin^{3} φ \cos φ \sin θ \cos θ \\ + \sin^{3} φ \cos φ \sin^{2} θ} \\ = 4 [\int_{0}^{π / 4} d φ \sin^{3} φ \cos φ] [\int_{0}^{2 π} d θ (1 + \sin θ \cos θ)] \\ = 4 {[\frac{\sin^{4} φ}{4}]}_{0}^{π / 4} {[θ + \frac{\sin^{2} θ}{2}]}_{0}^{2 π} \\ = 4 \times \frac{1}{16} \times (2 π) = \frac{π}{2} \end{aligned}

Solution 2 — parametrizing by $x$ and $z$ .

We can also parametrize

S

by using

x

and

z

as parameters. On

S,

$y = \sqrt{2 - x^{2} - z^{2}}$ and
$y$ runs over the range $1 \leq y \leq \sqrt{2} .$ Correspondingly, $x^{2} + z^{2} = 2 - y^{2}$ runs over $0 \leq x^{2} + z^{2} \leq 1$

So we can use the parametrization

r (x, z) = x \hat{ı ı} + \sqrt{2 - x^{2} - z^{2}} \hat{ȷ ȷ} + z \hat{k} 0 \leq x^{2} + z^{2} \leq 1

Since

\begin{aligned} \frac{\partial r}{\partial x} & = \hat{ı ı} - \frac{x}{\sqrt{2 - x^{2} - z^{2}}} \hat{ȷ ȷ} \\ \frac{\partial r}{\partial z} & = - \frac{z}{\sqrt{2 - x^{2} - z^{2}}} \hat{ȷ ȷ} + \hat{k} \end{aligned}

so that

\begin{aligned} \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial z} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 1 & - \frac{x}{\sqrt{2 - x^{2} - z^{2}}} & 0 \\ 0 & - \frac{z}{\sqrt{2 - x^{2} - z^{2}}} & 1 \end{array}] \\ = - \frac{x}{\sqrt{2 - x^{2} - z^{2}}} \hat{ı ı} - \hat{ȷ ȷ} - \frac{z}{\sqrt{2 - x^{2} - z^{2}}} \hat{k} \end{aligned}

(3.3.1) gives

\begin{aligned} \hat{n} d S & = \pm \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial z} d x d z \\ = \mp (\frac{x}{\sqrt{2 - x^{2} - z^{2}}}, 1, \frac{z}{\sqrt{2 - x^{2} - z^{2}}}) d x d z \\ d S & = | \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial z} | d x d z \\ = \sqrt{1 + \frac{x^{2} + z^{2}}{2 - x^{2} - z^{2}}} d x d z = \frac{\sqrt{2}}{\sqrt{2 - x^{2} - z^{2}}} d x d z \end{aligned}

Choose the plus sign to give the outward pointing normal.

(a) The specified integral is

\begin{aligned} \iint_{S} y^{3} d S & = \iint_{x^{2} + z^{2} \leq 1} d x d z \frac{\sqrt{2}}{\sqrt{2 - x^{2} - z^{2}}} \overset{y^{3}}{\overset{⏞}{(2 - x^{2} - z^{2})^{3 / 2}}} \\ = \sqrt{2} \iint_{x^{2} + z^{2} \leq 1} d x d z (2 - x^{2} - z^{2}) \end{aligned}

Switching to polar coordinates with

x = r \cos θ

and

z = r \sin θ,

\begin{aligned} \iint_{S} y^{3} d S & = \sqrt{2} \int_{0}^{2 π} d θ \int_{0}^{1} d r r (2 - r^{2}) \\ = \sqrt{2} (2 π) {[r^{2} - \frac{r^{4}}{4}]}_{0}^{1} = \sqrt{2} (2 π) [\frac{3}{4}] = \frac{3}{\sqrt{2}} π \end{aligned}

(b) The specified integral is

\begin{aligned} \iint_{S} (x y \hat{ı ı} + x z \hat{ȷ ȷ} + z y \hat{k}) \cdot \hat{n} d S \\ = \iint_{x^{2} + z^{2} \leq 1} d x d z (x \sqrt{2 - x^{2} - z^{2}}, x z, z \sqrt{2 - x^{2} - z^{2}}) \cdot \\ (\frac{x}{\sqrt{2 - x^{2} - z^{2}}}, 1, \frac{z}{\sqrt{2 - x^{2} - z^{2}}}) \\ = \iint_{x^{2} + z^{2} \leq 1} d x d z {x^{2} + x z + z^{2}} \end{aligned}

Switching to polar coordinates with

x = r \cos θ

and

z = r \sin θ,

\begin{aligned} \iint_{S} (x y \hat{ı ı} + x z \hat{ȷ ȷ} + z y \hat{k}) \cdot \hat{n} d S \\ = \int_{0}^{2 π} d θ \int_{0}^{1} d r r (r^{2} \cos^{2} θ + r^{2} \sin θ \cos θ + r^{2} \sin^{2} θ) \\ = [\int_{0}^{1} d r r^{3}] [\int_{0}^{2 π} d θ (1 + \sin θ \cos θ)] \\ = {[\frac{r^{4}}{4}]}_{0}^{1} {[θ + \frac{\sin^{2} θ}{2}]}_{0}^{2 π} \\ = \frac{1}{4} \times (2 π) = \frac{π}{2} \end{aligned}

3.3.6.42. (✳).

Solution.

First observe that,

because $(x + y + 1)^{2} \geq 0,$ all points on $(x + y + 1)^{2} + z^{2} = 4$ have $| z | \leq 2$ and that,
for $| z_{0} | \leq 2,$ the surface $(x + y + 1)^{2} + z^{2} = 4$ intersects the horizontal plane $z = z_{0}$ on $(x + y + 1)^{2} = 4 - z_{0}^{2},$ i.e. on the two lines $x + y = \pm \sqrt{4 - z_{0}^{2}} - 1,$ $z = z_{0} .$
The line $x + y = \pm \sqrt{4 - z_{0}^{2}} - 1,$ $z = z_{0}$ intersects the first octant if and only if $z_{0} \geq 0$ and $\pm \sqrt{4 - z_{0}^{2}} - 1 \geq 0 .$
Thus $x + y = - \sqrt{4 - z_{0}^{2}} - 1,$ $z = z_{0}$ never intersects the first octant and
$x + y = \sqrt{4 - z_{0}^{2}} - 1,$ $z = z_{0}$ intersects the first octant if and only if $0 \leq z_{0} \leq \sqrt{3} .$
When $z_{0} = 0,$ the line $x + y = \sqrt{4 - z_{0}^{2}} - 1,$ $z = z_{0}$ is $x + y = 1,$ $z = 0 .$
When $z_{0} = \sqrt{3},$ the line $x + y = \sqrt{4 - z_{0}^{2}} - 1,$ $z = z_{0}$ is $x + y = 0,$ $z = 2 .$
So as $(x, y, z)$ runs over $S,$ $(x, y)$ runs over the triangle $x \geq 0,$ $y \geq 0,$ $x + y \leq 1 .$

Let

G (x, y, z) = (x + y + 1)^{2} + z^{2} .

Then

\hat{n} d S = \pm \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} d x d y = \pm \frac{2 (x + y + 1) \hat{ı ı} + 2 (x + y + 1) \hat{ȷ ȷ} + 2 z \hat{k}}{2 z} d x d y

For the downward normal, we need the minus sign, so

\begin{aligned} F \cdot \hat{n} d S & = - [x y \hat{ı ı} + (z - x y) \hat{ȷ ȷ}] \cdot [\frac{(x + y + 1) \hat{ı ı} + (x + y + 1) \hat{ȷ ȷ} + z \hat{k}}{z}] d x d y \\ = - \frac{1}{z} [x y (x + y + 1) + (z - x y) (x + y + 1)] d x d y \\ = - \frac{1}{z} [z (x + y + 1)] d x d y \\ = - (x + y + 1) d x d y \end{aligned}

The domain of integration is

x \geq 0, y \geq 0, x + y \leq 1,

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = - \int_{0}^{1} d x \int_{0}^{1 - x} d y (x + y + 1) \\ = - \int_{0}^{1} d x [(1 + x) (1 - x) + \frac{1}{2} (1 - x)^{2}] \\ = - \int_{0}^{1} d x [\frac{3}{2} - x - \frac{1}{2} x^{2}] = - [\frac{3}{2} - \frac{1}{2} - \frac{1}{6}] = - \frac{5}{6} \end{aligned}

4 Integral Theorems
4.1 Gradient, Divergence and Curl
4.1.6 Exercises

4.1.6.1. (✳).

Solution.

(a) A. The angle between

F

and

d r

is less than

90^{\circ}

along the entire path. So

F \cdot d r > 0

along the entire path and the work is positive.

(b) B.

F

is perpendicular to

d r

along all of

C_{2} .

\int_{C_{2}} F \cdot d r = 0 .

P_{x} = Q_{y} = 0

N .

\nabla \nabla \cdot F = 0

N .

(d) A. At

Q,

the vertical component of

F

is increasing from left to right (so that

Q_{x} > 0

) and the horizontal component of

F

is decreasing from bottom to top (so that

P_{y} < 0

). So

Q_{x} - P_{y} > 0

N .

(e) B. At

D,

the horizontal component of

F

is increasing from left to right, so that

P_{x} > 0 .

4.1.6.2.

Solution.

No. The vector field

F (x, y, z) = \hat{ı ı} + y \hat{k}

has

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 1 & 0 & y \end{array}] \\ = \hat{ı ı} \end{aligned}

has dot product

1

with

F (x, y, z)

(for all

x,

y,

z

) and so is not perpendicular to it.

4.1.6.3.

Solution.

(a) By the product rule

\begin{aligned} \nabla \nabla \cdot (f F) & = \frac{\partial}{\partial x} (f F_{1}) + \frac{\partial}{\partial y} (f F_{2}) + \frac{\partial}{\partial z} (f F_{3}) \\ = f \frac{\partial F_{1}}{\partial x} + f \frac{\partial F_{2}}{\partial y} + f \frac{\partial F_{3}}{\partial z} \\ + F_{1} \frac{\partial f}{\partial x} + F_{2} \frac{\partial f}{\partial y} + F_{3} \frac{\partial f}{\partial z} \\ = f \nabla \nabla \cdot F + F \cdot \nabla \nabla f \end{aligned}

(b) Again by the product rule

\begin{aligned} \nabla \nabla \cdot (F \times G) \\ = \frac{\partial}{\partial x} (F_{2} G_{3} - F_{3} G_{2}) + \frac{\partial}{\partial y} (F_{3} G_{1} - F_{1} G_{3}) + \frac{\partial}{\partial z} (F_{1} G_{2} - F_{2} G_{1}) \\ = \frac{\partial F_{2}}{\partial x} G_{3} - \frac{\partial F_{3}}{\partial x} G_{2} + \frac{\partial F_{3}}{\partial y} G_{1} - \frac{\partial F_{1}}{\partial y} G_{3} + \frac{\partial F_{1}}{\partial z} G_{2} - \frac{\partial F_{2}}{\partial z} G_{1} \\ + F_{2} \frac{\partial G_{3}}{\partial x} - F_{3} \frac{\partial G_{2}}{\partial x} + F_{3} \frac{\partial G_{1}}{\partial y} - F_{1} \frac{\partial G_{3}}{\partial y} + F_{1} \frac{\partial G_{2}}{\partial z} - F_{2} \frac{\partial G_{1}}{\partial z} \\ = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) G_{1} + (\frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}) G_{2} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) G_{3} \\ - F_{1} (\frac{\partial G_{3}}{\partial y} - \frac{\partial G_{2}}{\partial z}) - F_{2} (\frac{\partial G_{1}}{\partial z} - \frac{\partial G_{3}}{\partial x}) - F_{3} (\frac{\partial G_{2}}{\partial x} - \frac{\partial G_{1}}{\partial y}) \\ = G \cdot (\nabla \nabla \times F) - F \cdot (\nabla \nabla \times G) \end{aligned}

{\nabla \nabla}^{2} (f g) = \nabla \nabla \cdot [\nabla \nabla (f g)] .

First

\begin{aligned} \nabla \nabla (f g) & = \hat{ı ı} \frac{\partial}{\partial x} (f g) + \hat{ȷ ȷ} \frac{\partial}{\partial y} (f g) + \hat{k} \frac{\partial}{\partial z} (f g) \\ = \hat{ı ı} g \frac{\partial f}{\partial x} + \hat{ȷ ȷ} g \frac{\partial f}{\partial y} + \hat{k} g \frac{\partial f}{\partial z} \\ + \hat{ı ı} f \frac{\partial g}{\partial x} + \hat{ȷ ȷ} f \frac{\partial g}{\partial y} + \hat{k} f \frac{\partial g}{\partial z} \\ = g \nabla \nabla f + f \nabla \nabla g \end{aligned}

So by part (a), twice,

\begin{aligned} {\nabla \nabla}^{2} (f g) & = \nabla \nabla \cdot (g \nabla \nabla f) + \nabla \nabla \cdot (f \nabla \nabla g) \\ = g (\nabla \nabla \cdot \nabla \nabla f) + (\nabla \nabla g) \cdot (\nabla \nabla f) + f (\nabla \nabla \cdot \nabla \nabla g) + (\nabla \nabla f) \cdot (\nabla \nabla g) \\ = f {\nabla \nabla}^{2} g + 2 \nabla \nabla f \cdot \nabla \nabla g + g {\nabla \nabla}^{2} f \end{aligned}

4.1.6.4.

Solution.

(a) By definition

\begin{aligned} \nabla \nabla \cdot (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}) & = \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial y} (y) + \frac{\partial}{\partial z} (z) = 3 \\ \nabla \nabla \times (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z \end{array}] = 0 \end{aligned}

(b) By definition

\begin{aligned} \nabla \nabla \cdot (x y^{2} \hat{ı ı} - y z^{2} \hat{ȷ ȷ} + z x^{2} \hat{k}) & = \frac{\partial}{\partial x} (x y^{2}) + \frac{\partial}{\partial y} (- y z^{2}) + \frac{\partial}{\partial z} (z x^{2}) \\ = y^{2} - z^{2} + x^{2} \\ \nabla \nabla \times (x y^{2} \hat{ı ı} - y z^{2} \hat{ȷ ȷ} + z x^{2} \hat{k}) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x y^{2} & - y z^{2} & z x^{2} \end{array}] \\ = 2 y z \hat{ı ı} - 2 x z \hat{ȷ ȷ} - 2 x y \hat{k} \end{aligned}

\begin{aligned} \nabla \nabla \cdot (\frac{x}{\sqrt{x^{2} + y^{2}}} \hat{ı ı} + \frac{y}{\sqrt{x^{2} + y^{2}}} \hat{ȷ ȷ}) \\ = \frac{\partial}{\partial x} (\frac{x}{\sqrt{x^{2} + y^{2}}}) + \frac{\partial}{\partial y} (\frac{y}{\sqrt{x^{2} + y^{2}}}) \\ = \frac{1}{\sqrt{x^{2} + y^{2}}} - \frac{x^{2}}{{[x^{2} + y^{2}]}^{3 / 2}} + \frac{1}{\sqrt{x^{2} + y^{2}}} - \frac{y^{2}}{{[x^{2} + y^{2}]}^{3 / 2}} \\ = \frac{x^{2} + y^{2} - x^{2} + x^{2} + y^{2} - y^{2}}{{[x^{2} + y^{2}]}^{3 / 2}} \\ = \frac{1}{\sqrt{x^{2} + y^{2}}} \\ \nabla \nabla \times (\frac{x}{\sqrt{x^{2} + y^{2}}} \hat{ı ı} + \frac{y}{\sqrt{x^{2} + y^{2}}} \hat{ȷ ȷ}) \\ = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{x}{\sqrt{x^{2} + y^{2}}} & \frac{y}{\sqrt{x^{2} + y^{2}}} & 0 \end{array}] \\ = (- \frac{x y}{{[x^{2} + y^{2}]}^{3 / 2}} + \frac{x y}{{[x^{2} + y^{2}]}^{3 / 2}}) \hat{k} \\ = 0 \end{aligned}

(d) By definition

\begin{aligned} \nabla \nabla \cdot (- \frac{y}{\sqrt{x^{2} + y^{2}}} \hat{ı ı} + \frac{x}{\sqrt{x^{2} + y^{2}}} \hat{ȷ ȷ}) \\ = \frac{\partial}{\partial x} (- \frac{y}{\sqrt{x^{2} + y^{2}}}) + \frac{\partial}{\partial y} (\frac{x}{\sqrt{x^{2} + y^{2}}}) \\ = \frac{x y}{{[x^{2} + y^{2}]}^{3 / 2}} - \frac{x y}{{[x^{2} + y^{2}]}^{3 / 2}} \\ = 0 \\ \nabla \nabla \times (- \frac{y}{\sqrt{x^{2} + y^{2}}} \hat{ı ı} + \frac{x}{\sqrt{x^{2} + y^{2}}} \hat{ȷ ȷ}) \\ = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ - \frac{y}{\sqrt{x^{2} + y^{2}}} & \frac{x}{\sqrt{x^{2} + y^{2}}} & 0 \end{array}] \\ = (\frac{1}{\sqrt{x^{2} + y^{2}}} - \frac{x^{2}}{{[x^{2} + y^{2}]}^{3 / 2}} + \frac{1}{\sqrt{x^{2} + y^{2}}} - \frac{y^{2}}{{[x^{2} + y^{2}]}^{3 / 2}}) \hat{k} \\ = \frac{x^{2} + y^{2} - x^{2} + x^{2} + y^{2} - y^{2}}{{[x^{2} + y^{2}]}^{3 / 2}} \hat{k} \\ = \frac{\hat{k}}{\sqrt{x^{2} + y^{2}}} \end{aligned}

4.1.6.5. (✳).

Solution.

(a) We are to compute the divergence of

\frac{r}{r} = \frac{x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}}{[x^{2} + y^{2} + z^{2}]^{1 / 2}} .

Since

\begin{aligned} \frac{\partial}{\partial x} \frac{x}{{[x^{2} + y^{2} + z^{2}]}^{1 / 2}} & = \frac{1}{{[x^{2} + y^{2} + z^{2}]}^{1 / 2}} - \frac{1}{2} \frac{x (2 x)}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \\ = \frac{y^{2} + z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \\ \frac{\partial}{\partial y} \frac{y}{{[x^{2} + y^{2} + z^{2}]}^{1 / 2}} & = \frac{1}{{[x^{2} + y^{2} + z^{2}]}^{1 / 2}} - \frac{1}{2} \frac{y (2 y)}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \\ = \frac{x^{2} + z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \\ \frac{\partial}{\partial z} \frac{z}{{[x^{2} + y^{2} + z^{2}]}^{1 / 2}} & = \frac{1}{{[x^{2} + y^{2} + z^{2}]}^{1 / 2}} - \frac{1}{2} \frac{z (2 z)}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \\ = \frac{x^{2} + y^{2}}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \end{aligned}

the specified divergence is

\begin{aligned} \nabla \nabla (\frac{r}{r}) & = \frac{2 x^{2} + 2 y^{2} + 2 z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} = \frac{2 r^{2}}{r^{3}} = \frac{2}{r} \end{aligned}

(b)

\begin{aligned} \nabla \nabla \times (y z \hat{ı ı} + 2 x z \hat{ȷ ȷ} + e^{x y} \hat{k}) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y z & 2 x z & e^{x y} \end{array}] \\ = (x e^{x y} - 2 x) \hat{ı ı} - (y e^{x y} - y) \hat{ȷ ȷ} + z \hat{k} \end{aligned}

4.1.6.6. (✳).

Solution.

(a) Since

r^{k} = (x^{2} + y^{2} + z^{2})^{k / 2},

\begin{aligned} \frac{\partial}{\partial x} r^{k} & = 2 x \frac{k}{2} (x^{2} + y^{2} + z^{2})^{\frac{k}{2} - 1} = k (r \cdot \hat{ı ı}) r^{k - 2} \\ \frac{\partial}{\partial y} r^{k} & = 2 y \frac{k}{2} (x^{2} + y^{2} + z^{2})^{\frac{k}{2} - 1} = k (r \cdot \hat{ȷ ȷ}) r^{k - 2} \\ \frac{\partial}{\partial z} r^{k} & = 2 z \frac{k}{2} (x^{2} + y^{2} + z^{2})^{\frac{k}{2} - 1} = k (r \cdot \hat{k}) r^{k - 2} \end{aligned}

We want

k = - 3 .

(b) Using the computation in part (a)

\begin{aligned} \nabla \nabla \cdot (r^{k} r) & = \frac{\partial}{\partial x} (x r^{k}) + \frac{\partial}{\partial y} (y r^{k}) + \frac{\partial}{\partial z} (z r^{k}) \\ = 3 r^{k} + x \frac{\partial}{\partial x} r^{k} + y \frac{\partial}{\partial y} r^{k} + z \frac{\partial}{\partial z} r^{k} \\ = 3 r^{k} + x (k x r^{k - 2}) + y (k y r^{k - 2}) + z (k z r^{k - 2}) \\ = (3 + k) r^{k} \end{aligned}

We want

k = 2 .

{\nabla \nabla}^{2} = \nabla \nabla \cdot \nabla \nabla,

\begin{aligned} {\nabla \nabla}^{2} (r^{k}) & = \nabla \nabla \cdot (\nabla \nabla (r^{k})) \\ = \nabla \nabla \cdot (k r^{k - 2} r) & by part (a) \\ = k (3 + k - 2) r^{k - 2} & by part (b),  but with k replaced by k - 2 \end{aligned}

We want

k = - 2 .

4.1.6.7. (✳).

Solution.

(a)

\nabla \nabla \cdot r = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3

(b)

\nabla \nabla (r^{2}) = (\hat{ı ı} \frac{\partial}{\partial x} + \hat{ȷ ȷ} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}) (x^{2} + y^{2} + z^{2}) = 2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + 2 z \hat{k} = 2 r

r \times a = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ x & y & z \\ a_{1} & a_{2} & a_{3} \end{matrix}] = \hat{ı ı} (a_{3} y - a_{2} z) + \hat{ȷ ȷ} (a_{1} z - a_{3} x) + \hat{k} (a_{2} x - a_{1} y)

we have

\begin{aligned} \nabla \nabla \times (r \times a) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ a_{3} y - a_{2} z & a_{1} z - a_{3} x & a_{2} x - a_{1} y \end{array}] \\ = - 2 a_{1} \hat{ı ı} - 2 a_{2} \hat{ȷ ȷ} - 2 a_{3} \hat{k} \\ = - 2 a \end{aligned}

(d) Since

\begin{aligned} \nabla \nabla (r) & = (\hat{ı ı} \frac{\partial}{\partial x} + \hat{ȷ ȷ} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}) (x^{2} + y^{2} + z^{2})^{1 / 2} \\ = \hat{ı ı} \frac{x}{(x^{2} + y^{2} + z^{2})^{1 / 2}} + \hat{ȷ ȷ} \frac{y}{(x^{2} + y^{2} + z^{2})^{1 / 2}} + \hat{k} \frac{x}{(x^{2} + y^{2} + z^{2})^{1 / 2}} \end{aligned}

we have

\begin{aligned} \nabla \nabla \cdot (\nabla \nabla (r)) & = \frac{\partial}{\partial x} \frac{x}{(x^{2} + y^{2} + z^{2})^{1 / 2}} + \frac{\partial}{\partial y} \frac{y}{(x^{2} + y^{2} + z^{2})^{1 / 2}} \\ + \frac{\partial}{\partial z} \frac{z}{(x^{2} + y^{2} + z^{2})^{1 / 2}} \\ = \frac{3}{(x^{2} + y^{2} + z^{2})^{1 / 2}} - \frac{1}{2} \frac{2 x^{2} + 2 y^{2} + 2 z^{2}}{(x^{2} + y^{2} + z^{2})^{3 / 2}} \\ = \frac{2}{(x^{2} + y^{2} + z^{2})^{1 / 2}} \\ = \frac{2}{r} \end{aligned}

4.1.6.8. (✳).

Solution.

(a) Since

\begin{aligned} \nabla \nabla (\frac{1}{r}) = (\hat{ı ı} \frac{\partial}{\partial x} + \hat{ȷ ȷ} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}) (x^{2} + y^{2} + z^{2})^{- 1 / 2} \\ = - \hat{ı ı} \frac{x}{(x^{2} + y^{2} + z^{2})^{3 / 2}} - \hat{ȷ ȷ} \frac{y}{(x^{2} + y^{2} + z^{2})^{3 / 2}} - \hat{k} \frac{z}{(x^{2} + y^{2} + z^{2})^{3 / 2}} \\ = - \hat{ı ı} \frac{x}{r^{3}} - \hat{ȷ ȷ} \frac{y}{r^{3}} - \hat{k} \frac{x}{r^{3}} \end{aligned}

we have

a = - 3 .

(b) Since

\begin{aligned} \nabla \nabla \cdot (r r) & = \frac{\partial}{\partial x} [(x^{2} + y^{2} + z^{2})^{1 / 2} x] + \frac{\partial}{\partial y} [(x^{2} + y^{2} + z^{2})^{1 / 2} y] \\ + \frac{\partial}{\partial z} [(x^{2} + y^{2} + z^{2})^{1 / 2} z] \\ = 3 (x^{2} + y^{2} + z^{2})^{1 / 2} + \frac{1}{2} \frac{2 x^{2} + 2 y^{2} + 2 z^{2}}{(x^{2} + y^{2} + z^{2})^{1 / 2}} \\ = 4 (x^{2} + y^{2} + z^{2})^{1 / 2} \\ = 4 r \end{aligned}

we have

a = 4 .

\begin{aligned} \nabla \nabla (r^{3}) = (\hat{ı ı} \frac{\partial}{\partial x} + \hat{ȷ ȷ} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}) (x^{2} + y^{2} + z^{2})^{3 / 2} \\ = \hat{ı ı} 3 x (x^{2} + y^{2} + z^{2})^{1 / 2} + \hat{ȷ ȷ} 3 y (x^{2} + y^{2} + z^{2})^{1 / 2} + \hat{k} 3 z (x^{2} + y^{2} + z^{2})^{1 / 2} \\ = 3 r r \end{aligned}

we have

\begin{aligned} \nabla \nabla \cdot (\nabla \nabla (r^{3})) & = \nabla \nabla \cdot (3 r r) = 3 \nabla \nabla \cdot (r r) = 3 (4 r) by part (b) \\ = 12 r \end{aligned}

so that

a = 12 .

4.1.6.9.

Solution.

(a) Since

\nabla \nabla \cdot F = \frac{\partial}{\partial x} (1 + y z) + \frac{\partial}{\partial y} (2 y + z x) + \frac{\partial}{\partial z} (3 z^{2} + x y) = 2 + 6 z \neq 0,

F

fails the screening test and cannot have a vector potential.

(b) The vector field

A = A_{1} \hat{ı ı} + A_{2} \hat{ȷ ȷ}

is a vector potential for

G

if and only if

G = \nabla \nabla \times A,

which is the case if and only if

\begin{aligned} - \frac{\partial A_{2}}{\partial z} & = y z & ⟺ & A_{2} = - \frac{1}{2} y z^{2} + B_{2} (x, y) \\ \frac{\partial A_{1}}{\partial z} & = z x & ⟺ & A_{1} = \frac{1}{2} x z^{2} + B_{1} (x, y) \\ \frac{\partial A_{2}}{\partial x} - \frac{\partial A_{1}}{\partial y} & = x y & ⟺ & \frac{\partial B_{2}}{\partial x} - \frac{\partial B_{1}}{\partial y} = x y \end{aligned}

There are infinitely many solutions to

\frac{\partial B_{2}}{\partial x} - \frac{\partial B_{1}}{\partial y} = x y .

In fact

B_{2}

is completely arbitrary. If one chooses

B_{2} = 0,

then

B_{1} = - \frac{1}{2} x y^{2}

does the job. If one chooses

B_{1} = 0,

then

B_{2} = \frac{1}{2} x^{2} y

does the job. Thus two solutions are

A = \frac{1}{2} (z^{2} - y^{2}) x \hat{ı ı} - \frac{1}{2} y z^{2} \hat{ȷ ȷ}

and

A = \frac{1}{2} x z^{2} \hat{ı ı} + \frac{1}{2} (x^{2} - z^{2}) y \hat{ȷ ȷ} .

4.1.6.10. (✳).

Solution.

(a)

F

is well-defined wherever the denominator

x^{2} + z^{2}

is nonzero. So the (largest possible) domain is

\begin{matrix} D = {(x, y, z) | x^{2} + z^{2} \neq 0} \end{matrix}

(b) As preliminary computations, let’s find

\begin{aligned} \frac{\partial}{\partial z} (\frac{- z}{x^{2} + z^{2}}) & = \frac{- 1}{x^{2} + z^{2}} - \frac{2 z (- z)}{{(x^{2} + z^{2})}^{2}} = \frac{- x^{2} + z^{2}}{{(x^{2} + z^{2})}^{2}} \\ \frac{\partial}{\partial x} (\frac{x}{x^{2} + z^{2}}) & = \frac{1}{x^{2} + z^{2}} - \frac{2 x (x)}{{(x^{2} + z^{2})}^{2}} = \frac{- x^{2} + z^{2}}{{(x^{2} + z^{2})}^{2}} \end{aligned}

So the curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{- z}{x^{2} + z^{2}} & y & \frac{x}{x^{2} + z^{2}} \end{array}] = - (\frac{- x^{2} + z^{2}}{{(x^{2} + z^{2})}^{2}} - \frac{- x^{2} + z^{2}}{{(x^{2} + z^{2})}^{2}}) \hat{ȷ ȷ} \\ = 0 \end{aligned}

on the domain of $F$ .

\begin{aligned} \frac{\partial}{\partial x} (\frac{- z}{x^{2} + z^{2}}) & = - \frac{2 x (- z)}{{(x^{2} + z^{2})}^{2}} = \frac{2 x z}{{(x^{2} + z^{2})}^{2}} \\ \frac{\partial}{\partial z} (\frac{x}{x^{2} + z^{2}}) & = - \frac{2 z (x)}{{(x^{2} + z^{2})}^{2}} = \frac{- 2 x z}{{(x^{2} + z^{2})}^{2}} \end{aligned}

So the divergence of

F

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (\frac{- z}{x^{2} + z^{2}}) + \frac{\partial}{\partial y} (y) + \frac{\partial}{\partial z} (\frac{x}{x^{2} + z^{2}}) = 1 \end{aligned}

(d) By part (b), the vector field passes the conservative field screening test

\nabla \nabla \times F = 0 .

But we should still be suspicious because of the similarity of

F

to the vector field of Examples 2.3.14 and 4.3.8.

So let’s compute the line integral of

F

around the (closed) circle

y = 0,

x^{2} + z^{2} = 1,

parametrized by

r (t) = \cos t \hat{ı ı} + \sin t \hat{k} r^{'} (t) = - \sin t \hat{ı ı} + \cos t \hat{k}

The line integral is

\begin{aligned} \int_{C} F \cdot d r & = \int_{0}^{2 π} {\overset{\frac{- z}{x^{2} + z^{2}}}{\overset{⏞}{- \sin t}} \hat{ı ı} + \overset{\frac{x}{x^{2} + y^{2}}}{\overset{⏞}{\cos t}} \hat{k}} \cdot {\overset{r^{'} (t)}{\overset{⏞}{- \sin t \hat{ı ı} + \cos t \hat{k}}}} d t \\ = \int_{0}^{2 π} d t = 2 π \end{aligned}

As the integral of

F

around the simple closed curve

C

is not zero,

F

cannot be conservative on

D .

See Theorem 2.4.7 and Examples 2.3.14 and 4.3.8.

4.1.6.11. (✳).

Solution.

(a) By the vector identity of Theorem 4.1.7.a,

\nabla \cdot F = \nabla \cdot \nabla \times G = 0

So we must have

\begin{aligned} 0 & = \nabla \cdot F = \nabla \cdot ((x z + x y) \hat{ı ı} + α (y z - x y) \hat{ȷ ȷ} + β (y z + x z) \hat{k}) \\ = (z + y) + α (z - x) + β (y + x) \end{aligned}

This is true for all

(x, y, z)

if and only if

α = β = - 1 .

(b) Since

\begin{aligned} \nabla \times G & = \nabla \times (x y z \hat{ı ı} - x y z \hat{ȷ ȷ} + g (x, y, z) \hat{k}) \\ = (g_{y} + x y) \hat{ı ı} - (g_{x} - x y) h j + (- y z - x z) \hat{k} \end{aligned}

we will have that

\nabla \times G = F

if and only if

\begin{aligned} (g_{y} + x y) \hat{ı ı} - (g_{x} - x y) \hat{ȷ ȷ} + (- y z - x z) \hat{k} \\ = (x z + x y) \hat{ı ı} - (y z - x y) \hat{ȷ ȷ} - (y z + x z) \hat{k} \end{aligned}

which is the case if and only if

g_{y} = x z, g_{x} = y z

The first equation,

g_{y} = x z,

is satisfied if and only if

g = x y z + h (x, z) .

The second equation is also satisfied if and only if

g_{x} = y z + h_{x} (x, z) = y z .

This is the case if and only if

h_{x} (x, z) = 0 .

That is, if and only if

h

is independent of

x .

Equivalently, if and only if

h (x, z) = w (z)

for some function

w (z) .

So, in fact, any function of the form

g (x, y, z) = x y z + w (z)

will work.

4.1.6.12.

Solution.

(a) Denote by

θ

the angle between

\hat{a}

and

r .

The point

r

is a distance

ℓ = | r | \sin θ

from the axis of rotation. So as the body rotates, the point sweeps out a circle of radius

ℓ

centred on the axis of rotation.

In one second the point sweeps out an arc of this circle that subtends an angle of

Ω

radians. This arc is the fraction

\frac{Ω}{2 π}

of a full circle and so has length

\frac{Ω}{2 π} 2 π ℓ = Ω ℓ = Ω | r | \sin θ .

Thus the point is moving with speed

Ω | r | \sin θ .

The velocity vector of the point must have length

Ω | r | \sin θ

and direction perpendicular to both

\hat{a}

and

r .

The vector

Ω Ω \times r

is perpendicular to both

r

and

Ω Ω = Ω \hat{a}

and has length

| Ω Ω | | r | \sin θ = Ω | r | \sin θ

as desired. So the velocity vector is either

Ω Ω \times r

or its negative. By the right hand rule it is

Ω Ω \times r .

(b) By vector identities

\begin{aligned} \nabla \nabla \cdot (F \times G) & = G \cdot (\nabla \nabla \times F) - F \cdot (\nabla \nabla \times G) \\ \nabla \nabla \times (F \times G) & = F (\nabla \nabla \cdot G) - (\nabla \nabla \cdot F) G + (G \cdot \nabla \nabla) F - (F \cdot \nabla \nabla) G \end{aligned}

(which are Theorems 4.1.4(d) and 4.1.5(d)) and the assumption that

Ω Ω

is constant

\begin{aligned} \nabla \nabla \times (Ω Ω \times r) & = Ω Ω (\nabla \nabla \cdot r) - (\nabla \nabla \cdot Ω Ω) r + (r \cdot \nabla \nabla) Ω Ω - (Ω Ω \cdot \nabla \nabla) r \\ = Ω Ω (\nabla \nabla \cdot r) - (Ω Ω \cdot \nabla \nabla) r \\ \nabla \nabla \cdot (Ω Ω \times r) & = r \cdot (\nabla \nabla \times Ω Ω) - Ω Ω \cdot (\nabla \nabla \times r) = - Ω Ω \cdot (\nabla \nabla \times r) \end{aligned}

Substituting in

\begin{aligned} \nabla \nabla \cdot r & = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3 \\ \nabla \nabla \times r & = (\frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}) \hat{ı ı} + (\frac{\partial x}{\partial z} - \frac{\partial z}{\partial x}) \hat{ȷ ȷ} + (\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}) \hat{k} = 0 \\ (Ω Ω \cdot \nabla \nabla) r & = (Ω_{1} \frac{\partial}{\partial x} + Ω_{2} \frac{\partial}{\partial y} + Ω_{3} \frac{\partial}{\partial z}) (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}) \\ = Ω_{1} \hat{ı ı} + Ω_{2} \hat{ȷ ȷ} + Ω_{3} \hat{k} = Ω Ω \end{aligned}

gives

\nabla \nabla \times (Ω Ω \times r) = 2 Ω Ω \nabla \nabla \cdot (Ω Ω \times r) = 0

6378 \sin (90^{\circ} - 49^{\circ}) = 6378 \cos (49^{\circ}) = 4184

km from the axis of rotation. The rate of rotation is

Ω = \frac{2 π}{24}

radians per hour. In each hour the students sweep out an arc of

\frac{2 π}{24}

radians from a circle of radius

4184

km. Their speed is

\frac{2 π}{24} \times 4184 = 1095

km/hr.

4.1.6.13.

Solution.

We shall show that

\frac{\partial G_{3}}{\partial y} - \frac{\partial G_{2}}{\partial z} = F_{1} .

The other components are similar. First we have

\begin{aligned} t F (r (t)) \times \frac{d r}{d t} (t) & = t F (t x, t y, t z) \times (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}) \\ = t det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ F_{1} & F_{2} & F_{3} \\ x & y & z \end{array}] \end{aligned}

Reading off the

\hat{k}

and

\hat{ȷ ȷ}

components of the determinant gives

\begin{aligned} G_{3} (x, y, z) & = \int_{0}^{1} t [F_{1} (t x, t y, t z) y - F_{2} (t x, t y, t z) x] d t \\ G_{2} (x, y, z) & = \int_{0}^{1} t [F_{3} (t x, t y, t z) x - F_{1} (t x, t y, t z) z] d t \end{aligned}

\begin{aligned} \frac{\partial G_{3}}{\partial y} & = \int_{0}^{1} t [F_{1} (t x, t y, t z) + \frac{\partial F_{1}}{\partial y} (t x, t y, t z) t y - \frac{\partial F_{2}}{\partial y} (t x, t y, t z) t x] d t \\ \frac{\partial G_{2}}{\partial z} & = \int_{0}^{1} t [\frac{\partial F_{3}}{\partial z} (t x, t y, t z) t x - \frac{\partial F_{1}}{\partial z} (t x, t y, t z) t z - F_{1} (t x, t y, t z)] d t \end{aligned}

so that

\begin{aligned} \frac{\partial G_{3}}{\partial y} - \frac{\partial G_{2}}{\partial z} & = \int_{0}^{1} [2 t F_{1} (t x, t y, t z) \\ + t^{2} y \frac{\partial F_{1}}{\partial y} (t x, t y, t z) + t^{2} z \frac{\partial F_{1}}{\partial z} (t x, t y, t z) \\ - t^{2} x \frac{\partial F_{2}}{\partial y} (t x, t y, t z) - t^{2} x \frac{\partial F_{3}}{\partial z} (t x, t y, t z)] d t \end{aligned}

Since, by hypothesis,

\nabla \nabla \cdot F = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z} = 0,

the last two terms

\begin{matrix} - t^{2} x {\frac{\partial F_{2}}{\partial y} (t x, t y, t z) + \frac{\partial F_{3}}{\partial z} (t x, t y, t z)} = - t^{2} x {- \frac{\partial F_{1}}{\partial x} (t x, t y, t z)} \end{matrix}

so that

\begin{aligned} \frac{\partial G_{3}}{\partial y} - \frac{\partial G_{2}}{\partial z} \\ = \int_{0}^{1} [2 t F_{1} (t x, t y, t z) + t^{2} x \frac{\partial F_{1}}{\partial x} (t x, t y, t z) + t^{2} y \frac{\partial F_{1}}{\partial y} (t x, t y, t z) \\ + t^{2} z \frac{\partial F_{1}}{\partial z} (t x, t y, t z)] d t \\ = \int_{0}^{1} \frac{d}{d t} [t^{2} F_{1} (t x, t y, t z)] d t = [t^{2} F_{1} (t x, t y, t z)]_{t = 0}^{t = 1} \\ = F_{1} (x, y, z) \end{aligned}

4.2 The Divergence Theorem
4.2.6 Exercises

4.2.6.1.

Solution.

(a) Expressing the left hand side as an iterated integral, with

z

as the innermost integration variable, we have

\begin{aligned} ∭_{V} \frac{\partial f}{\partial z} (x, y, z) d x d y d z & = \int_{0}^{1} d x \int_{0}^{1} d y [\int_{0}^{1} d z \frac{\partial f}{\partial z} (x, y, z)] \\ = \int_{0}^{1} d x \int_{0}^{1} d y [f (x, y, 1) - f (x, y, 0)] \\ by the fundamental theorem of calculus \\ = \iint_{R} [f (x, y, 1) - f (x, y, 0)] d x d y \\ = \iint_{R} f (x, y, 1) d x d y - \iint_{R} f (x, y, 0) d x d y \end{aligned}

(b) Define the vector field

F (x, y, z) = f (x, y, z) \hat{k} .

Then the divergence of

F

\nabla \nabla \cdot F (x, y, z) = \frac{\partial f}{\partial z} (x, y, z) .

The boundary of the cube

V

is the union of six faces

\begin{aligned} S_{1} & = {(x, y, z) | 0 \leq x \leq 1, 0 \leq y \leq 1, z = 1} with outward normal \hat{n} = \hat{k} \\ S_{2} & = {(x, y, z) | 0 \leq x \leq 1, 0 \leq y \leq 1, z = 0} with outward normal \hat{n} = - \hat{k} \\ S_{3} & = {(x, y, z) | 0 \leq x \leq 1, 0 \leq z \leq 1, y = 1} with outward normal \hat{n} = \hat{ȷ ȷ} \\ S_{4} & = {(x, y, z) | 0 \leq x \leq 1, 0 \leq z \leq 1, y = 0} with outward normal \hat{n} = - \hat{ȷ ȷ} \\ S_{5} & = {(x, y, z) | 0 \leq y \leq 1, 0 \leq z \leq 1, x = 1} with outward normal \hat{n} = \hat{ı ı} \\ S_{6} & = {(x, y, z) | 0 \leq y \leq 1, 0 \leq z \leq 1, x = 0} with outward normal \hat{n} = - \hat{ı ı} \end{aligned}

Observe that

\begin{array}{r} F \cdot \hat{n} = f \hat{k} \cdot \hat{n} = {\begin{cases} + f & on S_{1} \\ - f & on S_{2} \\ 0 & on S_{3}, S_{4}, S_{5}, S_{6} \end{cases} \end{array}

So the divergence theorem gives

\begin{aligned} ∭_{V} \frac{\partial f}{\partial z} (x, y, z) d x d y d z & = ∭_{V} \nabla \nabla \cdot F (x, y, z) d x d y d z \\ = \iint_{\partial V} F \cdot \hat{n} d S = \sum_{j = 1}^{6} \iint_{S_{j}} F \cdot \hat{n} d S \\ = \iint_{S_{1}} f d S - \iint_{S_{2}} f d S \\ = \iint_{R} f (x, y, 1) d x d y - \iint_{R} f (x, y, 0) d x d y \end{aligned}

4.2.6.2.

Solution.

(a) The divergence of

ϕ a

\nabla \nabla \cdot (ϕ a) = \nabla \nabla ϕ \cdot a + ϕ \nabla \nabla \cdot a = \nabla \nabla ϕ \cdot a,

since

a

is constant. So, by the divergence theorem,

\begin{aligned} \iint_{\partial V} ϕ a \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot (ϕ a) d V = ∭_{V} \nabla \nabla ϕ \cdot a d V \\ ⟹ [\iint_{\partial V} ϕ \hat{n} d S - ∭_{V} \nabla \nabla ϕ d V] \cdot a = 0 \end{aligned}

This is true for all vectors

a .

In particular, applying this with

a = \hat{ı ı}, \hat{ȷ ȷ}, \hat{k},

we have that all three components of

[\iint_{\partial V} ϕ \hat{n} d S - ∭_{V} \nabla \nabla ϕ d V]

are zero. So

\iint_{\partial V} ϕ \hat{n} d S - ∭_{V} \nabla \nabla ϕ d V = 0

(b) By part (a), with

ϕ = x^{2} + y^{2} + z^{2}

and

\nabla \nabla ϕ = 2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + 2 z \hat{k},

\begin{aligned} \frac{1}{2 | V |} \iint_{\partial V} (x^{2} + y^{2} + z^{2}) \hat{n} d S & = \frac{1}{2 | V |} ∭_{V} (2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + 2 z \hat{k}) d V \\ = (\bar{x}, \bar{y}, \bar{z}) \end{aligned}

4.2.6.3.

Solution.

(a) We’ll parametrize the sphere using the spherical coordinates

θ

and

φ .

\begin{aligned} x & = \sin φ \cos θ \\ y & = \sin φ \sin θ \\ z & = \cos φ \end{aligned}

with

0 \leq θ \leq 2 π,

0 \leq φ \leq π .

Since

\begin{aligned} (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) & = (- \sin φ \sin θ, \sin φ \cos θ, 0) \\ (\frac{\partial x}{\partial φ}, \frac{\partial y}{\partial φ}, \frac{\partial z}{\partial φ}) & = (\cos φ \cos θ, \cos φ \sin θ, - \sin φ) \end{aligned}

(3.3.1) yields

\begin{aligned} \hat{n} d S = \pm (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) \times (\frac{\partial x}{\partial φ}, \frac{\partial y}{\partial φ}, \frac{\partial z}{\partial φ}) d θ d φ \\ = \pm (- \sin φ \sin θ, \sin φ \cos θ, 0) \times (\cos φ \cos θ, \cos φ \sin θ, - \sin φ) d θ d φ \\ = \pm (- \sin^{2} φ \cos θ, - \sin^{2} φ \sin θ, - \sin φ \cos φ) d θ d φ \\ = \mp \sin φ (\sin φ \cos θ, \sin φ \sin θ, \cos φ) d θ d φ \\ = \mp \sin φ (x (θ, φ), y (θ, φ), z (θ, φ)) d θ d φ \end{aligned}

To get an outward pointing normal we need the

+

sign, since then

\hat{n} (θ, φ)

is a positive multiple, namely

\sin φ,

times

r (θ, φ) .

So, on

S,

\begin{aligned} F \cdot \hat{n} d S & = \sin φ \overset{F}{\overset{⏞}{(\sin φ \cos θ, \sin φ \sin θ, \cos^{2} φ)}} \\ \cdot (\sin φ \cos θ, \sin φ \sin θ, \cos φ) d θ d φ \\ = \sin φ (\sin^{2} φ \cos^{2} θ + \sin^{2} φ \sin^{2} θ + \cos^{3} φ) \end{aligned}

and

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{0}^{π} d φ \int_{0}^{2 π} d θ \sin φ (\sin^{2} φ \cos^{2} θ + \sin^{2} φ \sin^{2} θ + \cos^{3} φ) \\ = \int_{0}^{π} d φ \int_{0}^{2 π} d θ (\sin^{3} φ + \sin φ \cos^{3} φ) \\ = 2 π {\int_{0}^{π} d φ \sin^{3} φ + {[- \frac{1}{4} \cos^{4} φ]}_{0}^{π}} \\ = 2 π \int_{0}^{π} d φ \sin φ (1 - \cos^{2} φ) = 2 π {[- \cos φ + \frac{1}{3} \cos^{3} φ]}_{0}^{π} \\ = 2 π [\frac{4}{3}] \\ = \frac{8 π}{3} \end{aligned}

(b) Let

V

be the interior of

S .

Then, by the divergence theorem,

\iint_{S} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} (1 + 1 + 2 z) d V

By oddness under

z \to - z,

the

z

integral vanishes, so that

\iint_{S} F \cdot \hat{n} d S = 2 ∭_{V} d V = 2 Volume (V) = 2 \frac{4 π}{3} = \frac{8 π}{3}

4.2.6.4.

Solution.

(a) Let’s use spherical coordinates. As

S

is the sphere of radius

a

centred on the origin, we can parametrize it by

\begin{aligned} r (θ, φ) & = a \sin φ \cos θ \hat{ı ı} + a \sin φ \sin θ \hat{ȷ ȷ} + a \cos φ \hat{k} \\ \frac{\partial r}{\partial θ} & = - a \sin φ \sin θ \hat{ı ı} + a \sin φ \cos θ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial φ} & = a \cos φ \cos θ \hat{ı ı} + a \cos φ \sin θ \hat{ȷ ȷ} - a \sin φ \hat{k} \\ \hat{n} d S & = \pm \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} d θ d φ \\ = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - a \sin φ \sin θ & a \sin φ \cos θ & 0 \\ a \cos φ \cos θ & a \cos φ \sin θ & - a \sin φ \end{array}] d θ d φ \\ = \pm (- a^{2} \sin^{2} φ \cos θ \hat{ı ı} - a^{2} \sin^{2} φ \sin θ \hat{ȷ ȷ} - a^{2} \sin φ \cos φ \hat{k}) d θ d φ \\ = \mp a^{2} \sin φ (\sin φ \cos θ \hat{ı ı} + \sin φ \sin θ \hat{ȷ ȷ} + \cos φ \hat{k}) d θ d φ \end{aligned}

For the outward normal, we want the

+

sign, so

\begin{aligned} \hat{n} d S & = a^{2} \sin φ (\sin φ \cos θ \hat{ı ı} + \sin φ \sin θ \hat{ȷ ȷ} + \cos φ \hat{k}) d θ d φ \\ F \cdot \hat{n} d S & = z (θ, φ) \hat{k} \cdot \hat{n} d S = a^{3} \sin φ \cos^{2} φ d θ d φ \end{aligned}

and

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = a^{3} \int_{0}^{π} d φ \int_{0}^{2 π} d θ \sin φ \cos^{2} φ \\ = 2 π a^{3} \int_{0}^{π} d φ \sin φ \cos^{2} φ = 2 π a^{3} {[- \frac{1}{3} \cos^{3} φ]}_{0}^{π} \\ = \frac{4}{3} π a^{3} \end{aligned}

(b) Call the solid

x^{2} + y^{2} + z^{2} \leq a^{2},

V .

\nabla \nabla \cdot F = \frac{\partial}{\partial x} (0) + \frac{\partial}{\partial y} (0) + \frac{\partial}{\partial z} (z) = 1

the divergence theorem gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} d V = Volume (V) = \frac{4}{3} π a^{3} \end{aligned}

4.2.6.5.

Solution.

(a) On

D,

z = 0

and

\hat{n} = - \hat{k} d S = d x d y F \cdot \hat{n} = - y^{2}

so that

\iint_{D} F \cdot \hat{n} d S = - \iint_{D} y^{2} d x d y

Switching to polar coordinates

\begin{aligned} \iint_{D} F \cdot \hat{n} d S & = - \int_{0}^{3} d r r \int_{0}^{2 π} d θ (r \sin θ)^{2} \\ = - [\int_{0}^{3} d r r^{3}] [\int_{0}^{2 π} d θ \sin^{2} θ] \\ = - \frac{3^{4}}{4} [\int_{0}^{2 π} d θ \frac{1 - \cos (2 θ)}{2}] \\ = - \frac{81}{4} [\frac{θ}{2} - \frac{\sin (2 θ)}{4}]_{0}^{2 π} \\ = - \frac{81}{4} π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \sin^{2} θ d θ,

see Example 2.4.4.

(b) Observe that

\nabla \nabla \cdot F = x + 2 .

Since

x

is odd and

V

is invariant under

x \to - x,

we have

∭_{V} x d V = 0

(more details below) so that

∭_{V} \nabla \nabla \cdot F d V = ∭_{V} (x + 2) d V = 2 ∭_{V} d V = 2 | V |

Here are two more detailed arguments showing that

∭_{V} x d V = 0 .

Argument 1: We may rewrite the equation

z = \frac{9 - x^{2} - y^{2}}{9 + x^{2} + y^{2}}

of the curved boundary of

V

z (9 + x^{2} + y^{2}) = 9 - x^{2} - y^{2} ⟺ x^{2} + y^{2} = \frac{9 (1 - z)}{1 + z}

This is the equation of the circle of radius

r (z) = \sqrt{\frac{9 (1 - z)}{1 + z}}

centred on

x = y = 0 .

z

runs from

0

1,

and for each fixed

0 \leq z \leq 1,

y

runs from

- r (z)

r (z)

and, for each fixed

y

and

z,

x

runs from

- \sqrt{r (z)^{2} - y^{2}}

\sqrt{r (z)^{2} - y^{2}} .

\begin{matrix} ∭_{V} x d V = \int_{0}^{1} d z \int_{- r (z)}^{r (z)} d y \int_{- \sqrt{r (z)^{2} - y^{2}}}^{\sqrt{r (z)^{2} - y^{2}}} d x x = \int_{0}^{1} d z \int_{- r (z)}^{r (z)} d y 0 = 0 \end{matrix}

since

\int_{- a}^{a} x d x = 0

for any

a > 0 .

Argument 2: As we have observed above, the curved boundary of

V

x^{2} + y^{2} = \frac{9 (1 - z)}{1 + z}

which is invariant under rotations about the

z

--axis. By that symmetry, the centroid of

V

lies on the

z

-axis. Recall that, for any solid

V,

the centroid of

V

(\bar{x}, \bar{y}, \bar{z})

with

\bar{x} = \frac{∭_{V} x d V}{∭_{V} d V} \bar{y} = \frac{∭_{V} y d V}{∭_{V} d V} \bar{z} = \frac{∭_{V} z d V}{∭_{V} d V}

∭_{V} x d V = \bar{x} Volume (V) = 0 and ∭_{V} y d V = \bar{y} Volume (V) = 0

∭_{V} \nabla \nabla \cdot F d V = \iint_{\partial V} F \cdot \hat{n} d S = \iint_{S} F \cdot \hat{n} d S + \iint_{D} F \cdot \hat{n} d S

so that

\iint_{S} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V - \iint_{D} F \cdot \hat{n} d S = 2 | V | + \frac{81}{4} π

4.2.6.6.

Solution.

(a) Let

G (x, y, z) = x^{2} + y^{2} + z .

Then the surface is

G (x, y, z) = 1

and

\nabla \nabla G (x, y, z) = 2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}

is upward pointing (i.e. the coefficient of

\hat{k}

is positive) so that, by (3.3.3),

\begin{aligned} \hat{n} d S & = \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} d x d y = \frac{2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}}{1} d x d y \\ = (2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}) d x d y \\ F \cdot \hat{n} d S & = [x \hat{ı ı} + y \hat{ȷ ȷ} + \hat{k}] \cdot [2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}] d x d y \\ = [2 x^{2} + 2 y^{2} + 1] d x d y \end{aligned}

Switching to polar coordinates

\iint_{S} F \cdot \hat{n} d S = \int_{0}^{1} d r r \int_{0}^{2 π} d θ (2 r^{2} + 1) = 2 π {[\frac{2}{4} r^{4} + \frac{1}{2} r^{2}]}_{0}^{1} = 2 π

(b) Call the solid

0 \leq z \leq 1 - x^{2} - y^{2},

V .

Let

D

denote the bottom surface of

V .

The disk

D

has radius

1,

area

π,

z = 0

and outward normal

- \hat{k},

so that

\iint_{D} F \cdot \hat{n} d S = - \iint_{D} F \cdot \hat{k} d x d y = - \iint_{D} d x d y = - π

\nabla \nabla \cdot F = \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial y} (y) + \frac{\partial}{\partial z} (1) = 2

the divergence theorem gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{D} F \cdot \hat{n} d S \\ = ∭_{V} 2 d V - (- π) \\ = π + 2 ∭_{V} d V \end{aligned}

To evaluate the volume

∭_{V} d V,

we slice the

V

into thin horizontal pancakes. Here is a sketch of the pancake at height

z .

Its cross-section is a circular disk of radius

\sqrt{1 - z},

and hence of area

π (1 - z) .

As the pancake has thickness

d z,

it has volume

π (1 - z) d z .

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = π + 2 \int_{0}^{1} d z \iint_{x^{2} + y^{2} \leq 1 - z} d x d y = π + 2 \int_{0}^{1} d z π (1 - z) \\ = π + 2 π {[z - \frac{1}{2} z^{2}]}_{0}^{1} = 2 π \end{aligned}

4.2.6.7. (✳).

Solution.

(a) The divergence is

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (z + \sin y) + \frac{\partial}{\partial y} (z y) + \frac{\partial}{\partial z} (\sin x \cos y) \\ = z \end{aligned}

(b) Let

\begin{matrix} V = {(x, y, z) | x^{2} + y^{2} + z^{2} \leq 9} \end{matrix}

By the divergence theorem (note that we are to find the outward flux),

\begin{aligned} \iint_{\partial V} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} z d V = 0 \end{aligned}

since

z

is odd.

4.2.6.8.

Solution.

Call the silo

V .

Call the sides and top of the silo

S .

Call the base of the silo (namely,

x^{2} + y^{2} \leq 1,

z = 0

)

B .

By the divergence theorem,

\begin{aligned} \iint_{S} V \cdot \hat{n} d S + \iint_{B} V \cdot (- \hat{k}) d S & = ∭_{V} \nabla \nabla \cdot V d V \\ \iint_{S} V \cdot \hat{n} d S - \iint_{x^{2} + y^{2} \leq 1} (x^{2} + y) d x d y & = ∭_{V} (2 x y z + z) d V \end{aligned}

By oddness under

Y \to - y,

\iint_{x^{2} + y^{2} \leq 1} y d x d y = ∭_{V} x y z d V = 0,

\begin{aligned} \iint_{S} V \cdot \hat{n} d S & = \iint_{x^{2} + y^{2} \leq 1} x^{2} d x d y + ∭_{V} z d V \\ = \int_{0}^{1} d r \int_{0}^{2 π} d θ r (\overset{x}{\overset{⏞}{r \cos θ}})^{2} + ∭_{V} z d V \end{aligned}

We can evaluate the volume integral by decomposing

V

into thin horizontal pancakes. See Section 1.6 in the CLP-2 text. For

0 \leq z \leq 1,

the horizontal cross-section of the silo at height

z

is a circle of radius

1

and hence of area

π .

For

z \geq 1,

the horizontal cross-section of the silo at height

z

is again a circle. Its radius is determined by the equation

x^{2} + y^{2} + z^{2} = 2

of the top of the silo. The radius is

\sqrt{2 - z^{2}},

so the cross-section has area

π (2 - z^{2}) .

The biggest that

z

can get is

\sqrt{2} .

Thus

\begin{aligned} \iint_{S} V \cdot \hat{n} d S & = \int_{0}^{1} d r \int_{0}^{2 π} d θ r (r \cos θ)^{2} + \int_{0}^{1} d z π z + \int_{1}^{\sqrt{2}} d z π (2 - z^{2}) z \\ = [\int_{0}^{1} d r r^{3}] [\int_{0}^{2 π} d θ \frac{\cos (2 θ) + 1}{2}] + \int_{0}^{1} d z π z \\ + \int_{1}^{\sqrt{2}} d z π (2 - z^{2}) z \\ = {[\frac{r^{4}}{4}]}_{0}^{1} π + {[π \frac{z^{2}}{2}]}_{0}^{1} + π {[z^{2} - \frac{z^{4}}{4}]}_{1}^{\sqrt{2}} \\ = \frac{π}{4} + \frac{π}{2} + π [1 - \frac{3}{4}] \\ = π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} θ d θ,

see Example 2.4.4.

4.2.6.9.

Solution.

Apply the divergence theorem. The divergence of

F

\nabla \nabla \cdot F = \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial y} (x y) + \frac{\partial}{\partial z} (3 z - y z) = 3 + 3 x - y

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = ∭_{B} \nabla \nabla \cdot F d V = ∭_{B} (3 + 3 x - y) d V \end{matrix}

To evaluate the integrals of

x

and

y

we use that, for any solid

V

R^{3},

\begin{matrix} ∭_{V} d V = Volume (V) \end{matrix}

and

\begin{matrix} \bar{x} = \frac{∭_{V} x d V}{Volume (V)} \bar{y} = \frac{∭_{V} y d V}{Volume (V)} \bar{z} = \frac{∭_{V} z d V}{Volume (V)} \end{matrix}

where

(\bar{x}, \bar{y}, \bar{z})

is the centroid of

V .

Our ball has volume

V

and centroid

(\bar{x}, \bar{y}, \bar{z}) = (x_{0}, y_{0}, z_{0}) .

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = V [3 + 3 \bar{x} - \bar{y}] = [3 + 3 x_{0} - y_{0}] V \end{matrix}

4.2.6.10. (✳).

Solution.

Let

V = {(x, y, z) | x^{2} + y^{2} \leq 1 - z^{4}, 0 \leq z \leq 1}

Then the boundary,

\partial V,

V,

with the orientation that is used in the divergence theorem, consists of two parts

the surface $S,$ but with the upward pointing normal, and
the disk $D = {(x, y, z) | x^{2} + y^{2} \leq 1, z = 0},$ with normal $- \hat{k} .$

So the divergence theorem gives

\begin{matrix} ∭_{V} \nabla \nabla \cdot F d V = \iint_{\partial V} F \cdot \hat{n} d S = - \iint_{S} F \cdot \hat{n} d S + \iint_{D} F \cdot (- \hat{k}) d S \end{matrix}

\nabla \nabla \cdot F = 0

and

F (x, y, 0) = (1, 1, 1)

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = \iint_{D} F \cdot (- \hat{k}) d S = - \iint_{D} d S = - π \end{matrix}

4.2.6.11. (✳).

Solution.

Let

V

be the solid

x^{2} + y^{2} + 2 z^{2} \leq 2,

z \geq 0 .

The surface of

V

consists of the half-ellipsoid

S = {(x, y, z) | x^{2} + y^{2} + 2 z^{2} = 2, z \geq 0},

on top with upward pointing normal, and the disk

D = {(x, y, z)} z = 0, x^{2} + y^{2} \leq 2,

on the bottom with normal

- \hat{k} .

Call the vector field

F .

By the divergence theorem

\iint_{S} F \cdot \hat{n} d S + \iint_{D} F \cdot (- \hat{k}) d S = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} 4 d V

The ellipsoid has

a = \sqrt{2},

b = \sqrt{2},

c = 1

and volume

\frac{4}{3} π a b c = \frac{8}{3} π .

∭_{V} 4 d V = 4 \times \frac{1}{2} (V o l u m e o f t h e e l l i p s o i d) = \frac{16 π}{3}

D,

z = 0

and

\iint_{D} x d S = \iint_{D} y d S = 0

because

x

and

y

are odd. So

\iint_{D} F \cdot (- \hat{k}) d S = \iint_{D} (x \hat{ı ı} + y \hat{ȷ ȷ} + 0 \hat{k}) \cdot (- \hat{k}) d S = 0

and the desired flux is

\iint_{S} F \cdot \hat{n} d S = ∭_{V} 4 d V = \frac{16}{3} π

4.2.6.12. (✳).

Solution.

(a) If

(x, y, z) \neq 0,

\begin{aligned} \nabla \nabla \cdot F (x, y, z) = \frac{\partial}{\partial x} \frac{x}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} + \frac{\partial}{\partial y} \frac{y}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \\ + \frac{\partial}{\partial z} \frac{z}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} \\ = \frac{[x^{2} + y^{2} + z^{2}] - x \frac{3}{2} (2 x)}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} + \frac{[x^{2} + y^{2} + z^{2}] - y \frac{3}{2} (2 y)}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ + \frac{[x^{2} + y^{2} + z^{2}] - z \frac{3}{2} (2 z)}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ = \frac{3 [x^{2} + y^{2} + z^{2}] - 3 x^{2} - 3 y^{2} - 3 z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ = 0 \end{aligned}

(x, y, z) = 0,

F (x, y, z)

is not defined and hence

\nabla \nabla \cdot F (x, y, z)

is also not defined.

(b) Let

a > 0 .

Write

σ_{a} = {(x, y, z) | x^{2} + y^{2} + z^{2} = a^{2}} .

The outward unit normal to

σ_{a}

\hat{n} = \frac{r}{| r |}

so that

\begin{aligned} \int_{σ_{a}} F \cdot \hat{n} d S & = \int_{| r | = a} \frac{r}{| r |^{3}} \cdot \frac{r}{| r |} d S = \int_{| r | = a} \frac{1}{| r |^{2}} d S = \frac{1}{a^{2}} \int_{| r | = a} d S \\ = \frac{1}{a^{2}} (4 π a^{2}) = 4 π \neq 0 \end{aligned}

(c) No, the results of (a) and (b) do not contradict the divergence theorem. One hypothesis of the divergence theorem is that

\nabla \nabla \cdot F

(in fact all first order derivatives of

F

) be defined and continuous throughout the solid that

\nabla \nabla \cdot F

is to be integrated over. That hypothesis is violated in this case.

(d) Let’s first figure out what the surface

z^{2} - x^{2} - y^{2} + 1 = 0,

i.e. the surface

x^{2} + y^{2} = 1 + z^{2},

looks like. For each

z_{0},

the

z = z_{0}

cross-section of this surface is the circle

x^{2} + y^{2} = 1 + z_{0}^{2} .

The radius of this circle is

1

when

z_{0} = 0

and grows as

| z_{0} |

increases. So the solid region

E

looks like an hourglass drum, as sketched in the figure on the left below.

We are going to use the divergence theorem to compute the flux of

F

out through the surface

σ

E .

However we cannot apply the divergence theorem using

E

as the solid, because

F

is not defined at the origin,

(0, 0, 0),

which is a point in

E .

So we pick any

0 < a < 1,

and define the auxiliary solid

E_{a} = {(x, y, z) | x^{2} + y^{2} + z^{2} \geq a^{2}, x^{2} + y^{2} \leq 1 + z^{2}, - 1 \leq z \leq 1}

The solid

E_{a}

is constructed from the solid

E

by removing the ball

x^{2} + y^{2} + z^{2} \leq a^{2}

from it. A side view of

E_{a}

is sketched in the figure on the right above. As in part (b), denote by

σ_{a}

the surface

x^{2} + y^{2} + z^{2} = a^{2}

with outward pointing normal. Then the boundary of

E_{a}

\partial E_{a} = σ - σ_{a},

meaning that it consists of two parts. One part is the boundary,

σ,

E,

with outward pointing normal. The other part is the surface

x^{2} + y^{2} + z^{2} = a^{2},

but with normal pointing into the sphere, opposite to the normals for

σ_{a} .

Consequently the divergence theorem gives

\begin{matrix} 0 = ∭_{E_{a}} \nabla \nabla \cdot F d V = \iint_{\partial E_{a}} F \cdot \hat{n} d S = \iint_{σ} F \cdot \hat{n} d S - \iint_{σ_{a}} F \cdot \hat{n} d S \end{matrix}

so that, by part (b)

\begin{matrix} \iint_{σ} F \cdot \hat{n} d S = \iint_{σ_{a}} F \cdot \hat{n} d S = 4 π \end{matrix}

(e) The equation

z^{2} - x^{2} - y^{2} + 4 y - 3 = 0

can be rewritten as

x^{2} + (y - 2)^{2} = 1 + z^{2}

As is part (d), for each

z_{0},

the

z = z_{0}

cross-section of this surface is a circle

x^{2} + (y - 2)^{2} = 1 + z_{0}^{2}

of radius

\sqrt{1 + z_{0}^{2}} .

But this circle is centred at

(0, 2, z_{0}),

whereas the corresponding circle in part (d) was centred at

(0, 0, z_{0}) .

The solid

R

again has the shape of an hourglass drum. But while the origin

(0, 0, 0)

was in

E,

it is not in

R = {(x, y, z) | x^{2} + (y - 2)^{2} \leq 1 + z^{2}, - 1 \leq z \leq 1}

\nabla \nabla \cdot F = 0

throughout all of

R

and the divergence theorem gives

\begin{matrix} \iint_{Σ} F \cdot \hat{n} d S = \iint_{\partial R} F \cdot \hat{n} d S = ∭_{R} \nabla \nabla \cdot F d V = 0 \end{matrix}

4.2.6.13. (✳).

Solution.

(a) If the surface were the sphere

x^{2} + y^{2} + z^{2} = 1,

we could parametrize it using the spherical coordinates

θ

and

φ

(with the radial spherical coordinate

ρ = 1

\begin{aligned} x & = \sin φ \cos θ \\ y & = \sin φ \sin θ \\ z & = \cos φ \end{aligned}

with

0 \leq θ < 2 π,

0 \leq φ \leq π .

Our surface is not a sphere, but the equation looks like the equation of the sphere with the units of the

y

- and

z

-coordinates changed. In particular, if we define

\tilde{y} = y / 2

and

\tilde{z} = z / 2,

so that

y = 2 \tilde{y}

and

z = 2 \tilde{z},

then on our surface

\begin{matrix} 1 = x^{2} + \frac{y^{2}}{4} + \frac{z^{2}}{4} = x^{2} + \frac{(2 \tilde{y})^{2}}{4} + \frac{(2 \tilde{z})^{2}}{4} = x^{2} + {\tilde{y}}^{2} + {\tilde{z}}^{2} \end{matrix}

and we can parametrize

\begin{aligned} x & = \sin φ \cos θ \\ \tilde{y} & = \sin φ \sin θ \\ \tilde{z} & = \cos φ \end{aligned}

and then

\begin{aligned} x & = \sin φ \cos θ \\ y & = 2 \tilde{y} = 2 \sin φ \sin θ \\ z & = 2 \tilde{z} = 2 \cos φ \end{aligned}

\begin{aligned} r (θ, φ) = \sin φ \cos θ \hat{ı ı} + 2 \sin φ \sin θ \hat{ȷ ȷ} + 2 \cos φ \hat{k} \\ 0 \leq θ < 2 π, 0 \leq φ \leq π \end{aligned}

(b) Considering part (c) in this question, we are presumably to evaluate the flux integral directly. Since

\begin{aligned} (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) & = (- \sin φ \sin θ, 2 \sin φ \cos θ, 0) \\ (\frac{\partial x}{\partial φ}, \frac{\partial y}{\partial φ}, \frac{\partial z}{\partial φ}) & = (\cos φ \cos θ, 2 \cos φ \sin θ, - 2 \sin φ) \end{aligned}

(3.3.1) yields

\begin{aligned} \hat{n} d S & = \pm (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) \times (\frac{\partial x}{\partial φ}, \frac{\partial y}{\partial φ}, \frac{\partial z}{\partial φ}) d θ d φ \\ = \pm (- \sin φ \sin θ, 2 \sin φ \cos θ, 0) \\ \times (\cos φ \cos θ, 2 \cos φ \sin θ, - 2 \sin φ) d θ d φ \\ = \pm (- 4 \sin^{2} φ \cos θ, - 2 \sin^{2} φ \sin θ, - 2 \sin φ \cos φ) d θ d φ \\ = \mp 2 \sin φ (2 \sin φ \cos θ, \sin φ \sin θ, \cos φ) d θ d φ \end{aligned}

To get an outward pointing normal we need the

+

sign. For example, with the

+

sign, the

z

-component is

2 \sin φ \cos φ = \sin (2 φ)

so that the normal is pointing upward when

0 < φ < \frac{π}{2},

i.e. in the northern hemisphere, and is pointing downward when

\frac{π}{2} < φ < π,

i.e. in the southern hemisphere. So

\begin{aligned} F \cdot \hat{n} d S & = {(\sin φ \cos θ) (4 \sin^{2} φ \cos θ) + (2 \sin φ \sin θ) (2 \sin^{2} φ \sin θ) \\ + (2 \cos φ) (2 \sin φ \cos φ)} d θ d φ \\ = {4 \sin^{3} φ \cos^{2} θ + 4 \sin^{3} φ \sin^{2} θ + 4 \sin φ \cos^{2} φ} d θ d φ \\ = 4 \sin φ (\sin^{2} φ + \cos^{2} φ) d θ d φ \\ = 4 \sin φ d θ d φ \end{aligned}

and the flux is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{0}^{π} d φ \int_{0}^{2 π} d θ 4 \sin φ = 8 π \int_{0}^{π} d φ \sin φ = 16 π \end{aligned}

V = {(x, y, z) | x^{2} + \frac{y^{2}}{4} + \frac{z^{2}}{4} \leq 1}

Since

\nabla \nabla \cdot F = 3,

the divergence theorem gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V = 3 Volume (V) \end{aligned}

The volume contained in the ellipsoid,

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1,

of semiaxes

a,

b

and

c

\frac{4}{3} π a b c .

In our case

a = 1,

b = c = 2,

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = 3 Volume (V) = 3 \times \frac{4}{3} π (1) (2) (2) = 16 π \end{aligned}

which is exactly what we found in part (b).

The volume of the ellipsoid

V

can also be found by observing that, in

V,

$x$ runs from $- 1$ to $1$ and
for each fixed $- 1 \leq x \leq 1,$ $(y, z)$ runs over the disk $y^{2} + z^{2} \leq 4 (1 - x^{2}),$ which has area $4 π (1 - x^{2}) .$

That is

V = {(x, y, z) | - 1 \leq x \leq 1, y^{2} + z^{2} \leq 4 (1 - x^{2})}

so that

\begin{aligned} Volume (V) & = \int_{- 1}^{1} d x \iint_{y^{2} + z^{2} \leq 4 (1 - x^{2})} d y d z \\ = \int_{- 1}^{1} d x 4 π (1 - x^{2}) = 2 \times 4 π \int_{0}^{1} d x (1 - x^{2}) = 8 π [1 - \frac{1}{3}] \\ = \frac{16 π}{3} \end{aligned}

4.2.6.14. (✳).

Solution.

Set

V = {(x, y, z) | x^{2} + y^{2} \leq 2, 0 \leq z \leq 2 x + 3}

Let’s try the divergence theorem. Since

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x^{3} + \cos (y^{2})) + \frac{\partial}{\partial y} (y^{3} + z e^{x}) + \frac{\partial}{\partial z} (z^{2} + \arctan (x y)) \\ = 3 x^{2} + 3 y^{2} + 2 z \end{aligned}

the divergence theorem (Theorem 4.2.2) gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V \\ = \int_{x^{2} + y^{2} \leq 2} d x d y \int_{0}^{2 x + 3} d z (3 x^{2} + 3 y^{2} + 2 z) \\ = \int_{x^{2} + y^{2} \leq 2} d x d y {3 (x^{2} + y^{2}) (2 x + 3) + (2 x + 3)^{2}} \\ = \int_{x^{2} + y^{2} \leq 2} d x d y {9 + 12 x + 13 x^{2} + 9 y^{2} + 6 x^{3} + 6 x y^{2}} \\ = 9 (2 π) + \int_{x^{2} + y^{2} \leq 2} d x d y {13 x^{2} + 9 y^{2}} \end{aligned}

because

12 x,

6 x^{3}

and

6 x y^{2}

are all odd under

x \to - x .

To evaluate the final remaining integral, let’s switch to polar coordinates.

\begin{aligned} \iint_{x^{2} + y^{2} \leq 2} {13 x^{2} + 9 y^{2}} d x d y & = \int_{0}^{\sqrt{2}} d r r \int_{0}^{2 π} d θ {13 (r \cos θ)^{2} + 9 (r \sin θ)^{2}} \\ = \int_{0}^{\sqrt{2}} d r r^{3} \int_{0}^{2 π} d θ {13 \cos^{2} θ + 9 \sin^{2} θ} \end{aligned}

Since

\begin{aligned} \int_{0}^{2 π} \cos^{2} θ d θ & = \int_{0}^{2 π} \frac{\cos (2 θ) + 1}{2} d θ = {[\frac{\sin (2 θ)}{4} + \frac{θ}{2}]}_{0}^{2 π} = π \\ \int_{0}^{2 π} \sin^{2} θ d θ & = \int_{0}^{2 π} \frac{1 - \cos (2 θ)}{2} d θ = {[\frac{θ}{2} - \frac{\sin (2 θ)}{4}]}_{0}^{2 π} = π \end{aligned}

we finally have

\iint_{S} F \cdot \hat{n} d S = 18 π + \frac{(\sqrt{2})^{4}}{4} π {13 + 9} = (18 + 22) π = 40 π

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} θ d θ

and

\int_{0}^{2 π} \sin^{2} θ d θ,

see Example 2.4.4.

4.2.6.15. (✳).

Solution 1. By divergence theorem

Set

F = (x + y, x + z, y + z) .

Then

\nabla \nabla \cdot F = 2 .

That’s really simple. So let’s try using the divergence theorem.

Set $S = {(x, y, z) | x^{2} + z^{2} = 4, 0 \leq y \leq 3} .$ We are to compute $\iint_{S} F \cdot \hat{n} d S,$ with $\hat{n}$ denoting the outward normal to $S .$ $S$ is not the boundary of a solid, so we cannot compute $\iint_{S} F \cdot \hat{n} d S$ by applying the divergence theorem directly. The figure on the left below shows the part of $S$ that is in the first octant.
On the other hand $S,$ is “almost” the boundary of
$V = {(x, y, z) | x^{2} + z^{2} \leq 4, 0 \leq y \leq 3}$
The boundary, $\partial V$ of $V$ consists of three pieces — $S$ and the two disks
$\begin{aligned} D_{l} & = {(x, y, z) | x^{2} + z^{2} \leq 4, y = 0} \\ D_{r} & = {(x, y, z) | x^{2} + z^{2} \leq 4, y = 3} \end{aligned}$
The figure on the right above shows the parts of $S,$ $V,$ $D_{l}$ and $D_{r}$ that are in the first octant.

The outward normal to

D_{r}

\hat{ȷ ȷ}

and the outward normal to

D_{l}

- \hat{ȷ ȷ},

to the divergence theorem gives

\begin{aligned} ∭_{V} \nabla \nabla \cdot F d V & = \iint_{\partial V} F \cdot \hat{n} d S \\ = \iint_{S} F \cdot \hat{n} d S + \iint_{D_{r}} F \cdot \hat{ȷ ȷ} d S + \iint_{D_{l}} F \cdot (- \hat{ȷ ȷ}) d S \end{aligned}

Since

\nabla \nabla \cdot F = 2

and

F \cdot \hat{ȷ ȷ} = x + z,

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} 2 d V - \iint_{x^{2} + z^{2} \leq 4} (x + z) d x d z - \iint_{x^{2} + z^{2} \leq 4} (- x - z) d x d z \\ = ∭_{V} 2 d V \\ = 2 volume (V) = 2 (π 2^{2}) 3 = 24 π \end{aligned}

Solution 2. By direct evaluation

Let’s parametrize the surface by

\begin{aligned} r (θ, y) & = 2 \cos θ \hat{ı ı} + y \hat{ȷ ȷ} + 2 \sin θ \hat{k} 0 \leq θ < 2 π, 0 \leq y \leq 3 \end{aligned}

Then

\begin{aligned} \frac{\partial r}{\partial θ} & = (- 2 \sin θ, 0, 2 \cos θ) \\ \frac{\partial r}{\partial y} & = (0, 1, 0) \\ \hat{n} d S & = \pm \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial y} d θ d y = \pm (- 2 \cos θ, 0, - 2 \sin θ) d θ d y \end{aligned}

To get the outward normal, we want the minus sign. So

\hat{n} d S = (2 \cos θ, 0, 2 \sin θ) d θ d y

and, since

F (r (θ, y)) = (2 \cos θ + y, 2 \cos θ + 2 \sin θ, y + 2 \sin θ)

the specified flux is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{0}^{2 π} d θ \int_{0}^{3} d y (2 \cos θ + y, 2 \cos θ + 2 \sin θ, y + 2 \sin θ) \\ \cdot (2 \cos θ, 0, 2 \sin θ) \\ = \int_{0}^{2 π} d θ \int_{0}^{3} d y (4 \cos^{2} θ + 2 y \cos θ + 2 y \sin θ + 4 \sin^{2} θ) \\ = \int_{0}^{2 π} d θ \int_{0}^{3} d y (4 + 2 y \cos θ + 2 y \sin θ) \end{aligned}

Since

\int_{0}^{2 π} d θ \cos θ = \int_{0}^{2 π} d θ \sin θ = 0,

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = 4 \int_{0}^{2 π} d θ \int_{0}^{3} d y = 4 (2 π) 3 = 24 π \end{aligned}

4.2.6.16. (✳).

Solution.

The question highlights that the vector field has divergence

0 .

That strongly suggests that we use the divergence theorem. Set

V = {(x, y, z) | 0 \leq z \leq 1 - (x^{2} + y^{2})^{2}}

Then the boundary,

\partial V,

V

consists of two parts, namely

S

(with normal pointing upwards) and the disk

D = {(x, y, 0) | x^{2} + y^{2} \leq 1}

(with normal pointing downwards). The divergence theorem (Theorem 4.2.2) gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{D} F \cdot (- \hat{k}) d S \\ = \iint_{D} (x^{2} + y^{2}) d x d y \end{aligned}

Switching to polar coordinates, the flux is

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = \int_{0}^{1} d r r \int_{0}^{2 π} d θ r^{2} = 2 π \int_{0}^{1} d r r^{3} = 2 π \frac{1}{4} = \frac{π}{2} \end{matrix}

4.2.6.17. (✳).

Solution.

F

looks complicated, we will probably want to avoid evaluating the flux integral directly. Let’s first compute the divergence of

F,

to see if it looks wise to use the divergence theorem instead.

\begin{matrix} \nabla \nabla \cdot F = \frac{\partial}{\partial x} (\tan \sqrt{z} + \sin (y^{3})) + \frac{\partial}{\partial y} (e^{- x^{2}}) + \frac{\partial}{\partial z} (z) = 1 \end{matrix}

Looks good! We cannot yet apply the divergence theorem, since

S

is not the boundary of a solid region

V .

To help us choose a solid

V

whose boundary at least includes

S,

here is a sketch.

S

is the top of the “ice cream cone”

Note that the paraboloid

z = 2 - x^{2} - y^{2}

and the cone

z = \sqrt{x^{2} + y^{2}}

intersect along the circle

x^{2} + y^{2} = 1,

z = 1 .

Probably the simplest solid whose boundary includes

S

V = {(x, y, z) | 1 \leq z \leq 2 - x^{2} - y^{2}, x^{2} + y^{2} \leq 1}

The boundary

\partial V

V

consists of

S

(with upward pointing normal) and the disk

D = {(x, y, z) | x^{2} + y^{2} \leq 1, z = 1}

with normal

- \hat{k} .

So the divergence theorem gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{D} F \cdot (- \hat{k}) d S \\ = ∭_{V} \overset{\nabla \nabla \cdot F}{\overset{⏞}{1}} d V + \iint_{D} \overset{F \cdot \hat{k} = z}{\overset{⏞}{1}} d S \end{aligned}

D

is a disk of radius 1,

\iint_{D} d S = π .

To compute the volume of

V,

we’ll slice it into a stack of horizontal “pancakes”. Since

z = 2 - x^{2} - y^{2}

is equivalent to

\sqrt{x^{2} + y^{2}} = \sqrt{2 - z},

the pancake at height

z

is a circular disk of radius

\sqrt{2 - z}

and hence of cross-sectional area

π (2 - z) .

So the volume of

V

\begin{aligned} ∭_{V} d V & = \int_{1}^{2} π (2 - z) d z = - \frac{π}{2} (2 - z)^{2} |_{1}^{2} = \frac{π}{2} \end{aligned}

and the flux

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = \frac{π}{2} + π = \frac{3}{2} π \end{matrix}

4.2.6.18. (✳).

Solution.

F

looks complicated, we will probably want to avoid evaluating the flux integral directly. Let’s first compute the divergence of

F,

to see if it looks wise to use the divergence theorem instead.

\begin{matrix} \nabla \nabla \cdot F = \frac{\partial}{\partial x} (\cos z + x y^{2}) + \frac{\partial}{\partial y} (x e^{- z}) + \frac{\partial}{\partial z} (\sin y + x^{2} z) = y^{2} + x^{2} \end{matrix}

Looks promising. Furthermore

S

is the boundary of the solid region

V = {(x, y, z) | x^{2} + y^{2} \leq z \leq 4}

So the divergence theorem gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} (x^{2} + y^{2}) d V \end{aligned}

To compute the triple integral, we’ll use the cylindrical coordinates

(r, θ, z) .

The

z

-coordinate runs from

0

4 .

For each fixed

0 \leq z \leq 4

(see the blue disk in the figure below — which shows the part of

V

in the first octant),

(x, y)

runs over

0 \leq x^{2} + y^{2} \leq z,

which in cylindrical coordinates is

0 \leq r^{2} \leq z

0 \leq r \leq \sqrt{z} .

So the flux and the triple integral are

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} (x^{2} + y^{2}) d V \\ = \int_{0}^{4} d z \int_{0}^{\sqrt{z}} d r r \int_{0}^{2 π} d θ r^{2} \\ = 2 π \int_{0}^{4} d z \int_{0}^{\sqrt{z}} d r r^{3} \\ = 2 π \int_{0}^{4} d z \frac{z^{2}}{4} = 2 π \frac{4^{3}}{3 \times 4} \\ = \frac{32}{3} π \end{aligned}

4.2.6.19. (✳).

Solution.

If we were to evaluate this integral directly using, for example, spherical coordinates, our integrand would contain

\sin (x) = \sin (2 \sin φ \cos θ)

That’s not very friendly looking. So let’s consider using the divergence theorem instead. To start,

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (e^{y} + x z) + \frac{\partial}{\partial y} (z y + \sin (x)) + \frac{\partial}{\partial z} (z^{2} - 1) = 4 z \end{aligned}

That’s nice and simple. So let’s move on to consideration of

S .

The part of

S

in the first octant is outlined in red in the figure on the left below.

The surface

S

is not closed, and so is not the boundary of a solid, so we cannot apply the divergence theorem directly. But we can easily come up with a solid whose boundary contains

S .

Let

V = {(x, y, z) | x^{2} + y^{2} + z^{2} \leq 4, 0 \leq z \leq 1}

The boundary

\partial V

V

consists of three parts —

S,

the bottom disk

D_{b} = {(x, y, z) | x^{2} + y^{2} \leq 4, z = 0}

and the top disk

D_{t} = {(x, y, z) | x^{2} + y^{2} \leq 3, z = 1}

The outward normal to

D_{t}

\hat{k}

and the outward normal to

D_{b}

- \hat{k} .

So the divergence theorem gives

\begin{aligned} ∭_{V} \nabla \nabla \cdot F d V & = \iint_{\partial V} F \cdot \hat{n} d S \\ = \iint_{S} F \cdot \hat{n} d S + \iint_{D_{t}} F \cdot \hat{k} d S + \iint_{D_{b}} F \cdot (- \hat{k}) d S \end{aligned}

D_{b},

z = 0

so that

F \cdot (- \hat{k}) = - (0^{2} - 1) = 1

and on

D_{t},

z = 1

so that

F \cdot \hat{k} = 1^{2} - 1 = 0 .

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = ∭_{V} \overset{\nabla \nabla \cdot F}{\overset{⏞}{4 z}} d V - \iint_{D_{b}} d S \end{matrix}

The constant

z

cross-section of

V

is a disk of radius

\sqrt{4 - z^{2}}

and hence of area

π (4 - z^{2})

and

D_{b}

is a disk of radius

2

and hence of area

4 π .

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{0}^{1} (4 z) π (4 - z^{2}) d z - 4 π = 4 π [2 z^{2} - \frac{z^{4}}{4}]_{0}^{1} - 4 π = 3 π \end{aligned}

4.2.6.20. (✳).

Solution.

The divergence of

F,

namely,

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x^{2} z + \cos π y) + \frac{\partial}{\partial y} (y z + \sin π z) + \frac{\partial}{\partial z} (x - y^{2}) \\ = 2 x z + z \end{aligned}

is a lot simpler than

F

itself. So let’s use the divergence theorem (Theorem 4.2.2).

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = ∭_{B} \nabla \nabla \cdot F d V = ∭_{B} (2 x z + z) d V \end{matrix}

B

is invariant under

x \to - x

while

2 x z

is odd under

x \to - x,

the integral

∭_{B} 2 x z d V

is zero. To help set up the limits of integration for

∭_{B} z d V,

note that, in

B,

$(x, y)$ runs over the rectangle $- 1 \leq x \leq 1,$ $0 \leq y \leq 2$ and
for each fixed $(x, y),$ $z$ runs over $0 \leq z \leq 3 - y .$

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{- 1}^{1} d x \int_{0}^{2} d y \int_{0}^{3 - y} d z z \\ = \frac{1}{2} \int_{- 1}^{1} d x \int_{0}^{2} d y (3 - y)^{2} \\ = - \frac{1}{2} \int_{- 1}^{1} d x \int_{3}^{1} d u u^{2} with u = 3 - y, d u = - d y \\ = - \frac{1}{2} \int_{- 1}^{1} d x [\frac{1^{3}}{3} - \frac{3^{3}}{3}]_{- 1}^{1} \\ = \frac{26}{3} \end{aligned}

4.2.6.21. (✳).

Solution.

The vector field

F

looks very complicated. That strongly suggests that we not evaluate the integral directly. So let’s start by computing

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x + \cos (z^{2})) + \frac{\partial}{\partial y} (y + \ln (x^{2} + z^{5})) + \frac{\partial}{\partial z} (\sqrt{x^{2} + y^{2}}) \\ = 2 \end{aligned}

That’s really simple, which suggest that we use the divergence theorem. But the surface

S

is not closed, and so is not the boundary of a solid. So we cannot apply the divergence theorem directly. But we can easily come up with a solid whose boundary contains

S .

Let

\begin{aligned} V & = {(x, y, z) | 0 \leq z \leq \sqrt{1 - x^{2} - y^{2}}, x^{2} + y^{2} \leq 1} \end{aligned}

Then the boundary,

\partial V,

V

consists of two parts, namely

S

(with normal pointing upwards) and the disk

D = {(x, y, 0) | x^{2} + y^{2} \leq 1}

(with normal

- \hat{k}

). The divergence theorem (Theorem 4.2.2) gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{D} F \cdot (- \hat{k}) d S \\ = ∭_{V} 2 d V + \iint_{D} \sqrt{x^{2} + y^{2}} d x d y \\ = 2 \frac{1}{2} \frac{4}{3} π 1^{3} + \iint_{D} \sqrt{x^{2} + y^{2}} d x d y \end{aligned}

Switching to polar coordinates, the flux is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \frac{4}{3} π + \int_{0}^{1} d r r \int_{0}^{2 π} d θ r = \frac{4}{3} π + 2 π \int_{0}^{1} d r r^{2} \\ = \frac{4}{3} π + 2 π \frac{1}{3} = 2 π \end{aligned}

4.2.6.22.

Solution.

(a) By the divergence theorem (Theorem 4.2.2), the outward flux of

F

through the boundary of

E

\begin{aligned} \iint_{\partial E} F \cdot \hat{n} d S & = ∭_{E} \nabla \nabla \cdot F d V \\ = ∭_{E} (- x^{2} - y^{2} + 4) d V \end{aligned}

To evaluate this integral we switch to cylindrical coordinates. In cylindrical coordinates

E = {(r \cos θ, r \sin θ, z) | 0 \leq z \leq 4, r^{2} \leq z}

\begin{aligned} \iint_{\partial E} F \cdot \hat{n} d S & = \int_{0}^{4} d z \int_{0}^{\sqrt{z}} d r r \int_{0}^{2 π} d θ (- r^{2} + 4) \\ = 2 π \int_{0}^{4} d z \int_{0}^{\sqrt{z}} d r (4 r - r^{3}) \\ = 2 π \int_{0}^{4} d z (2 z - \frac{z^{2}}{4}) \\ = 2 π [z^{2} - \frac{z^{3}}{12}]_{0}^{4} = 2 π [16 - \frac{16}{3}] = \frac{64}{3} π \end{aligned}

(b) The boundary of

E

consists of two parts —

S,

but with downward pointing normal

\hat{n} = - \hat{N},

on the bottom and the disk

D = {(x, y, z) | z = 4, x^{2} + y^{2} \leq 4}

with normal

\hat{k},

on top.

So, by part (a),

\begin{aligned} \frac{64}{3} π = \iint_{\partial E} F \cdot \hat{n} d S & = - \iint_{S} F \cdot \hat{N} d S + \iint_{D} F \cdot \hat{k} d S \\ = - \iint_{S} F \cdot \hat{N} d S + \iint_{D} \overset{F \cdot \hat{k}}{\overset{⏞}{4 z}} d S \end{aligned}

Since

z = 4

D,

and

D

is a disk of radius 2,

\begin{matrix} \iint_{S} F \cdot \hat{N} d S = - \frac{64}{3} π + 16 \iint_{D} d S = - \frac{64}{3} π + 16 (4 π) = \frac{128}{3} π \end{matrix}

4.2.6.23. (✳).

Solution.

(a) Since

\begin{aligned} \frac{\partial}{\partial x} \frac{x}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} & = \frac{1}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} - \frac{3}{2} \frac{x (2 x)}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ = \frac{- 2 x^{2} + y^{2} + z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ \frac{\partial}{\partial y} \frac{y}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} & = \frac{1}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} - \frac{3}{2} \frac{y (2 y)}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ = \frac{x^{2} - 2 y^{2} + z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ \frac{\partial}{\partial z} \frac{z}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} & = \frac{1}{{[x^{2} + y^{2} + z^{2}]}^{3 / 2}} - \frac{3}{2} \frac{z (2 z)}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \\ = \frac{x^{2} + y^{2} - 2 z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} \end{aligned}

the specified divergence is

\begin{aligned} \nabla \nabla \cdot F & = \frac{(- 2 x^{2} + y^{2} + z^{2}) + (x^{2} - 2 y^{2} + z^{2}) + (x^{2} + y^{2} - 2 z^{2})}{{[x^{2} + y^{2} + z^{2}]}^{5 / 2}} = 0 \end{aligned}

(x, y, z) \neq 0

and is not defined if

(x, y, z) = 0 .

(b), (c) Set

\begin{aligned} V_{1} & = {(x, y, z) | x^{2} + (y - 2)^{2} + z^{2} \leq 9} \\ V_{2} & = {(x, y, z) | x^{2} + (y - 2)^{2} + z^{2} \leq 1} \end{aligned}

Here are side views of both

V_{1}

and

V_{2} .

Both

V_{1}

and

V_{2}

are spherical balls centred on

(0, 2, 0) .

The difference between them is that

V_{1}

has radius

3

while

V_{2}

has radius

1 .

In particular

(0, 0, 0)

is not in

V_{2} .

\nabla \nabla \cdot F

is well-defined and zero throughout

V_{2}

and, by the divergence theorem (Theorem 4.2.2),

\begin{matrix} \iint_{S_{2}} F \cdot \hat{n} d S = ∭_{V_{2}} \nabla \nabla \cdot F d V = 0 \end{matrix}

On the other hand,

(0, 0, 0)

is in

V_{1} .

We cannot blindly apply the divergence theorem to

V_{1}

—

\nabla \nabla \cdot F (x, y, z)

is not defined at the point

(x, y, z) = (0, 0, 0)

V_{1} .

We can work around this obstruction by

choosing a number $ρ > 0$ that is small enough that the sphere
$S_{ρ} = {(x, y, z) | x^{2} + y^{2} + z^{2} = ρ^{2}}$
is completely contained inside $V_{1}$ (for example, $ρ = \frac{1}{2}$ is fine)
and then removing the interior of $S_{ρ}$ from $V_{1} .$

This produces

V_{3} = {(x, y) | x^{2} + (y - 2)^{2} + z^{2} \leq 9, x^{2} + y^{2} + z^{2} \geq ρ^{2}}

whose side view is sketched below.

The boundary of

V_{3}

consists of two parts

the sphere $S_{1},$ with outward normal and
the sphere $S_{ρ}$ with inward normal $\hat{n} = - \frac{r}{| r |}$

The divergence

\nabla \nabla \cdot F

is well-defined and zero throughout

V_{3}

so that, by the divergence theorem,

\begin{matrix} 0 = ∭_{V_{3}} \nabla \nabla \cdot F d V = \iint_{S_{1}} F \cdot \hat{n} d S + \iint_{S_{ρ}} F \cdot (- \frac{r}{| r |}) d S \end{matrix}

\begin{aligned} \iint_{S_{1}} F \cdot \hat{n} d S & = \iint_{S_{ρ}} F \cdot (\frac{r}{| r |}) d S = \iint_{S_{ρ}} (\frac{r}{| r |^{3}}) \cdot (\frac{r}{| r |}) d S \\ = \iint_{S_{ρ}} \frac{1}{| r |^{2}} d S = \iint_{S_{ρ}} \frac{1}{ρ^{2}} d S \\ = \frac{1}{ρ^{2}} 4 π ρ^{2} = 4 π \end{aligned}

since

S_{ρ}

is a sphere of radius

ρ

and hence of surface area

4 π ρ^{2} .

(d) The flux integrals

\iint_{S_{1}} F \cdot \hat{n} d S

and

\iint_{S_{2}} F \cdot \hat{n} d S

are different, because the one point,

(0, 0, 0),

where

\nabla \nabla \cdot F

fails to be well-defined and zero, is contained inside

S_{1}

but is not contained inside

S_{2} .

4.2.6.24. (✳).

Solution.

The vector field

F

looks pretty complicated. But its divergence

\nabla \nabla \cdot F = 2 + 3 + 1 = 6

is very simple. So let’s use the divergence theorem (Theorem 4.2.9). It says

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{E} \nabla \nabla \cdot F d V = ∭_{E} 6 d V = 6 Volume (E) \end{aligned}

For any fixed

0 \leq X \leq 2,

the cross-section of

E

with

x = X

has side view

That cross-section has area

2 \times \frac{2 + 4}{2} = 6 .

Consequently the volume of

E

2 \times 6 = 12

and

\iint_{S} F \cdot \hat{n} d S = 6 \times 12 = 72

4.2.6.25. (✳).

Solution.

(a) The divergence is

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (z \arctan (y^{2})) + \frac{\partial}{\partial y} (z^{3} \ln (x^{2} + 1)) + \frac{\partial}{\partial z} (3 z) = 3 \end{aligned}

(b) The complexity of

F

and the simplicity of

\nabla \nabla \cdot F

strongly suggest that we use the divergence theorem to evaluate

\iint_{S} F \cdot \hat{n} d S .

However,

S

is not a closed surface and is not the boundary of a solid. The figure on the left below is a sketch of the part of

S

in the first octant.

On the other hand

S

is part of the surface of the solid

V = {(x, y, z) | x^{2} + y^{2} + z^{2} \leq 4, z \geq 1}

which is sketched on the right above. The boundary of

V

consists of two parts:

the original surface $S,$ but with upward, rather than downward, normal and
the disk $D = {(x, y, z) | x^{2} + y^{2} \leq 3, z = 1}$ with normal $- \hat{k} .$

So the divergence theorem (Theorem 4.2.9) gives

\begin{aligned} \iint_{\partial V} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V = 3 ∭_{V} d V \\ ⟹ - \iint_{S} F \cdot \hat{n} d S + \iint_{D} F \cdot (- \hat{k}) d S & = 3 Volume (V) \end{aligned}

Thus

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = - 3 Volume (V) + \iint_{D} \overset{- F \cdot \hat{k}}{\overset{⏞}{- 3}} d S \\ = - 3 Volume (V) - 3 Area (D) \\ = - 3 Volume (V) - 9 π \end{aligned}

since

D

is a circular disk of radius

\sqrt{3} .

To compute the volume of

V,

we slice

V

into thin horizontal pancakes each of thinkness

d z .

The pancake at height

z

has cross-section the circular disk

x^{2} + y^{2} \leq 4 - z^{2} .

As this disk has area

π (4 - z^{2}),

the pancake has volume

π (4 - z^{2}) d z .

All together

\begin{aligned} Volume (V) & = \int_{1}^{2} d z π (4 - z^{2}) = π {[4 z - \frac{z^{3}}{3}]}_{1}^{2} = π [4 - \frac{7}{3}] = \frac{5 π}{3} \end{aligned}

and

\iint_{S} F \cdot \hat{n} d S = - 3 \frac{5 π}{3} - 9 π = - 14 π

4.2.6.26. (✳).

Solution.

Let’s try the divergence theorem. Set

V = {(x, y, z) | x^{2} + y^{2} + z^{2} \leq 3}

Then the boundary of

V

S,

but with outward pointing normal. Since

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x y^{2} + y^{4} z^{6}) + \frac{\partial}{\partial y} (y z^{2} + x^{4} z) + \frac{\partial}{\partial z} (z x^{2} + x y^{4}) \\ = y^{2} + z^{2} + x^{2} \end{aligned}

and because

S

is oriented inward, the divergence theorem (Theorem 4.2.2) gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = - ∭_{V} \nabla \nabla \cdot F d V = - ∭_{V} (x^{2} + y^{2} + z^{2}) d V \end{aligned}

Switching to spherical coordinates (see Appendix A.6.3)

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = - \int_{0}^{\sqrt{3}} d ρ \int_{0}^{π} d φ \int_{0}^{2 π} d θ ρ^{4} \sin φ \\ = - 2 π [\int_{0}^{\sqrt{3}} d ρ ρ^{4}] [\int_{0}^{π} d φ \sin φ] \\ = - 2 π {[\frac{ρ^{5}}{5}]}_{0}^{\sqrt{3}} [- \cos φ]_{0}^{π} \\ = - \frac{36 \sqrt{3}}{5} π \end{aligned}

4.2.6.27. (✳).

Solution.

(a) The divergence of

F

\begin{aligned} \nabla \cdot F & = \frac{\partial}{\partial x} (- 2 x y) + \frac{\partial}{\partial y} (y^{2} + \sin (x z)) + \frac{\partial}{\partial z} (x^{2} + y^{2}) \\ = - 2 y + 2 y + 0 = 0 \end{aligned}

(b) Call the specified surface

S

and set

V = {(x, y, z) | x^{2} + y^{2} + (z - 12)^{2} \leq 13^{2}, z \geq 0}

The boundary,

\partial V,

V

consists of two parts —

S,

with outward normal, and the disk

D = {(x, y, z) | x^{2} + y^{2} \leq 13^{2} - 12^{2} = 5^{2}, z = 0}

with normal

- \hat{k} .

By the divergence theorem, the desired flux is

\begin{aligned} \iint_{S} F \cdot \hat{n} d s & = ∭_{V} \nabla \cdot F d V - \iint_{D} F \cdot (- \hat{k}) d S \\ = ∭_{V} 0 d V + \iint_{D} (x^{2} + y^{2}) d x d y \\ = 0 + \int_{0}^{5} d r r \int_{0}^{2 π} d θ r^{2} \\ = 2 π \frac{5^{4}}{4} = \frac{625}{2} π \end{aligned}

4.2.6.28. (✳).

Solution.

The boundary of the solid

V

enclosed by

S

and

z = \pm 1

consists of three pieces:

S,

the top disk

S_{1} = {(x, y, z) | x^{2} + y^{2} \leq 2, z = 1}

and the bottom disk

S_{2} = {(x, y, z) | x^{2} + y^{2} \leq 2, z = - 1}

S_{1},

\hat{n} = \hat{k}

and

F \cdot \hat{n} = F \cdot \hat{k} = x y - z - z^{2} |_{z = 1} = x y - 2

so that, denoting

D = {(x, y) | x^{2} + y^{2} \leq 2},

\iint_{S_{1}} F \cdot \hat{n} d S = \iint_{D} (x y - 2) d x d y = - 2 Area (D) = - 4 π

Here we have used that the integral

\iint_{D} x y d x d y = 0

because

x y

is odd under

x \to - x .

S_{2},

\hat{n} = - \hat{k}

and

F \cdot \hat{n} = - F \cdot \hat{k} = - (x y - z - z^{2}) |_{z = - 1} = - x y

so that

\iint_{S_{2}} F \cdot \hat{n} d S = \iint_{D} (- x y) d x d y = 0

By the divergence theorem (Theorem 4.2.2),

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \iint_{V} \nabla \nabla \cdot F d V - \iint_{S_{1}} F \cdot \hat{n} d S - \iint_{S_{2}} F \cdot \hat{n} d S \\ = 0 - (- 4 π) - 0 = 4 π \end{aligned}

since

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x + e^{y z}) + \frac{\partial}{\partial y} (2 y z + \sin (x z)) + \frac{\partial}{\partial z} (x y - z - z^{2}) \\ = 1 + 2 z - 1 - 2 z \\ = 0 \end{aligned}

4.2.6.29. (✳).

Solution.

Direct Solution. The surface is given by the implicit equation

f (x, y, z) = 0

with

f (x, y, z) = x^{2} + y^{2} + 2 z^{2} - 1 .

Hence, by (3.3.3),

\hat{n} d S = \frac{\nabla \nabla f}{\nabla \nabla f \cdot \hat{k}} d x d y = \frac{2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + 4 z \hat{k}}{4 z} d x d y

This

\hat{n}

has positive

\hat{k}

component. Assume that it is the desired

\hat{n},

though this was not specified in the question. Since

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{2} - y - 1 & e^{\cos y} + z^{3} & 2 x z + z^{5} \end{array}] \\ = - 3 z^{2} \hat{ı ı} - 2 z \hat{ȷ ȷ} + \hat{k} \end{aligned}

we have

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{x^{2} + y^{2} \leq 1} (- 3 z (x, y)^{2} \hat{ı ı} - 2 z (x, y) \hat{ȷ ȷ} + \hat{k}) \cdot \frac{2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + 4 z (x, y) \hat{k}}{4 z (x, y)} d x d y \\ = \iint_{x^{2} + y^{2} \leq 1} (- \frac{3}{2} x z (x, y) - y + 1) d x d y \end{aligned}

Since

y

is an odd function of

y

and

x z (x, y) = x \sqrt{\frac{1}{2} (1 - x^{2} - y^{2})}

is an odd function of

x,

they both integrate to zero. Hence

\iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{x^{2} + y^{2} \leq 1} 1 d x d y = π

Tricky Solution. Let

V

be the solid

x^{2} + y^{2} + 2 z^{2} \leq 1,

z \geq 0 .

The surface of

V

consists of

S

with upward pointing normal and the disk

D = {(x, y, z) | z = 0, x^{2} + y^{2} \leq 1}

with normal

- \hat{k} .

By the divergence theorem, Theorem 4.2.2,

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S + \iint_{D} \nabla \nabla \times F \cdot (- \hat{k}) d S & = ∭_{V} \nabla \nabla \cdot \nabla \nabla \times F d V \\ = ∭_{V} 0 d V = 0 \end{aligned}

Hence

\iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{D} \nabla \nabla \times F \cdot \hat{k} d S = \iint_{D} d S = π

4.2.6.30. (✳).

Solution.

Let

S^{'}

be the disk

x^{2} + y^{2} \leq 3, z = 0

(with

\hat{n}

the downward pointing normal) and let

V

be the portion of the ball

x^{2} + y^{2} + (z - 1)^{2} \leq 4

with

z \geq 0 .

Then, by the divergence theorem,

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{S^{'}} F \cdot (- \hat{k}) d S \\ = ∭_{V} (2 x + 2 y) d V + \iint_{S^{'}} (4 + 5 x) d x d y \end{aligned}

Because

x

is odd under

x \to - x

and

y

is odd under

y \to - y,

∭_{V} x d V = ∭_{V} y d V = \iint_{S^{'}} x d x d y = 0

so that

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = 4 \iint_{S^{'}} d x d y = 4 Area (S^{'}) = 4 \times π (\sqrt{3})^{2} = 12 π \end{matrix}

4.2.6.31. (✳).

Solution.

Call the hemisphere

0 \leq z \leq \sqrt{4 - x^{2} - y^{2}},

H .

Call the bottom surface of the hemisphere

D

and the top surface

S .

The disk

D

has radius

2,

area

4 π,

z = 0

and the outward normal

- \hat{k},

so that

\iint_{D} F \cdot \hat{n} d S = - \iint_{D} F \cdot \hat{k} d x d y = - \iint_{D} d x d y = - 4 π

\nabla \nabla \cdot F = \frac{\partial}{\partial x} (x y^{2}) + \frac{\partial}{\partial y} (x^{2} y) + \frac{\partial}{\partial z} (1) = x^{2} + y^{2}

the divergence theorem (Theorem 4.2.2) gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{H} \nabla \nabla \cdot F d V - \iint_{D} F \cdot \hat{n} d S \\ = ∭_{R} (x^{2} + y^{2}) d V - (- 4 π) \end{aligned}

To evaluate the remaining integral, let’s switch to the cylindrical coordinates

(r, θ, z) .

In cylindrical coordinates, the equation

x^{2} + y^{2} + z^{2} = 4

becomes

r^{2} + z^{2} = 4 .

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = 4 π + \int_{0}^{2} d z \int_{0}^{\sqrt{4 - z^{2}}} d r r \int_{0}^{2 π} d θ r^{2} \\ = 4 π + 2 π \int_{0}^{2} d z \frac{1}{4} (\sqrt{4 - z^{2}})^{4} \\ = 4 π + \frac{π}{2} \int_{0}^{2} d z (16 - 8 z^{2} + z^{4}) \\ = 4 π + \frac{π}{2} [16 z - \frac{8}{3} z^{3} + \frac{1}{5} z^{5}]_{0}^{2} \\ = \frac{188}{15} π \approx 39.37 \end{aligned}

4.2.6.32. (✳).

Solution.

Let

S_{t},

S_{b}

and

S_{c}

denote the top, bottom and curved surfaces of

D

respectively. On the top surface,

z = 5

and the outward normal to

D

\hat{k},

so that

\iint_{S_{t}} F \cdot \hat{n} d S = \iint_{x^{2} + y^{2} \leq 1} (15 - 5 y e^{5}) d x d y = 15 \iint_{x^{2} + y^{2} \leq 1} d x d y = 15 π

The integral over

y

was zero because

y

is odd under

y \to - y .

On the bottom surface,

z = 0

and the outward normal to

D

- \hat{k},

so that

\iint_{S_{b}} F \cdot \hat{n} d S = - \iint_{x^{2} + y^{2} \leq 1} (3 \times 0 - 0 \times y e^{0}) d x d y = 0

Again, the integral over

y

was zero because

y

is odd under

y \to - y .

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x + x y e^{z}) + \frac{1}{2} \frac{\partial}{\partial y} (y^{2} z e^{z}) + \frac{\partial}{\partial z} (3 z - y z e^{z}) \\ = (1 + y e^{z}) + y z e^{z} + (3 - y z e^{z} - y e^{z}) = 4 \end{aligned}

the divergence theorem gives

\begin{aligned} \iint_{S_{c}} F \cdot \hat{n} d S & = ∭_{D} \nabla \nabla \cdot F d V - \iint_{S_{t}} F \cdot \hat{n} d S - \iint_{S_{b}} F \cdot \hat{n} d S \\ = ∭_{D} 4 d V - 15 π - 0 \\ = 4 \times π 1^{2} \times 5 - 15 π = 5 π \end{aligned}

4.2.6.33.

Solution.

Let

V = {(x, y, z) | x^{2} + y^{2} + z^{2} \leq a^{2}, x \geq 0, y \geq 0, z \geq 0} .

Then

\partial V

consists of

the $x = 0$ face ${(x, y, z) | y^{2} + z^{2} \leq a^{2}, x = 0, y \geq 0, z \geq 0}$ with normal $\hat{n} = - \hat{ı ı},$
the $y = 0$ face ${(x, y, z) | x^{2} + z^{2} \leq a^{2}, x \geq 0, y = 0, z \geq 0}$ with normal $\hat{n} = - \hat{ȷ ȷ},$
the $z = 0$ face ${(x, y, z) | x^{2} + y^{2} \leq a^{2}, x \geq 0, y \geq 0, z = 0}$ with normal $\hat{n} = - \hat{k},$
and the first octant part of the sphere. Call it $S .$

Then

\begin{aligned} ∭_{V} \nabla \nabla \cdot F d V & = ∭_{V} [z + 1 + z - 2 z] d V = ∭_{V} d V = \frac{1}{8} \frac{4}{3} π a^{3} \\ = \frac{1}{6} π a^{3} \\ \iint_{\binom{z = 0}{f a c e}} F \cdot (- \hat{k}) d x d y & = \iint_{\binom{z = 0}{f a c e}} (2 x + 0^{2}) d x d y = 2 \int_{0}^{a} d r r \int_{0}^{π / 2} d θ r \cos θ \\ = 2 \int_{0}^{a} r^{2} d r = \frac{2 a^{3}}{3} \\ \iint_{\binom{y = 0}{f a c e}} F \cdot (- \hat{ȷ ȷ}) d x d z & = - \iint_{\binom{y = 0}{f a c e}} (0 + 0 z) d x d z = 0 \\ \iint_{\binom{x = 0}{f a c e}} F \cdot (- \hat{ı ı}) d y d z & = \iint_{\binom{x = 0}{f a c e}} - (y + 0 z) d y d z = - \int_{0}^{a} d r r \int_{0}^{π / 2} d θ r \sin θ \\ = - \int_{0}^{a} r^{2} d r = - \frac{a^{3}}{3} \end{aligned}

By the divergence theorem

\begin{aligned} \iint_{S} F \cdot \hat{n} d x d y & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{\binom{x = 0}{f a c e}} F \cdot (- \hat{ı ı}) d y d z \\ - \iint_{\binom{y = 0}{f a c e}} F \cdot (- \hat{ȷ ȷ}) d x d z - \iint_{\binom{z = 0}{f a c e}} F \cdot (- \hat{k}) d x d y \\ = [\frac{π}{6} - \frac{1}{3}] a^{3} \end{aligned}

4.2.6.34.

Solution.

(a) On the cylindrical surface

S_{1},

use (surprise!) cylindrical coordinates. Since the cylinder has radius

\sqrt{2},

we may parametrize it by

\begin{aligned} r (θ, z) & = \sqrt{2} \cos θ \hat{ı ı} + \sqrt{2} \sin θ \hat{ȷ ȷ} + z \hat{k} \\ \frac{\partial r}{\partial θ} (θ, z) & = - \sqrt{2} \sin θ \hat{ı ı} + \sqrt{2} \cos θ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial z} (θ, z) & = \hat{k} \\ \hat{n} d S & = \pm \frac{\partial r}{\partial θ} (θ, z) \times \frac{\partial r}{\partial z} (θ, z) d θ d z \\ = \pm det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - \sqrt{2} \sin θ \hat{ı ı} & \sqrt{2} \cos θ & 0 \\ 0 & 0 & 1 \end{array}] d θ d z \\ = \pm (\sqrt{2} \cos θ \hat{ı ı} + \sqrt{2} \sin θ \hat{ȷ ȷ}) d θ d z \end{aligned}

To get the inward pointing normal, choose the minus sign. So

\begin{aligned} F \cdot \hat{n} d S & = [\sqrt{2} (\cos θ - z \sin θ) \hat{ı ı} + \sqrt{2} (\sin θ + z \cos θ) \hat{ȷ ȷ} + (\dots) \hat{k}] \\ \cdot [- \sqrt{2} \cos θ \hat{ı ı} - \sqrt{2} \sin θ \hat{ȷ ȷ}] d θ d z \\ = - 2 [(\cos θ - z \sin θ) \cos θ + (\sin θ + z \cos θ) \sin θ] d θ d z \\ = - 2 d θ d z \end{aligned}

On the intersection of the sphere and cylinder

\begin{matrix} z^{2} = 4 - x^{2} - y^{2} = 4 - 2 = 2 \end{matrix}

z

runs from

- \sqrt{2}

\sqrt{2}

(see the figure below) and

\begin{matrix} \iint_{S_{1}} F \cdot \hat{n} d S = - 2 \int_{- \sqrt{2}}^{\sqrt{2}} d z \int_{0}^{2 π} d θ = - 8 \sqrt{2} π \end{matrix}

(b) Observe that

\nabla \nabla \cdot F = 3 .

∭_{V} \nabla \nabla \cdot F d V = ∭_{V} 3 d V

The horizontal cross-section of

V

at height

z

is a washer with outer radius

\sqrt{4 - z^{2}}

(determined by the equation of the sphere) and inner radius

\sqrt{2}

(determined by the equation of the cylinder).

So the cross-section has area

π (\sqrt{4 - z^{2}})^{2} - π (\sqrt{2})^{2} = π (2 - z^{2})

and

\begin{aligned} ∭_{V} \nabla \nabla \cdot F d V & = 3 ∭_{V} d V = 3 \int_{- \sqrt{2}}^{\sqrt{2}} π (2 - z^{2}) d z \\ = 6 π \int_{0}^{\sqrt{2}} (2 - z^{2}) d z = 6 π (2 \sqrt{2} - \frac{2^{3 / 2}}{3}) \\ = 8 \sqrt{2} π \end{aligned}

\iint_{S_{2}} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V - \iint_{S_{1}} F \cdot \hat{n} d S = 16 \sqrt{2} π

4.2.6.35.

Solution.

By the divergence theorem

\iint_{\partial V} E \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot E d V

So by Gauss’ law

∭_{V} \nabla \nabla \cdot E d V = 4 π ∭_{V} ρ d V ⟹ ∭_{V} [\nabla \nabla \cdot E - 4 π ρ] d V = 0

This is true for all solids

V

for which the divergence theorem applies. If there were some point in

R^{3}

for which

\nabla \nabla \cdot E - 4 π ρ

were, say, strictly bigger than zero, then, by continuity, we could find a ball

B_{ϵ}

centered on that point with

\nabla \nabla \cdot E - 4 π ρ > 0

everywhere on

B_{ϵ} .

This would force

∭_{B_{ϵ}} [\nabla \nabla \cdot E - 4 π ρ] d V > 0,

which violates

∭_{V} [\nabla \nabla \cdot E - 4 π ρ] d V = 0

with

V

set equal to

B_{ϵ} .

Hence

\nabla \nabla \cdot E - 4 π ρ

must be zero everywhere.

4.2.6.36.

Solution.

By the divergence theorem

\begin{aligned} \iint_{\partial V} r \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot r d V = ∭_{V} \nabla \nabla \cdot (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}) d V \\ = ∭_{V} 3 d V = 3 Volume (V) \end{aligned}

Our geometric explanation starts with the observation that the volume of the cone with vertex

(0, 0, 0)

and base a tiny piece of surface

d S

\frac{1}{3}

times the area of the base times the height of the cone. The height of the cone is

| \hat{n} \cdot r |,

where

r

is a point in

d S .

So the volume of the cone is

\frac{1}{3} | \hat{n} \cdot r | d S .

First assume that

(0, 0, 0)

is in

V

and

V

is convex. Then

$\hat{n} \cdot r > 0,$ and the volume is $\frac{1}{3} \hat{n} \cdot r d S .$
the cone is contained in $V$ and
$V$ is the union of all the tiny conical pieces with $d S$ running over $\partial V .$

Volume (V) = \frac{1}{3} \iint_{\partial V} r \cdot \hat{n} d S

To generalise to the case that

V

is not convex or

(0, 0, 0)

is not in

V,

write

V

as the difference between a large convex solid and one or more smaller convex solids.

4.2.6.37. (✳).

Solution.

(a) We’ll parametrize the sphere using the spherical coordinates

θ

and

φ .

\begin{aligned} x & = 3 \sin φ \cos θ \\ y & = 3 \sin φ \sin θ \\ z & = 3 \cos φ \end{aligned}

with

0 \leq θ \leq 2 π,

0 \leq φ \leq π .

Since

\begin{aligned} (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) & = (- 3 \sin φ \sin θ, 3 \sin φ \cos θ, 0) \\ (\frac{\partial x}{\partial φ}, \frac{\partial y}{\partial φ}, \frac{\partial z}{\partial φ}) & = (3 \cos φ \cos θ, 3 \cos φ \sin θ, - 3 \sin φ) \end{aligned}

(3.3.1) yields

\begin{aligned} \hat{n} d S & = \pm (\frac{\partial x}{\partial θ}, \frac{\partial y}{\partial θ}, \frac{\partial z}{\partial θ}) \times (\frac{\partial x}{\partial φ}, \frac{\partial y}{\partial φ}, \frac{\partial z}{\partial φ}) d θ d φ \\ = \pm (- 3 \sin φ \sin θ, 3 \sin φ \cos θ, 0) \\ \times (3 \cos φ \cos θ, 3 \cos φ \sin θ, - 3 \sin φ) d θ d φ \\ = \pm (- 9 \sin^{2} φ \cos θ, - 9 \sin^{2} φ \sin θ, - 9 \sin φ \cos φ) d θ d φ \\ = \mp 9 \sin φ (\sin φ \cos θ, \sin φ \sin θ, \cos φ) d θ d φ \end{aligned}

To get an outward pointing normal we need the

+

sign. For example, with the

+

sign, the

z

-component is

9 \sin φ \cos φ = \frac{9}{2} \sin (2 φ)

so that the normal is pointing upward when

0 < φ < \frac{π}{2},

i.e. in the northern hemisphere, and is pointing downward when

\frac{π}{2} < φ < π,

i.e. in the southern hemisphere. (As a further consistency check, note that

\hat{n} (θ, φ)

is parallel to

r (θ, φ) .

) So

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = 9 \int_{0}^{2 π} d θ \int_{0}^{π} d φ \sin φ (0, 0, 3 \sin φ \cos θ + 3 \cos φ) \\ \cdot (\sin φ \cos θ, \sin φ \sin θ, \cos φ) \\ = 27 \int_{0}^{2 π} d θ \int_{0}^{π} d φ (\sin^{2} φ \cos φ \cos θ + \sin φ \cos^{2} φ) \\ = 54 π \int_{0}^{π} d φ \sin φ \cos^{2} φ since \int_{0}^{2 π} \cos θ d θ = 0 \\ = - 18 π [\cos^{3} φ]_{0}^{π} \\ = 36 π \end{aligned}

(b) Set

V = {(x, y, z) \in R^{3} | x^{2} + y^{2} + z^{2} \leq 9}

Since

\nabla \nabla \cdot F = \frac{\partial}{\partial z} (x + z) = 1

the divergence theorem (Theorem 4.2.2) gives

\iint_{S} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} d V = \frac{4}{3} π 3^{3} = 36 π

4.2.6.38. (✳).

Solution.

Denote by

V

the cube specified in the problem. Then

\partial V

consists of

S

together with the face

F

in the plane

z = 0,

oriented with the normal being

- \hat{k} .

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (y \cos (y^{2}) + z - 1) + \frac{\partial}{\partial y} (\frac{z}{x + 1} + 1) + \frac{\partial}{\partial z} (x y e^{z^{2}}) \\ = \frac{\partial}{\partial z} (x y e^{z^{2}}) \end{aligned}

the divergence theorem (Theorem 4.2.2) gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{F} F \cdot (- \hat{k}) d S \\ = \int_{0}^{1} d x \int_{0}^{1} d y \int_{0}^{1} d z \frac{\partial}{\partial z} (x y e^{z^{2}}) + \int_{0}^{1} d x \int_{0}^{1} d y x y e^{z^{2}} |_{z = 0} \\ = \int_{0}^{1} d x \int_{0}^{1} d y x y e^{z^{2}} |_{z = 0}^{z = 1} + \int_{0}^{1} d x \int_{0}^{1} d y x y e^{z^{2}} |_{z = 0} \\ = \int_{0}^{1} d x \int_{0}^{1} d y x y e^{z^{2}} |_{z = 1} \\ = e {[\frac{x^{2}}{2}]}_{0}^{1} {[\frac{y^{2}}{2}]}_{0}^{1} \\ = \frac{e}{4} \end{aligned}

4.2.6.39. (✳).

Solution.

(a) The equation of the surface is

G (x, y, z) = z - x y = 0 .

So one normal to the surface at

(1, 1, 1)

(\nabla \nabla G) (1, 1, 1) = (- y, - x, 1) |_{(x, y, z) = (1, 1, 1)} = (- 1, - 1, 1)

and a unit upward pointing normal at

(1, 1, 1)

\frac{(- 1, - 1, 1)}{| (- 1, - 1, 1) |} = \frac{1}{\sqrt{3}} (- 1, - 1, 1) .

(b) For the surface

G (x, y, z) = z - x y,

so that, by (3.3.3),

\begin{matrix} \hat{n} d S = \pm \frac{\nabla \nabla G (x, y, z)}{\nabla \nabla G (x, y, z) \cdot \hat{k}} d x d y = \pm (- y, - x, 1) d x d y \end{matrix}

The “

+

” sign gives the upward normal, so the specified upward flux is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \iint_{x^{2} + y^{2} \leq 9} (y, x, 3) \cdot (- y - x, 1) d x d y \\ = \iint_{x^{2} + y^{2} \leq 9} (3 - x^{2} - y^{2}) d x d y \end{aligned}

Switching to polar coordinates, the flux is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{0}^{3} d r r \int_{0}^{2 π} d θ (3 - r^{2}) = 2 π \int_{0}^{3} d r (3 r - r^{3}) \\ = 2 π (\frac{3}{2} 3^{2} - \frac{1}{4} 3^{4}) = - \frac{27 π}{2} \end{aligned}

θ

and

z .

\begin{aligned} x & = 3 \cos θ \\ y & = 3 \sin θ \\ z & = z \end{aligned}

with

0 \leq θ \leq 2 π

and

9 \sin θ \cos θ \leq z \leq 10 .

Then, using (3.3.1),

\begin{aligned} \frac{\partial r}{\partial θ} & = (- 3 \sin θ, 3 \cos θ, 0) \\ \frac{\partial r}{\partial z} & = (0, 0, 1) \\ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} & = 3 (\cos θ, \sin θ, 0) \\ \hat{n} d S & = \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} d θ d z = 3 (\cos θ, \sin θ, 0) d θ d z \end{aligned}

(We have taken the

+

\hat{n} d S = \pm \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} d θ d z

to give the outward pointing normal.) So the specified flux is

\begin{aligned} \iint F \cdot \hat{n} d S & = 3 \int_{0}^{2 π} d θ \int_{9 \cos θ \sin θ}^{10} d z \overset{F = (y, x, 3)}{\overset{⏞}{(3 \sin θ, 3 \cos θ, 3)}} \cdot (\cos θ, \sin θ, 0) \\ = 18 \int_{0}^{2 π} d θ \int_{9 \cos θ \sin θ}^{10} d z \sin θ \cos θ \\ = 18 \int_{0}^{2 π} d θ [10 - 9 \cos θ \sin θ] \sin θ \cos θ \\ = - 9 \times 18 \int_{0}^{2 π} d θ \sin^{2} θ \cos^{2} θ \\ since \int_{0}^{2 π} \sin θ \cos θ d θ = \frac{1}{2} \int_{0}^{2 π} \sin (2 θ) d θ = 0 \\ = - 9 \times 18 \times \frac{1}{4} \int_{0}^{2 π} d θ \sin^{2} (2 θ) \\ = - \frac{81}{2} \int_{0}^{2 π} d θ \frac{1 - \cos (4 θ)}{2} \\ = - \frac{81}{2} {[\frac{θ}{2} - \frac{\sin (4 θ)}{8}]}_{0}^{2 π} \\ = - \frac{81}{2} π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} d θ \sin^{2} (2 θ)

see Example 2.4.4.

x^{2} + y^{2} \leq 9,

then

| x | \leq 3

and

y \leq 3

so that

| x y | \leq 9 < 10 .

Set

\begin{aligned} \tilde{S} & = {(x, y, z) | x^{2} + y^{2} = 9, x y \leq z \leq 10} \\ V & = {(x, y, z) | x^{2} + y^{2} \leq 9, x y \leq z \leq 10} \end{aligned}

Note that the boundary,

\partial V,

V

consists of three parts:

the side $\tilde{S},$ with outward pointing normal (which is the surface and the normal specified in part (c) of the question)
the bottom, which is the surface $S$ of part (b), with downward pointing normal (which is opposite the normal specified in part (b)) and
the top, which is the surface $S_{T} = {(x, y, z) | x^{2} + y^{2} \leq 9, z = 10},$ with normal $\hat{n} = \hat{k} .$

Here is a sketch of the part of

\partial V

that is in the first octant.

Note that

\nabla \nabla \cdot F = 0 .

So the divergence theorem yields

\begin{aligned} 0 & = ∭_{V} \nabla \nabla \cdot F d V \\ = \iint_{\tilde{\partial} V} F \cdot \hat{n} d S \\ = \iint_{\tilde{S}} F \cdot \hat{n} d S - \iint_{S} F \cdot \hat{n} d S + \iint_{S_{T}} F \cdot \hat{k} d S \end{aligned}

This implies

\begin{aligned} \iint_{\tilde{S}} F \cdot \hat{n} d S & = \iint_{S} F \cdot \hat{n} d S - \iint_{S_{T}} F \cdot \hat{k} d S \\ = - \frac{27 π}{2} - \iint_{x^{2} + y^{2} \leq 9} 3 d S \\ = - \frac{27 π}{2} - 3 π 3^{2} = - \frac{81 π}{2} \end{aligned}

4.2.6.40. (✳).

Solution.

(a) The divergence of

F

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x + \sin y) + \frac{\partial}{\partial y} (z + y) + \frac{\partial}{\partial z} (z^{2}) \\ = 2 + 2 z \end{aligned}

(b) Set

\begin{aligned} V & = {(x, y, z) | x^{2} + y^{2} \leq 25, 0 \leq z \leq \sqrt{25 - x^{2} - y^{2}}} \\ S_{T} & = {(x, y, z) | x^{2} + y^{2} + z^{2} = 25, z \geq 0} \\ S_{B} & = {(x, y, z) | x^{2} + y^{2} \leq 25, z = 0} \end{aligned}

Note that the boundary,

\partial V,

V

consists ot two parts —

S_{T}

with upward normal, and

S_{B}

with normal

- \hat{k} .

We are to find the flux through

S_{T}

with upward normal. By the divergence theorem, it is

\begin{aligned} \iint_{S_{T}} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V - \iint_{S_{B}} F \cdot (- \hat{k}) d S \\ = ∭_{V} (2 + 2 z) d V \end{aligned}

since

F \cdot \hat{k} = z^{2} = 0

S_{B} .

We’ll compute the volume integral by expressing it as an iterated integral, with the

z

integration on the outside. In

V,

z

ranges for

0

5 .

The set of points at exactly height

z

V

{(x, y, z) | x^{2} + y^{2} \leq 25 - z^{2}} .

\begin{aligned} \iint_{S_{T}} F \cdot \hat{n} d S & = \int_{0}^{5} d z \iint_{x^{2} + y^{2} \leq 25 - z^{2}} d x d y (2 + 2 z) \\ = \int_{0}^{5} d z (2 + 2 z) \iint_{x^{2} + y^{2} \leq 25 - z^{2}} d x d y \\ = \int_{0}^{5} d z π (25 - z^{2}) (2 + 2 z) \end{aligned}

since

\iint_{x^{2} + y^{2} \leq 25 - z^{2}} d x d y

is the area of a disk of radius

\sqrt{25 - z^{2}} .

Continuing,

\begin{aligned} \iint_{S_{T}} F \cdot \hat{n} d S & = π \int_{0}^{5} d z (50 - 2 z^{2} + 50 z - 2 z^{3}) \\ = π (50 \times 5 - 2 \frac{5^{3}}{3} + 50 \frac{5^{2}}{2} - 2 \frac{5^{4}}{4}) \\ = π 5^{3} (2 - \frac{2}{3} + 5 - \frac{5}{2}) \\ = π 5^{3} (\frac{4}{3} + \frac{5}{2}) \\ = π \frac{23}{6} 5^{3} = \frac{2875}{6} π \end{aligned}

S

that is the boundary of a solid

V .

Use

the outward pointing normal for $S,$
$| V |$ to denote the volume of $V,$ and
$\bar{z} = \frac{1}{| V |} ∭_{V} z d V$ to denote the $z$ -component of the centroid (i.e. centre of mass with constant density) of $V .$

Then, by the divergence theorem

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} (2 + 2 z) d V \\ = 2 ∭_{V} d V + 2 ∭_{V} z d V \\ = 2 | V | + 2 | V | \bar{z} \end{aligned}

This takes the value

- 9

if and only if

\begin{matrix} 2 | V | \bar{z} = - 9 - 2 | V | ⟺ \bar{z} = - \frac{9}{2 | V |} - 1 \end{matrix}

One surface which obeys this condition is the unit cube (with outward normal) centred on

(0, 0, - \frac{11}{2}) .

4.2.6.41. (✳).

Solution.

(a) The constant

z

cross-section of the cone at height

0 \leq z \leq 1

is a circle of radius

2 z .

So we may parametrize the cone by

r (θ, z) = 2 z \cos θ \hat{ı ı} + 2 z \sin θ \hat{ȷ ȷ} + z \hat{k} 0 \leq θ < 2 π, 0 \leq z \leq 1

Since

\begin{aligned} \frac{\partial r}{\partial θ} & = (- 2 z \sin θ, 2 z \cos θ, 0) \\ \frac{\partial r}{\partial z} & = (2 \cos θ, 2 \sin θ, 1) \\ \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} & = (2 z \cos θ, 2 z \sin θ, - 4 z) \end{aligned}

(3.3.1) yields that the element of surface area for this parametrization is

\begin{aligned} d S & = | \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial z} | d θ d z = 2 z | (\cos θ, \sin θ, - 2) | d θ d z \\ = 2 \sqrt{5} z d θ d z \end{aligned}

In our parametrization the condition

x \leq y

becomes

2 z \cos θ \leq 2 z \sin θ,

which, for

z > 0,

is equivalent to

\tan θ \geq 1 .

So the specified integral is

\begin{aligned} \iint_{S} z^{2} d S & = 2 \sqrt{5} \int_{0}^{1} d z \int_{π / 4}^{π / 2} d θ z^{3} = \frac{\sqrt{5} π}{2} \int_{0}^{1} d z z^{3} = \frac{\sqrt{5} π}{8} \end{aligned}

(b) Let’s first do some strategizing. We have to compute a flux integral over a surface that is not closed. There are two potential sneaky attacks that come to mind.

The first uses Stokes’ theorem. But the flux integral in Stokes’ theorem is of the form $\iint_{S} \nabla \nabla \times A \cdot \hat{n} d S .$ So to be able to apply Stokes’ theorem in the current problem, $F$ has to be of the form $\nabla \nabla \times A .$ That is, $F$ has to have a vector potential. We know that in order for $F$ to have a vector potential, it must pass the screening test $\nabla \nabla \cdot F = 0 .$ Our $F = z \hat{k}$ fails this screening test. So we can’t use Stokes’ theorem.
The second uses the divergence theorem. But the flux integral in the divergence theorem is over the boundary of a solid. That is not the case for our $S .$ So in order to apply the divergence theorem in the current problem, we have to enlarge $S$ to the boundary of a solid. There are many ways to do this. But they all appear fairly complicated. So it does not seem wise to use the divergence theorem.

So it looks like we have to evaluate the flux integral directly. To do so, we have to determine

\hat{n} d S

for the specified rectangle. Look at the sketch of

S

below. It is part of a plane,

and that plane is invariant under translations parallel to the

x

axis. As the plane does not pass through the origin, the equation of the plane has to be of the form

b y + c z = 1 .

For

(0, 0, 4)

to be on the plane, we need

c = \frac{1}{4} .

For

(0, 2, 0)

to be on the plane, we need

b = \frac{1}{2} .

S

is contained in the plane

G (x, y, z) = \frac{y}{2} + \frac{z}{4} = 1

and equation (3.3.3) gives that

\hat{n} d S = \pm \frac{\nabla \nabla G (x, y, z)}{\nabla \nabla G (x, y, z) \cdot \hat{k}} d x d y = \pm \frac{(0, \frac{1}{2}, \frac{1}{4})}{\frac{1}{4}} d x d y = \pm (0, 2, 1) d x d y

The problem specifies that the normal is to be upward, i.e. is to have a positive

z

-component. So

\hat{n} d S = (0, 2, 1) d x d y

Again looking at the sketch of

S

above we see, as

(x, y, z)

runs over

S,

(x, y)

runs over

R = {(x, y) | 0 \leq x \leq 5, 0 \leq y \leq 2}

Thus our flux integral is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \iint_{R} \overset{z}{\overset{⏞}{(4 - 2 y)}} \hat{k} \cdot \overset{\hat{n} d S}{\overset{⏞}{(0, 2, 1) d x d y}} = \int_{0}^{2} d y \int_{0}^{5} d x (4 - 2 y) \\ = \int_{0}^{2} d y 5 (4 - 2 y) = 5 [4 y - y^{2}]_{0}^{2} = 20 \end{aligned}

\nabla \nabla \cdot F = 2 z,

which is pretty simple. So let’s use the divergence theorem. If

V = {(x, y, z) | 0 \leq x \leq 1, 0 \leq y \leq 2, 0 \leq z \leq 3},

the divergence theorem says that

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V = 2 ∭_{V} z d V \end{aligned}

This integral would be easy enough to evaluate directly, but we don’t need to. The average value of

z

(i.e. the

z

-coordinate of the centre of mass with constant density) is

\frac{3}{2},

by symmetry. Since

V

has volume

6,

that average value of

z

is also

\begin{matrix} \bar{z} = \frac{1}{6} ∭_{V} z d V = \frac{3}{2} \end{matrix}

∭_{V} z d V = 9

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = 2 ∭_{V} z d V = 18 \end{matrix}

4.2.6.42. (✳).

Solution.

(a) For the surface

z = f (x, y) = 1 - x^{2} - y^{2},

with an upwards pointing normal,

\hat{n} d S = [- f_{x} (x, y) d x - f_{y} (x, y) + \hat{k}] d x d y = [2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}] d x d y

by (3.3.2). So the specified upward flux is

\begin{aligned} \iint_{σ_{1}} F \cdot \hat{n} d S \\ = \iint_{x^{2} + y^{2} \leq 1} {[a (y^{2} + z^{2}) + b x z] \hat{ı ı} + [c (x^{2} + z^{2}) + d y z] \hat{ȷ ȷ} + x^{2} \hat{k}} \\ \cdot {2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}}_{z = 1 - x^{2} - y^{2}} d x d y \\ = \iint_{x^{2} + y^{2} \leq 1} {[2 a x (y^{2} + z^{2}) + 2 b x^{2} z] \\ + [2 c y (x^{2} + z^{2}) + 2 d y^{2} z] + x^{2}}_{z = 1 - x^{2} - y^{2}} d x d y \end{aligned}

Now

\iint_{x^{2} + y^{2} \leq 1} {2 a x (y^{2} + z^{2})}_{z = 1 - x^{2} - y^{2}} d x d y = 0

because the integrand is odd under

x \to - x

and

\iint_{x^{2} + y^{2} \leq 1} {2 c y (x^{2} + z^{2})}_{z = 1 - x^{2} - y^{2}} d x d y = 0

because the integrand is odd under

y \to - y .

So that leaves

\iint_{σ_{1}} F \cdot \hat{n} d S = \iint_{x^{2} + y^{2} \leq 1} {2 b x^{2} z + 2 d y^{2} z + x^{2}}_{z = 1 - x^{2} - y^{2}} d x d y

We’ll switch to polar coordinates to evaluate the remaining integral.

\begin{aligned} \iint_{σ_{1}} F \cdot \hat{n} d S & = \int_{0}^{1} d r r \int_{0}^{2 π} d θ {2 b r^{2} z \cos^{2} θ + 2 d r^{2} z \sin^{2} θ + r^{2} \cos^{2} θ}_{z = 1 - r^{2}} \end{aligned}

Now

\begin{aligned} \int_{0}^{2 π} \cos^{2} θ d θ & = \int_{0}^{2 π} \frac{\cos (2 θ) + 1}{2} d θ = {[\frac{\sin (2 θ)}{4} + \frac{θ}{2}]}_{0}^{2 π} = π \\ \int_{0}^{2 π} \sin^{2} θ d θ & = \int_{0}^{2 π} \frac{1 - \cos (2 θ)}{2} d θ = {[\frac{θ}{2} - \frac{\sin (2 θ)}{4}]}_{0}^{2 π} = π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} θ d θ

and

\int_{0}^{2 π} \sin^{2} θ d θ,

see Example 2.4.4. So, we finally have

\begin{aligned} \iint_{σ_{1}} F \cdot \hat{n} d S & = \int_{0}^{1} d r {2 π b r^{3} (1 - r^{2}) + 2 π d r^{3} (1 - r^{2}) + π r^{3}} \\ = 2 π b [\frac{1}{4} - \frac{1}{6}] + 2 π d [\frac{1}{4} - \frac{1}{6}] + π \frac{1}{4} = \frac{π}{4} + \frac{π (b + d)}{6} \end{aligned}

(b), (c) Here is a side view of

σ_{1},

σ_{2}

and

σ_{3} .

Set

\begin{aligned} V_{b} & = {(x, y, z) | 0 \leq z \leq 1 - x^{2} - y^{2}, x^{2} + y^{2} \leq 1} \\ V_{c} & = {(x, y, z) | x^{2} + y^{2} - 1 \leq z \leq 1 - x^{2} - y^{2}, x^{2} + y^{2} \leq 1} \end{aligned}

Then

\partial V_{b} = σ_{1} \cup σ_{3}

and

\partial V_{c} = σ_{1} \cup σ_{2},

all with outward pointing normals. Since the divergence of

F

\nabla \nabla \cdot F = \frac{\partial}{\partial x} [a (y^{2} + z^{2}) + b x z] + \frac{\partial}{\partial y} [c (x^{2} + z^{2}) + d y z] + \frac{\partial}{\partial z} [x^{2}] = (b + d) z

the divergence theorem gives

\begin{aligned} \iint_{σ_{1} \cup σ_{3}} F \cdot \hat{n} d S & = ∭_{V_{b}} \nabla \nabla \cdot F d V = (b + d) ∭_{V_{b}} z d V \\ \iint_{σ_{1} \cup σ_{2}} F \cdot \hat{n} d S & = ∭_{V_{c}} \nabla \nabla \cdot F d V = (b + d) ∭_{V_{c}} z d V \end{aligned}

Now on

V_{b},

z \geq 0

and

z > 0

except on

σ_{3} .

∭_{V_{b}} z d V > 0

and

\iint_{σ_{1} \cup σ_{3}} F \cdot \hat{n} d S

is zero if and only if

d = - b .

That’s the answer to part (b).

On the other hand,

V_{c}

is even under

z \to - z

so that

∭_{V_{c}} z d V = 0 .

Consequently

\iint_{σ_{1} \cup σ_{3}} F \cdot \hat{n} d S

is zero for all

a,

b,

c,

d .

That’s the answer to part (c).

4.2.6.43. (✳).

Solution.

We will be using the divergence theorem in both parts (a) and

b .

So as a prelimary calculation, let’s find the divergence of

H (x, y, z) = \frac{(x, y, z) - (a, b, c)}{{[(x - a)^{2} + (y - b)^{2} + (z - b)^{2}]}^{3 / 2}}

for any

(a, b, c) .

(x, y, z) \neq (a, b, c),

\begin{aligned} \nabla \nabla \cdot H (x, y, z) \\ = \frac{\partial}{\partial x} \frac{x - a}{{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}]}^{3 / 2}} \\ + \frac{\partial}{\partial y} \frac{y - b}{{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}]}^{3 / 2}} \\ + \frac{\partial}{\partial z} \frac{z - c}{{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}]}^{3 / 2}} \\ = \frac{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}] - (x - a) \frac{3}{2} (2 (x - a))}{{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}]}^{5 / 2}} \\ + \frac{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}] - (y - b) \frac{3}{2} (2 y)}{{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}]}^{5 / 2}} \\ + \frac{[(x - a)^{2} + (y - b)^{2} + z^{2}] - (z - c) \frac{3}{2} (2 (z - c))}{{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}]}^{5 / 2}} \\ = \frac{3 [(x - a)^{2} + (y - b)^{2} + (z - c)^{2}] - 3 (x - a)^{2} - 3 (y - b)^{2} - 3 (z - c)^{2}}{{[(x - a)^{2} + (y - b)^{2} + (z - c)^{2}]}^{5 / 2}} \\ = 0 \end{aligned}

(x, y, z) = (a, b, c),

H (x, y, z)

is not defined and hence

\nabla \nabla \cdot H (x, y, z)

is also not defined.

(b) By the above preliminary computation with

(a, b, c) = (3, 2, 2),

\nabla \nabla \cdot G

is defined and zero for all

(x, y, z) \neq (3, 2, 2),

and, in particular for all

(x, y, z)

V = {(x, y, z) | x^{2} + 2 y^{2} + 3 z^{2} \leq 16}

So, by the divergence theorem,

\begin{aligned} \iint_{S} G \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot G d V = 0 \end{aligned}

(a) Because

(1, 1, 2)

is inside

V,

we cannot use the argument of part (b), to conclude that the integral is zero. Let

ε > 0

be small enough that

S_{ε} = {(x, y, z) | (x - 2)^{2} + (y - 1)^{2} + (z - 1)^{2} = ε^{2}}

is completely contained inside

V,

as in the sketch below.

Set

V_{ε} = {(x, y, z) | x^{2} + 2 y^{2} + 3 z^{2} \leq 16, (x - 2)^{2} + (y - 1)^{2} + (z - 1)^{2} \geq ε^{2}}

The boundary,

\partial V_{ε},

V

consists of two parts —

S

and

S_{ε},

with the normals as in the figure above. The divergence

\nabla \nabla \cdot F

F

is well-defined and zero throughout

V_{ε} .

Consequently, the divergence theorem gives

\begin{aligned} 0 & = ∭_{V} \nabla \nabla \cdot F d V = \iint_{S} F \cdot \hat{n} d S + \iint_{S_{ε}} F \cdot \hat{n} d S \end{aligned}

\iint_{S} F \cdot \hat{n} d S = - \iint_{S_{ε}} F \cdot \hat{n} d S

The unit normal to

S_{ε}

at the point

(x, y, z)

S_{ε}

\hat{n} = - \frac{1}{ε} [(x - 2) \hat{ı ı} + (y - 1) \hat{ȷ ȷ} + (z - 1) \hat{k}]

(Recall that

| (2 - 1) \hat{ı ı} + (y - 1) \hat{ȷ ȷ} + (z - 1) \hat{k} | = ε

S_{ε} .

So, on

S_{ε},

\begin{aligned} F \cdot \hat{n} & = - \frac{1}{ε} (\frac{(x, y, z) - (2, 1, 1)}{{[(x - 2)^{2} + (y - 1)^{2} + (z - 1)^{2}]}^{3 / 2}}) \\ \cdot [(x - 2) \hat{ı ı} + (y - 1) \hat{ȷ ȷ} + (z - 1) \hat{k}] \\ = - \frac{1}{ε} (\frac{(x - 2)^{2} + (y - 1)^{2} + (z - 1)^{2}}{{[(x - 2)^{2} + (y - 1)^{2} + (z - 1)^{2}]}^{3 / 2}}) \\ = - \frac{1}{ε^{2}} \end{aligned}

Hence

\begin{matrix} \iint_{S} F \cdot \hat{n} d S = - \iint_{S_{ε}} (- \frac{1}{ε^{2}}) d S = \frac{1}{ε^{2}} (4 π ε^{2}) = 4 π \end{matrix}

4.2.6.44. (✳).

Solution.

This was part of Theorem 4.2.9. To prove it apply the divergence theorem, but with

F

replaced by

a \times F,

where

a

is any constant vector.

\begin{aligned} \iint_{\partial Ω} (a \times F) \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot (a \times F) d V \\ = ∭_{Ω} [F \cdot \underset{= 0}{\underset{⏟}{(\nabla \nabla \times a)}} - a \cdot (\nabla \nabla \times F)] d V \\ = - ∭_{Ω} a \cdot (\nabla \nabla \times F) d V = - a \cdot ∭_{Ω} \nabla \nabla \times F d V \end{aligned}

To get the second line we used the vector identity Theorem 4.1.4.d. To get the third line, we used that

a

is a constant, so that all of its derivatives are zero. For all vectors

a \cdot (b \times c) = (a \times b) \cdot c

(in case you don’t remember this, it was Lemma 4.1.8.a) so that

(a \times F) \cdot \hat{n} = a \cdot (F \times \hat{n})

and

\begin{aligned} a \cdot \iint_{\partial Ω} F \times n d S = - a \cdot ∭_{Ω} \nabla \nabla \times F d V \\ ⟹ & a \cdot {\iint_{\partial Ω} F \times n d S + ∭_{Ω} \nabla \nabla \times F d V} = 0 \end{aligned}

In particular, choosing

a = \hat{ı ı},

\hat{ȷ ȷ}

and

\hat{k},

we see that all three components of the vector

\iint_{\partial Ω} F \times n d S + ∭_{Ω} \nabla \nabla \times F d V

are zero. So

∭_{Ω} \nabla \nabla \times F d V = - \iint_{\partial Ω} F \times n d S = \iint_{\partial Ω} \hat{n} \times F d S

which is what we wanted show.

4.2.6.45. (✳).

Solution.

Pressure is force per unit surface area acting normally into a surface. So the force per unit surface area is

- p \hat{n} .

The total force acting on

S

- \iint_{S} p \hat{n} d S = - ∭_{E} \nabla p d V

We are assuming that

p

is a constant, so that

\nabla p = 0

and the total force is zero.

4.2.6.46. (✳).

Solution.

Let

S_{a}

denote the sphere

x^{2} + y^{2} + z^{2} = a^{2}

and

V_{a}

denote the solid inside it, which is the ball

x^{2} + y^{2} + z^{2} \leq a^{2} .

Then, by the divergence theorem, Theorem 4.2.2,

π (a^{3} + 2 a^{4}) = \iint_{S_{a}} F \cdot \hat{n} d S = ∭_{V_{a}} \nabla \nabla \cdot F d V

Now, for very small

a,

\nabla \nabla \cdot F

is almost equal to

\nabla \nabla \cdot F (0, 0, 0)

on all of

V_{a},

and the integral

∭_{V_{a}} \nabla \nabla \cdot F d V

will be

\nabla \nabla \cdot F (0, 0, 0) V o l u m e (V_{a}) + O (a^{4}) = \frac{4}{3} π a^{3} \nabla \nabla \cdot F (0, 0, 0) + O (a^{4})

Here

O (a^{4})

is an error term that is bounded by a constant times

a^{4} .

This is consistent with the above equation if and only if

\nabla \nabla \cdot F (0, 0, 0) = \frac{3}{4} .

4.2.6.47. (✳).

Solution.

Note that, since

z^{2} - 2 a z = (z - a)^{2} - a^{2},

S = {(x, y, z | x^{2} + y^{2} + (z - a)^{2} = 4 a^{2}, z \geq 0}

Let

V

be the solid

V = {(x, y, z) | x^{2} + y^{2} + (z - a)^{2} \leq 4 a^{2}, z \geq 0}

It is the interior of the sphere of radius

2 a

centred on

(0, 0, a) .

The surface of

V

(with outward normal) is the union of

S

(with normal pointing away from the origin) and the disk

B = {(x, y, 0) | x^{2} + y^{2} \leq 3 a^{2}}

with normal

- \hat{k} .

Hence, by the Divergence Theorem

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} n \cdot F d V - \iint_{B} F \cdot (- \hat{k}) d S \\ = ∭_{V} (2 x + 2 y + 1) d V - \iint_{B} (- 3 - x) d S \end{aligned}

Both

V

and

B

are invariant under

x \to - x

and under

y \to - y,

∭_{V} x d V = ∭_{V} y d V = \iint_{B} x d S = 0

and

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} d V + 3 \iint_{B} d S \end{aligned}

To evaluate the integral over

V,

we note that

z

runs from

0

3 a

and that the cross section of

V = {(x, y, z) | 0 \leq z \leq 3 a, x^{2} + y^{2} \leq 4 a^{2} - (z - a)^{2}, z \geq 0}

with fixed

z

is the circular disk

x^{2} + y^{2} \leq 4 a^{2} - (z - a)^{2} = 3 a^{2} + 2 a z - z^{2},

which has area

π (\sqrt{3 a^{2} + 2 a z - z^{2}})^{2} .

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{0}^{3 a} π (\sqrt{3 a^{2} + 2 a z - z^{2}})^{2} d z + 3 Area (B) \\ = π \int_{0}^{3 a} (3 a^{2} + 2 a z - z^{2}) d z + 3 π (3 a^{2}) \\ = π (3 a^{2} \times 3 a + 2 a \times \frac{9 a^{2}}{2} - \frac{27 a^{3}}{3}) + 9 π a^{2} \\ = 9 π a^{3} + 9 π a^{2} \end{aligned}

4.2.6.48. (✳).

Solution.

(a) Let

S

denote the boundary of

R .

Then “the total flux of

F = \nabla u

out through the boundary of

R

” is given by the integral

I = \iint_{S} F \cdot \hat{n} d S

Thanks to the divergence theorem,

\begin{aligned} I & = ∭_{R} \nabla \cdot F d V = ∭_{R} \nabla \cdot \nabla u d V = ∭_{R} (\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} + \frac{\partial^{2} u}{\partial z^{2}}) d V \\ = 0 \end{aligned}

(b) Similarly, “the total flux of

G = u \nabla u

out through the boundary of

R

” equals

J = \iint_{S} G \cdot \hat{n} d S = ∭_{R} \nabla \cdot G d V

Here

G = u F

(using the notation from part (a)), so by the vector identity of Theorem 4.1.4.c,

\nabla \cdot G = \nabla \cdot (u F) = (\nabla u) \cdot F + u (\nabla \cdot F)

But

F = \nabla u,

\nabla \cdot F = Δ u = 0

as in part (a), giving

\nabla \cdot G = | \nabla u |^{2} + 0

In conclusion,

J = ∭_{R} \nabla \cdot G d V = ∭_{R} [(\frac{\partial u}{\partial x})^{2} + (\frac{\partial u}{\partial y})^{2} + (\frac{\partial u}{\partial z})^{2}] d V

4.2.6.49. (✳).

Solution.

(a) This is a classic case for the divergence theorem. The flux we want equals

I = \iint_{S} F \cdot \hat{n} d S = ∭_{R} \nabla \cdot F d V = ∭_{R} (2 x + 2 - 2) d V = 2 ∭_{R} x d V

The solid

R

clearly has reflection symmetry across the plane

x = 0 .

So the

x

-coordinate of the centre of mass of

R,

i.e. the average value of

x

over

R,

i.e.

\bar{x} = \frac{∭_{R} x d V}{∭_{R} d V} = \frac{∭_{R} x d V}{V o l (R)}

is zero. Hence

I = 2 \bar{x} V o l (R) = 0

Alternatively, here is a direct evaluation of

2 ∭_{R} x d V .

The base region

x^{2} + (y - 1)^{2} \leq 1

is the circular disk of radius

1

centred on

(0, 1) .

In polar coordinates it is

r^{2} \cos^{2} θ + (r \sin θ - 1)^{2} \leq 1 or r^{2} - 2 r \sin θ + 1 \leq 1 or r \leq 2 \sin θ

Because the disk is contained in the upper half plane, the polar angle

θ

is restricted to

0 \leq θ \leq π .

So, in cylindrical coordinates, the solid

R

is described by

0 \leq θ \leq π, 0 \leq r \leq 2 \sin θ, 0 \leq z \leq r^{2} \sin^{2} θ

Hence

\begin{aligned} I & = 2 \int_{θ = 0}^{π} \int_{r = 0}^{2 \sin θ} \int_{z = 0}^{r^{2} \sin^{2} θ} (r \cos θ) d z r d r d θ \\ = 2 \int_{θ = 0}^{π} \int_{r = 0}^{2 \sin θ} r^{4} \sin^{2} θ \cos θ d r d θ \\ = 2 \int_{θ = 0}^{π} \sin^{2} θ \cos θ [\frac{2^{5} \sin^{5} θ}{5}] d θ \\ = \frac{64}{5} \int_{θ = 0}^{π} \sin^{7} θ \cos θ d θ \\ = \frac{64}{5} [\frac{\sin^{8} θ}{8}]_{θ = 0}^{π} \\ = 0 \end{aligned}

(b) using part (a): We have

\iint_{S} F \cdot \hat{n} d S = \iint_{S_{b o t t o m}} F \cdot \hat{n} d S + \iint_{S_{t o p}} F \cdot \hat{n} d S + \iint_{S_{s i d e}} F \cdot \hat{n} d S

S_{b o t t o m},

z = 0

and the outward unit normal is

\hat{n} = - \hat{k},

F \cdot \hat{n} = 0 .

Hence

\iint_{S_{b o t t o m}} F \cdot \hat{n} d S = \iint_{S_{b o t}} 0 d S = 0

S_{t o p},

z = y^{2},

F = (2 x, 2 y, - 2 y^{2})

and, by (3.3.2),

\hat{n} d S = (0, - 2 y, 1) d x d y

Hence (by the Hint)

\begin{aligned} \iint_{S_{b o t}} F \cdot \hat{n} d S & = \iint_{D} [- 4 y^{2} - 2 y^{2}] d x d y \\ = - 6 \int_{θ = 0}^{π} \int_{r = 0}^{2 \sin θ} (r^{2} \sin^{2} θ) r d r d θ \\ = - 6 \frac{2^{4}}{4} \int_{θ = 0}^{π} \sin^{6} θ d θ \\ = - 24 \frac{5}{6} \int_{θ = 0}^{π} \sin^{4} θ d θ \\ = - 24 \frac{5}{6} \frac{3}{4} \int_{θ = 0}^{π} \sin^{2} θ d θ \\ = - 24 \frac{5}{6} \frac{3}{4} \frac{1}{2} \int_{θ = 0}^{π} d θ \\ = - 24 [\frac{5}{6} \frac{3}{4} \frac{1}{2} π] \\ = - \frac{15}{2} π \end{aligned}

The conclusion is

\iint_{S_{s i d e}} F \cdot \hat{n} d S = \iint_{S} F \cdot \hat{n} d S - \iint_{S_{t o p}} F \cdot \hat{n} d S - \iint_{S_{b o t}} F \cdot \hat{n} d S = \frac{15}{2} π

(b) by direct evaluation: Use the polar equation

r = 2 \sin θ

to parametrize

S_{s i d e} :

\begin{aligned} r (θ, t) = (r \cos θ, r \sin θ, t) = (2 \sin θ \cos θ, 2 \sin^{2} θ, t), \\ 0 \leq θ \leq π, 0 \leq t \leq y^{2} = 4 \sin^{4} θ \end{aligned}

Then using (3.3.1),

\begin{aligned} F \cdot \hat{n} d S & = F \cdot (\frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial t}) d θ d t \\ = det [\begin{array}{c} 4 \sin^{2} θ \cos^{2} θ & 4 \sin^{2} θ & - 2 t \\ 2 (\cos^{2} θ - \sin^{2} θ) & 4 \sin θ \cos θ & 0 \\ 0 & 0 & 1 \end{array}] d θ d t \\ = det [\begin{array}{c} 4 \sin^{2} θ \cos^{2} θ & 4 \sin^{2} θ \\ 2 (\cos^{2} θ - \sin^{2} θ) & 4 \sin θ \cos θ \end{array}] d θ d t \\ = [16 \sin^{3} θ \cos^{3} θ - 8 \sin^{2} θ (\cos^{2} θ - \sin^{2} θ)] d θ d t \\ = [16 \sin^{3} θ (1 - \sin^{2} θ) \cos θ - 8 \sin^{2} θ (1 - 2 \sin^{2} θ)] d θ d t \\ = 8 [2 \sin^{3} θ \cos θ - 2 \sin^{5} θ \cos θ - \sin^{2} θ + 2 \sin^{4} θ] d θ d t \end{aligned}

\begin{aligned} \iint_{S_{s i d e}} F \cdot \hat{n} d S \\ = 8 \int_{θ = 0}^{π} \int_{t = 0}^{4 \sin^{4} θ} [2 \sin^{3} θ \cos θ - 2 \sin^{5} θ \cos θ - \sin^{2} θ + 2 \sin^{4} θ] d t d θ \\ = 32 \int_{θ = 0}^{π} [2 \sin^{7} θ \cos θ - 2 \sin^{9} θ \cos θ - \sin^{6} θ + 2 \sin^{8} θ] d θ \\ = 32 {[2 \frac{\sin^{8} θ}{8} - 2 \frac{\sin^{10} θ}{10}]}_{0}^{π} - 32 \int_{θ = 0}^{π} \sin^{6} θ d θ + 64 \int_{θ = 0}^{π} \sin^{8} θ d θ \\ = - 32 \frac{5}{6} \frac{3}{4} \frac{1}{2} π + 64 \frac{7}{8} \frac{5}{6} \frac{3}{4} \frac{1}{2} π (by the Hint as above) \\ = \frac{15}{2} π . \end{aligned}

(b) Offset polar alternative: We can also parametrize

S

using cylindrical coordinates translated so that the centre of the base of the cylinder, namely

(0, 1, 0),

plays the role of the origin. Then, looking at the figure

we see that

x = r \cos θ y = 1 + r \sin θ z = z

In these coordinates, the base region,

x^{2} + (y - 1)^{2} \leq 1,

z = 0,

of the cylinder is

0 \leq r \leq 1,

z = 0 .

So we can parametrize

S

x = \cos θ, y = 1 + \sin θ, z = t, 0 \leq θ \leq 2 π, 0 \leq t \leq (1 + \sin θ)^{2}

By (3.3.1),

\begin{aligned} \frac{\partial r}{\partial t} \times \frac{\partial r}{\partial θ} = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 0 & 0 & 1 \\ - \sin θ & \cos θ & 0 \end{array}] = (- \cos θ, - \sin θ, 0), \\ \hat{n} d S & = - \frac{\partial r}{\partial t} \times \frac{\partial r}{\partial θ} d t d θ = (\cos θ, \sin θ, 0) d t d θ \end{aligned}

where we have chosen the sign to give the outward pointing normal. So

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \int_{θ = 0}^{2 π} \int_{t = 0}^{(1 + \sin θ)^{2}} [\cos^{3} θ + 2 (1 + \sin θ) \sin θ] d t d θ \\ = \int_{0}^{2 π} [(1 + \sin θ)^{2} \cos^{3} θ + 2 (1 + \sin θ)^{3} \sin θ] d θ \\ = \int_{0}^{2 π} [2 \sin θ \cos^{3} θ + 6 \sin^{2} θ + 2 \sin^{4} θ] d θ \\ = - \frac{1}{2} \cos^{4} θ |_{0}^{2 π} + 12 \int_{0}^{π} \sin^{2} θ d θ + 4 \int_{0}^{π} \sin^{4} θ d θ \\ = 0 + 12 \frac{π}{2} + 4 \frac{3}{4} \frac{π}{2} = \frac{15}{2} π \end{aligned}

To get the third line, we used that the integral over

0 \leq θ \leq 2 π

of any odd power of

\sin θ

\cos θ

is zero.

4.2.6.50. (✳).

Solution.

The circle

x^{2} + y^{2} = 4 y,

or equivalently,

x^{2} + (y - 2)^{2} = 4,

has radius

2

and centre

(0, 2) .

On the bottom surface,

z = 0

and the outward normal is

- \hat{k},

so that

\iint_{D} F \cdot \hat{n} d S = - \iint_{D} F \cdot \hat{k} d x d y = - \iint_{D} (2 x + 3 y) d x d y

By symmetry, the centre of mass,

(\bar{x}, \bar{y}),

of the circle is

(0, 2) .

Here

\bar{x}

and

\bar{y}

are the average values

\begin{matrix} \bar{x} = \frac{\iint_{D} x d x d y}{\iint_{D} d x d y} \bar{y} = \frac{\iint_{D} y d x d y}{\iint_{D} d x d y} \end{matrix}

x

and

y

over

D .

As the disk

D

has area

4 π,

\iint_{D} x d x d y = 4 π \bar{x} = 0 \iint_{D} y d x d y = 4 π \bar{y} = 8 π

and

\iint_{D} F \cdot \hat{n} d S = - 4 π (2 \bar{x} + 3 \bar{y}) = - 4 π (2 \times 0 + 3 \times 2) = - 24 π

\begin{aligned} \nabla \nabla \cdot F & = \frac{\partial}{\partial x} (x + x^{2} y) + \frac{\partial}{\partial y} (y - x y^{2}) + \frac{\partial}{\partial z} (z + 2 x + 3 y) \\ = (1 + 2 x y) + (1 - 2 x y) + (1) \\ = 3 \end{aligned}

the divergence theorem gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{R} \nabla \nabla \cdot F d V - \iint_{D} F \cdot \hat{n} d S \\ = ∭_{R} 3 d V - (- 24 π) = 3 Vol (R) + 24 π = 3 \times 10 + 24 π \\ = 30 + 24 π \end{aligned}

4.2.6.51. (✳).

Solution.

(a) This question is very similar to question 4.2.6.22 above. The only difference is that the term

x \sin z

in the

F

of question 4.2.6.22 has been replaced with

x \sqrt[3]{\tan z}

in this question. But that is a significant change, because

\tan z

is singular (i.e. becomes infinite) at

z = \frac{π}{2}

and

E

includes many points with

z = \frac{π}{2} .

Consequently, we may not simply apply the divergence theorem (Theorem 4.2.2) to

E .

Instead we treat the integrals in this question as improper integrals. To be precise define, for each

ϵ > 0,

\begin{aligned} E_{ϵ}^{+} & = {(x, y, z | x^{2} + y^{2} \leq z, \frac{π}{2} + ϵ \leq z \leq 4} \\ E_{ϵ}^{-} & = {(x, y, z | x^{2} + y^{2} \leq z, 0 \leq z \leq \frac{π}{2} - ϵ} \end{aligned}

Here are sketches of the parts of

E_{ϵ}^{+}

and

E_{ϵ}^{-}

that lie in first octant.

The union of

E_{ϵ}^{+}

and

E_{ϵ}^{-}

is exactly

E

with a thin disk (of width

2 ϵ

) removed. We are allowed to apply the divergence theorem (Theorem 4.2.2) to both

E_{ϵ}^{+}

and

E_{ϵ}^{-} .

It gives

\begin{aligned} \iint_{\partial E_{ϵ}^{+}} F \cdot \hat{n} d S & = ∭_{E_{ϵ}^{+}} \nabla \nabla \cdot F d V = ∭_{E_{ϵ}^{+}} (- x^{2} - y^{2} + 4) d V \\ \iint_{\partial E_{ϵ}^{-}} F \cdot \hat{n} d S & = ∭_{E_{ϵ}^{-}} \nabla \nabla \cdot F d V = ∭_{E_{ϵ}^{-}} (- x^{2} - y^{2} + 4) d V \end{aligned}

The boundary

\partial E_{ϵ}^{+}

E_{ϵ}^{+}

consists of

a horizontal top disk
$D = {(x, y, z) | z = 4, x^{2} + y^{2} \leq 4}$
(which happens to be independent of $ϵ$ ) with outward normal $\hat{k},$
a horizontal bottom disk
$D_{ϵ}^{+} = {(x, y, z) | z = \frac{π}{2} + ϵ, x^{2} + y^{2} \leq z}$
with outward normal $- \hat{k},$ and
the part
$S_{ϵ}^{+} = {(x, y, z) | x^{2} + y^{2} = z, \frac{π}{2} + ϵ \leq z \leq 4}$
of the paraboloid $z = x^{2} + y^{2},$ with outward normal $\hat{n} = - \hat{N},$ in terms of the $\hat{N}$ specified in part (b) of this question

and the boundary

\partial E_{ϵ}^{-}

E_{ϵ}^{-}

consists of

a horizontal top disk
$D_{ϵ}^{-} = {(x, y, z) | z = \frac{π}{2} - ϵ, x^{2} + y^{2} \leq z}$
with outward normal $\hat{k},$ and
the part
$S_{ϵ}^{-} = {(x, y, z) | x^{2} + y^{2} = z, 0 \leq z \leq \frac{π}{2} - ϵ}$
of the paraboloid $z = x^{2} + y^{2},$ with outward normal $\hat{n} = - \hat{N},$ in terms of the $\hat{N}$ specified in part (b) of this question.

So the two divergence theorem equations above are

\begin{aligned} ∭_{E_{ϵ}^{+}} (- x^{2} - y^{2} + 4) d V & = - \iint_{S_{ϵ}^{+}} F \cdot \hat{N} d S + \iint_{D} \overset{F \cdot \hat{k}}{\overset{⏞}{4 z}} d S - \iint_{D_{ϵ}^{+}} \overset{F \cdot \hat{k}}{\overset{⏞}{4 z}} d S \\ ∭_{E_{ϵ}^{-}} (- x^{2} - y^{2} + 4) d V & = - \iint_{S_{ϵ}^{-}} F \cdot \hat{N} d S + \iint_{D_{ϵ}^{-}} \overset{F \cdot \hat{k}}{\overset{⏞}{4 z}} d S \end{aligned}

Adding these two equations together,

\begin{aligned} ∭_{E_{ϵ}^{+}} (- x^{2} - y^{2} + 4) d V + ∭_{E_{ϵ}^{-}} (- x^{2} - y^{2} + 4) d V \\ = - \iint_{S_{ϵ}^{+}} F \cdot \hat{N} d S - \iint_{S_{ϵ}^{-}} F \cdot \hat{N} d S + \iint_{D} 4 z d S \\ (*) & - \iint_{D_{ϵ}^{+}} 4 z d S + \iint_{D_{ϵ}^{-}} 4 z d S \end{aligned}

In the limit

ϵ \to 0,

the union of $E_{ϵ}^{+}$ and $E_{ϵ}^{-}$ becomes exactly $E,$
the union of $S_{ϵ}^{+}$ and $S_{ϵ}^{-}$ becomes exactly
$S = {(x, y, z) | x^{2} + y^{2} = z, 0 \leq z \leq 4}$
both $D_{ϵ}^{+}$ and $D_{ϵ}^{-}$ become
${(x, y, z) | z = \frac{π}{2}, x^{2} + y^{2} \leq z}$

lim_{ϵ \to 0} [- \iint_{D_{ϵ}^{+}} 4 z d S + \iint_{D_{ϵ}^{-}} 4 z d S] = 0

and taking the limit

ϵ \to 0

(*)

gives

\begin{aligned} ∭_{E} (- x^{2} - y^{2} + 4) d V & = - \iint_{S} F \cdot \hat{N} d S + \iint_{D} 4 z d S \\ = \iint_{\partial E} F \cdot \hat{n} d S \end{aligned}

To evaluate the integral on the left hand side we switch to cylindrical coordinates. In cylindrical coordinates

E = {(r \cos θ, r \sin θ, z) | 0 \leq z \leq 4, r^{2} \leq z}

\begin{aligned} \iint_{\partial E} F \cdot \hat{n} d S & = \int_{0}^{4} d z \int_{0}^{\sqrt{z}} d r r \int_{0}^{2 π} d θ (- r^{2} + 4) \\ = 2 π \int_{0}^{4} d z \int_{0}^{\sqrt{z}} d r (4 r - r^{3}) \\ = 2 π \int_{0}^{4} d z (2 z - \frac{z^{2}}{4}) \\ = 2 π [z^{2} - \frac{z^{3}}{12}]_{0}^{4} = 2 π [16 - \frac{16}{3}] = \frac{64}{3} π \end{aligned}

(b) The boundary of

E

consists of two parts —

S,

but with downward pointing normal, on the bottom and the disk

D = {(x, y, z) | z = 4, x^{2} + y^{2} \leq 4}

with normal

\hat{k},

on top.

So, by part (a),

\begin{aligned} \frac{64}{3} π = \iint_{\partial E} F \cdot \hat{n} d S & = - \iint_{S} F \cdot \hat{N} d S + \iint_{D} F \cdot \hat{k} d S \\ = - \iint_{S} F \cdot \hat{N} d S + \iint_{D} \overset{F \cdot \hat{k}}{\overset{⏞}{4 z}} d S \end{aligned}

Since

z = 4

D,

and

D

is a disk of radius 2,

\begin{matrix} \iint_{S} F \cdot \hat{N} d S = - \frac{64}{3} π + 16 \iint_{D} d S = - \frac{64}{3} π + 16 (4 π) = \frac{128}{3} π \end{matrix}

4.3 Green’s Theorem

Exercises

4.3.1.

Solution.

(a) Expressing the left hand side as an iterated integral, with

y

as the inner integration variable, we have

\begin{aligned} \iint_{R} \frac{\partial f}{\partial y} (x, y) d x d y & = \int_{0}^{1} d x [\int_{0}^{1} d y \frac{\partial f}{\partial y} (x, y)] \\ = \int_{0}^{1} d x [f (x, 1) - f (x, 0)] \\ by the fundamental theorem of calculus \\ = \int_{0}^{1} f (x, 1) d x - \int_{0}^{1} f (x, 0) d x \end{aligned}

(b) Define

F_{1} (x, y) = f (x, y)

and

F_{2} (x, y) = 0 .

Then. by Green’s theorem

\begin{aligned} \iint_{R} \frac{\partial f}{\partial y} (x, y) d x d y & = - \iint_{R} [\frac{\partial F_{2}}{\partial x} (x, y) - \frac{\partial F_{1}}{\partial y} (x, y)] d x d y \\ = - \int_{\partial R} [F_{1} (x, y) d x + F_{2} (x, y) d y] \\ = - \int_{\partial R} f (x, y) d x \end{aligned}

The boundary of

R,

oriented counterclockwise, is the union of four line segments.

\begin{matrix} C_{1} from (0, 0) to (1, 0) \\ C_{2} from (1, 0) to (1, 1) \\ C_{3} from (1, 1) to (0, 1) \\ C_{4} from (0, 1) to (0, 0) \end{matrix}

Now

x

is constant on

C_{2}

and

C_{4}

so that

\int_{C_{2}} f (x, y) d x = \int_{C_{4}} f (x, y) d x = 0

So, using

- C_{3}

to denote the line segment from

(0, 1)

(1, 1)

\begin{aligned} \iint_{R} \frac{\partial f}{\partial y} (x, y) d x d y & = - [\int_{C_{1}} f (x, y) d x + \int_{C_{3}} f (x, y) d x] \\ = \int_{- C_{3}} f (x, y) d x - \int_{C_{1}} f (x, y) d x \\ = \int_{0}^{1} f (x, 1) d x - \int_{0}^{1} f (x, 0) d x \end{aligned}

4.3.2.

Solution.

Let

r (s) = x (s) \hat{ı ı} + y (s) \hat{ȷ ȷ}

be a counterclockwise parametrization of

C

by arc length. Then

\hat{T} (s) = r^{'} (s) = x^{'} (s) \hat{ı ı} + y^{'} (s) \hat{ȷ ȷ}

is the forward pointing unit tangent vector to

C

r (s)

and

\hat{n} (s) = r^{'} (s) \times \hat{k} = y^{'} (s) \hat{ı ı} - x^{'} (s) \hat{ȷ ȷ} .

To see that

r^{'} (s) \times \hat{k}

really is

\hat{n} (s),

note that

y^{'} (s) \hat{ı ı} - x^{'} (s) \hat{ȷ ȷ}

has the same length, namely $1,$ as $r^{'} (s)$ (recall that $r (s)$ is a parametrization by arc length),
lies in the $x y$ -plane and
is perpendicular to $r^{'} (s) .$ (Check that $r^{'} (s) \cdot [y^{'} (s) \hat{ı ı} - x^{'} (s) \hat{ȷ ȷ}] = 0 .$ )
Use the right hand rule to check that $r^{'} (s) \times \hat{k}$ is $\hat{n}$ rather than $- \hat{n} .$

So, by Green’s theorem,

\begin{aligned} \oint_{C} F \cdot \hat{n} d s & = \oint_{C} [F_{1} \frac{d y}{d s} - F_{2} \frac{d x}{d s}] d s = \oint_{C} [- F_{2} d x + F_{1} d y] \\ = \iint_{R} [\frac{\partial}{\partial x} F_{1} - \frac{\partial}{\partial y} (- F_{2})] d x d y \\ = \iint_{R} \nabla \nabla \cdot F d x d y \end{aligned}

4.3.3.

Solution.

(a) Parametrize the circle by

x = a \cos θ, y = a \sin θ,

0 \leq θ \leq 2 π .

Then

d x = - a \sin θ d θ

and

d y = a \cos θ d θ

so that

\begin{aligned} \frac{1}{2 π} \oint_{C} \frac{x d y - y d x}{x^{2} + y^{2}} & = \frac{1}{2 π} \int_{0}^{2 π} \frac{a^{2} \cos^{2} θ d θ + a^{2} \sin^{2} θ d θ}{a^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = \frac{1}{2 π} \int_{0}^{2 π} d θ \\ = 1 \end{aligned}

(b) The boundary of the square has four sides — one with

y = - 1,

one with

x = 1,

one with

y = 1

and one with

x = - 1 .

To evaluate the integrals over the four sides

parametrize the $y = - 1$ part by $x$ so that $r (x) = x \hat{ı ı} - \hat{ȷ ȷ},$ $r^{'} (x) = \hat{ı ı},$ with $x$ running from $- 1$ to $1,$
parametrize the $x = + 1$ part by $y$ so that $r (y) = \hat{ı ı} + y \hat{ȷ ȷ},$ $r^{'} (y) = \hat{ȷ ȷ},$ with $y$ running from $- 1$ to $1,$
parametrize the $y = + 1$ part by $x$ so that $r (x) = x \hat{ı ı} + \hat{ȷ ȷ},$ $r^{'} (x) = \hat{ı ı},$ with $x$ running from $1$ to $- 1,$ and
parametrize the $x = - 1$ part by $y$ so that $r (y) = - \hat{ı ı} + y \hat{ȷ ȷ},$ $r^{'} (y) = \hat{ȷ ȷ},$ with $y$ running from $1$ to $- 1,$

so that the integral

\begin{aligned} \frac{1}{2 π} \oint_{C} \frac{x d y - y d x}{x^{2} + y^{2}} & = \frac{1}{2 π} \overset{y = - 1 p a r t}{\overset{⏞}{\int_{- 1}^{1} \frac{- (- 1) d x}{x^{2} + 1}}} + \frac{1}{2 π} \overset{x = + 1 p a r t}{\overset{⏞}{\int_{- 1}^{1} \frac{(1) d y}{1 + y^{2}}}} \\ + \frac{1}{2 π} \overset{y = + 1 p a r t}{\overset{⏞}{\int_{1}^{- 1} \frac{- (1) d x}{x^{2} + 1}}} + \frac{1}{2 π} \overset{x = - 1 p a r t}{\overset{⏞}{\int_{1}^{- 1} \frac{(- 1) d y}{1 + y^{2}}}} \\ = 4 \frac{1}{2 π} \arctan x |_{- 1}^{1} = \frac{2}{π} [\frac{π}{4} + \frac{π}{4}] = 1 \end{aligned}

a = \sqrt{2},

but with

θ

running from

0

π,

the outer semicircle gives

\frac{1}{2 π} \int_{0}^{π} \frac{a^{2} \cos^{2} θ d θ + a^{2} \sin^{2} θ d θ}{a^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = \frac{1}{2 π} \int_{0}^{π} d θ = \frac{1}{2}

As in part (a) with

a = 1,

but with

θ

running from

π

0,

the inner semicircle gives

\frac{1}{2 π} \int_{π}^{0} \frac{a^{2} \cos^{2} θ d θ + a^{2} \sin^{2} θ d θ}{a^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = \frac{1}{2 π} \int_{π}^{0} d θ = - \frac{1}{2}

The two flat pieces each give zero, since on them

y = 0

and

d y = 0 .

\frac{1}{2 π} \oint_{C} \frac{x d y - y d x}{x^{2} + y^{2}} = \frac{1}{2} + 0 - \frac{1}{2} + 0 = 0

An alternate solution, that uses Green’s theorem, is given in the solution to the next question.

4.3.4.

Solution.

The two partial derivatives

\begin{aligned} \frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}}) & = \frac{(x^{2} + y^{2}) - x (2 x)}{{(x^{2} + y^{2})}^{2}} & = \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} \\ \frac{\partial}{\partial y} (\frac{- y}{x^{2} + y^{2}}) & = \frac{- (x^{2} + y^{2}) - (- y) (2 y)}{{(x^{2} + y^{2})}^{2}} & = \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} \end{aligned}

are well-defined and equal everywhere except at the origin

(0, 0) .

Short discussion: Were it not for the singularity at

(0, 0),

the vector field of the last problem would be conservative and the integral

\int F \cdot d r

around any closed curve would be zero. But as we saw in parts (a) and (b) of Q[4.3.3], this is not the case. On the other hand, by Green’s theorem (Theorem 4.3.2), the integral around the boundary of any region that does not contain

(0, 0)

is zero, as happened in part (c) of Q[4.3.3].

Long discussion: First consider part (c) of Q[4.3.3]. The curve

C

is the boundary of the region

R = {(x, y) | 1 \leq x^{2} + y^{2} \leq 2, y \geq 0}

The partial derivatives

\frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}})

and

\frac{\partial}{\partial y} (\frac{- y}{x^{2} + y^{2}})

are well-defined and equal everywhere in

R .

So by Green’s theorem

\begin{aligned} \frac{1}{2 π} \oint_{C} \frac{x d y - y d x}{x^{2} + y^{2}} & = \frac{1}{2 π} \iint_{R} [\frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}}) - \frac{\partial}{\partial y} (\frac{- y}{x^{2} + y^{2}})] d x d y \\ = 0 \end{aligned}

which is the answer we got before.

We cannot apply Green’s theorem in this way for parts (a) and (b) of Q[4.3.3] because the singularity at

(0, 0)

is inside the curve

C

for both parts (a) and (b). On the other hand suppose, for simplicity, that

0 < a < 1 .

Denote by

C_{a},

C_{b}

the curves of parts (a) and (b), respectively. Define

R

to be the set of points that are inside

C_{b}

and outside

C_{a} .

That is,

R = {(x, y) | - 1 \leq x \leq 1, - 1 \leq y \leq 1, x^{2} + y^{2} \geq a^{2}}

Then the boundary,

\partial R,

R

consists of two parts. One part is

C_{b} .

The other part is

C_{a},

but oriented clockwise rather than counterclockwise. We’ll call it

- C_{a} .

Again the partial derivatives

\frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}})

and

\frac{\partial}{\partial y} (\frac{- y}{x^{2} + y^{2}})

are well-defined and equal everywhere in

R .

So by Green’s theorem

\begin{aligned} \frac{1}{2 π} \oint_{\partial R} \frac{x d y - y d x}{x^{2} + y^{2}} & = \frac{1}{2 π} \iint_{R} [\frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}}) - \frac{\partial}{\partial y} (\frac{- y}{x^{2} + y^{2}})] d x d y \\ = 0 \end{aligned}

Consequently

\begin{aligned} 0 & = \frac{1}{2 π} \oint_{\partial R} \frac{x d y - y d x}{x^{2} + y^{2}} = \frac{1}{2 π} \oint_{C_{b}} \frac{x d y - y d x}{x^{2} + y^{2}} + \frac{1}{2 π} \oint_{- C_{a}} \frac{x d y - y d x}{x^{2} + y^{2}} \\ = \frac{1}{2 π} \oint_{C_{b}} \frac{x d y - y d x}{x^{2} + y^{2}} - \frac{1}{2 π} \oint_{C_{a}} \frac{x d y - y d x}{x^{2} + y^{2}} \end{aligned}

and we conclude that the answers to parts (a) and (b) should be the same.We did indeed see that in Q[4.3.3].

4.3.5.

Solution 1. Direct evaluation

Here is a sketch of

C .

The square consists of four line segments.

The bottom line segment may be parametrized $r (x) = (x, 0), 0 \leq x \leq 3 .$ So the line integral along this segment is
$\int_{0}^{3} F (r (x)) \cdot \frac{d r}{d x} d x = \int_{0}^{3} (0, 0) \cdot (1, 0) d x = 0$
The second line segment may be parametrized $r (y) = (3, y), 0 \leq y \leq 3 .$ So the line integral along this segment is
$\int_{0}^{3} F (r (y)) \cdot \frac{d r}{d y} d y = \int_{0}^{3} (9 y^{2}, 6 y) \cdot (0, 1) d y = \int_{0}^{3} 6 y d y = 27$
The third line segment may be parametrized $r (t) = (3 - t, 3), 0 \leq t \leq 3 .$ So the line integral along this segment is
$\begin{aligned} \int_{0}^{3} F (r (t)) \cdot \frac{d r}{d t} d t & = \int_{0}^{3} (9 (3 - t)^{2}, 6 (3 - t)) \cdot (- 1, 0) d t \\ = - \int_{0}^{3} 9 (3 - t)^{2} d t = - 81 \end{aligned}$
The final line segment may be parametrized $r (t) = (0, 3 - t), 0 \leq t \leq 3 .$ So the line integral along this segment is
$\int_{0}^{3} F (r (t)) \cdot \frac{d r}{d t} d t = \int_{0}^{3} (0, 0) \cdot (0, - 1) d t = 0$

The full line integral is

\oint_{C} F \cdot d r = 0 + 27 - 81 + 0 = - 54

Solution 2. By Green’s theorem

We apply Green’s Theorem.

\begin{aligned} \oint_{C} x^{2} y^{2} d x + 2 x y d y & = \int_{0}^{3} d x \int_{0}^{3} d y [\frac{\partial}{\partial x} (2 x y) - \frac{\partial}{\partial y} (x^{2} y^{2})] \\ = \int_{0}^{3} d x \int_{0}^{3} d y [2 y - 2 x^{2} y] \\ = \int_{0}^{3} d x [9 - 9 x^{2}] \\ = 27 - 9 \frac{3^{3}}{3} = - 54 \end{aligned}

4.3.6.

Solution.

Call the trapezoid

T .

By Green’s theorem,

\begin{aligned} \oint_{C} (x \sin y^{2} - y^{2}) d x + (x^{2} y \cos y^{2} + 3 x) d y \\ = \iint_{T} {\frac{\partial}{\partial x} (x^{2} y \cos y^{2} + 3 x) - \frac{\partial}{\partial y} (x \sin y^{2} - y^{2})} d x d y \\ = \iint_{T} (2 x y \cos y^{2} + 3 - 2 x y \cos y^{2} + 2 y) d x d y \\ = \iint_{T} (3 + 2 y) d x d y \end{aligned}

The integral

\iint_{T} (2 y) d x d y

vanishes because

2 y

changes sign under

y \to - y

while the domain of integration is invariant under

y \to - y .

The integral

\iint_{T} 3 d x d y

3

times the area of the trapezoid, which is its width (1) times the average of its heights

(\frac{1}{2} [2 + 4]) = 3 .

\oint_{C} (x \sin y^{2} - y^{2}) d x + (x^{2} y \cos y^{2} + 3 x) d y = 3 \times 1 \times 3 = 9

4.3.7. (✳).

Solution 1. Using Green’s theorem

By Green’s theorem (Theorem 4.3.2), using

D

to denote the half-disk

0 \leq y \leq \sqrt{4 - x^{2}},

\begin{aligned} \oint_{C} (\frac{1}{3} x^{2} y^{3} - x^{4} y) d x + (x y^{4} + x^{3} y^{2}) d y \\ = \iint_{D} [\frac{\partial}{\partial x} (x y^{4} + x^{3} y^{2}) - \frac{\partial}{\partial y} (\frac{1}{3} x^{2} y^{3} - x^{4} y)] d x d y \\ = \iint_{D} (x^{4} + 2 x^{2} y^{2} + y^{4}) d x d y \\ = \iint_{D} (x^{2} + y^{2})^{2} d x d y \end{aligned}

Switching to polar coordinates

\begin{aligned} \oint_{C} (\frac{1}{3} x^{2} y^{3} - x^{4} y) d x + (x y^{4} + x^{3} y^{2}) d y & = \int_{0}^{2} d r r \int_{0}^{π} d θ r^{4} = {π \frac{r^{6}}{6} |}_{0}^{2} \\ = \frac{32}{3} π \end{aligned}

Solution 2. By direct evaluation

Write

C

as the union of

C_{1},

the straight line from

(- 2, 0)

(2, 0),

and

C_{2},

the half-circle

r (θ) = x (θ) \hat{ı ı} + y (θ) \hat{ȷ ȷ} = 2 \cos θ \hat{ı ı} + 2 \sin θ \hat{ȷ ȷ},

0 \leq θ \leq π .

y = 0

at every point of

C_{1},

\int_{C_{1}} (\frac{1}{3} x^{2} y^{3} - x^{4} y) d x + (x y^{4} + x^{3} y^{2}) d y = 0

and

\begin{aligned} I & = \int_{C_{2}} (\frac{1}{3} x^{2} y^{3} - x^{4} y) d x + (x y^{4} + x^{3} y^{2}) d y \\ = \int_{0}^{π} [(\frac{1}{3} x (θ)^{2} y (θ)^{3} - x (θ)^{4} y (θ)) x^{'} (θ) \\ + (x (θ) y (θ)^{4} + x (θ)^{3} y (θ)^{2}) y^{'} (θ)] d θ \\ = \int_{0}^{π} [(\frac{1}{3} 2^{5} \cos^{2} θ \sin^{3} θ - 2^{5} \cos^{4} θ \sin θ) (- 2 \sin θ) \\ + (2^{5} \cos θ \sin^{4} θ + 2^{5} \cos^{3} θ \sin^{2} θ) (2 \cos θ)] d θ \\ = 2^{5} \int_{0}^{π} (\frac{4}{3} \cos^{2} θ \sin^{4} θ + 4 \cos^{4} θ \sin^{2} θ) d θ \\ = 2^{5} \int_{0}^{π} \sin^{2} (2 θ) (\frac{1}{3} \sin^{2} θ + \cos^{2} θ) d θ \\ = 2^{4} \int_{0}^{π} \sin^{2} (2 θ) (\frac{1}{3} [1 - \cos (2 θ)] + [1 + \cos (2 θ)]) d θ \\ since \cos (2 θ) = 2 \cos^{2} θ - 1 = 1 - 2 \sin^{2} θ \\ = \frac{2^{5}}{3} \int_{0}^{π} \sin^{2} (2 θ) [2 + \cos (2 θ)] d θ \\ = \frac{2^{5}}{3} \int_{0}^{π} [1 - \cos (4 θ) + \sin^{2} (2 θ) \cos (2 θ)] d θ \\ = \frac{2^{5}}{3} [θ - \frac{1}{4} \sin (4 θ) + \frac{1}{6} \sin^{3} (2 θ)]_{0}^{π} = \frac{32}{3} π \end{aligned}

4.3.8. (✳).

Solution.

Let’s use Green’s theorem. The rectangle, which we shall denote

R,

R = {(x, y)} 1 \leq x \leq 3, 0 \leq y \leq 1

So Green’s theorem gives

\begin{aligned} \oint_{C} (3 y^{2} + 2 x e^{y^{2}}) d x + (2 y x^{2} e^{y^{2}}) d y \\ = \iint_{R} [\frac{\partial}{\partial x} (2 y x^{2} e^{y^{2}}) - \frac{\partial}{\partial y} (3 y^{2} + 2 x e^{y^{2}})] d x d y \\ = \iint_{R} [4 x y e^{y^{2}} - 6 y - 4 x y e^{y^{2}}] d x d y \\ = - 6 \int_{1}^{3} d x \int_{0}^{1} d y y = - 6 \int_{1}^{3} d x \frac{1}{2} \\ = - 6 \end{aligned}

4.3.9. (✳).

Solution.

(a) The curves

y = x^{2} + 4 x + 4

and

y = 4 - x^{2}

meet when

x^{2} + 4 x + 4 = 4 - x^{2} ⟺ 2 x^{2} + 4 x = 2 x (x + 2) = 0

So the curves intersect at

(0, 4)

and

(- 2, 0) .

Here is a sketch.

(b) Let

R = {(x, y) \in R^{2} | x^{2} + 4 x + 4 \leq y \leq 4 - x^{2}, - 2 \leq x \leq 0}

By Green’s theorem (Theorem 4.3.2)

\begin{aligned} \oint_{C} x y d x + (e^{y} + x^{2}) d y & = \iint_{R} {\frac{\partial}{\partial x} (e^{y} + x^{2}) - \frac{\partial}{\partial y} (x y)} d x d y \\ = \int_{- 2}^{0} d x \int_{x^{2} + 4 x + 4}^{4 - x^{2}} d y x \\ = \int_{- 2}^{0} d x (- 2 x^{2} - 4 x) x \\ = {[- \frac{x^{4}}{2} - \frac{4 x^{3}}{3}]}_{- 2}^{0} \\ = - \frac{8}{3} \end{aligned}

4.3.10. (✳).

Solution.

The integral that would be used for direct evaluation looks very complicated. So let’s try Green’s theorem. The curve

C

is the boundary of the triangle

T = {(x, y) | 0 \leq x \leq 1, 0 \leq y \leq 2 x}

\begin{aligned} \int_{C} F \cdot d r & = \int_{C} {(y^{2} - e^{- y^{2}} + \sin x) d x + (2 x y e^{- y^{2}} + x) d y} \\ = \iint_{T} {\frac{\partial}{\partial x} (2 x y e^{- y^{2}} + x) - \frac{\partial}{\partial y} (y^{2} - e^{- y^{2}} + \sin x)} d x d y \\ = \iint_{T} {(2 y e^{- y^{2}} + 1) - (2 y + 2 y e^{- y^{2}})} d x d y \\ = \int_{0}^{1} d x \int_{0}^{2 x} d y {1 - 2 y} \\ = \int_{0}^{1} d x {2 x - 4 x^{2}} \\ = 1 - \frac{4}{3} = - \frac{1}{3} \end{aligned}

4.3.11. (✳).

Solution.

Here is a sketch of the two curves in question.

Note that the curves

y = x^{2} - 4 x + 3

and

y = 3 - x^{2} + 2 x

intersect when

x^{2} - 4 x + 3 = 3 - x^{2} + 2 x

2 x^{2} - 6 x = 2 x (x - 3) = 0

x = 0,

3 .

The integrand for direct evaluation looks complicated. So let’s use Green’s theorem with

F_{1} (x, y) = 2 x e^{y} + \sqrt{2 + x^{2}},

F_{2} (x, y) = x^{2} (2 + e^{y})

and

R = {(x, y) | x^{2} - 4 x + 3 \leq y \leq 3 - x^{2} + 2 x, 0 \leq x \leq 3}

By Green’s theorem, which is Theorem 4.3.2,

\begin{aligned} \int_{C} (2 x e^{y} + \sqrt{2 + x^{2}}) d x + x^{2} (2 + e^{y}) d y & = \iint_{R} {\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}} d x d y \\ = \iint_{R} {2 x (2 + e^{y}) - 2 x e^{y}} d x d y \\ = 4 \int_{0}^{3} d x \int_{x^{2} - 4 x + 3}^{3 - x^{2} + 2 x} d y x \\ = 4 \int_{0}^{3} d x (6 x - 2 x^{2}) x \\ = 4 {[2 x^{3} - \frac{1}{2} x^{4}]}_{0}^{3} \\ = 54 \end{aligned}

4.3.12. (✳).

Solution.

Direct evaluation will lead to three integrals, one for each side of the triangle. The integral from

(0, 0)

and

(1, - 2)

and the integral from

(1, 2)

(0, 0)

will each contain six (nonconstant) terms. This does not look very efficient. So let’s try Green’s theorem. Denote by

T,

the triangle

T = {(x, y) | 0 \leq x \leq 1, - 2 x \leq y \leq 2 x}

It has boundary

\partial T = C,

oriented counterclockwise as desired. So, by Green’s theorem,

\begin{aligned} \int_{C} F \cdot d r & = \int_{\partial T} {(\frac{3}{2} y^{2} + e^{- y} + \sin x) d x + (\frac{1}{2} x^{2} + x - x e^{- y}) d y} \\ = \iint_{T} {\frac{\partial}{\partial x} (\frac{1}{2} x^{2} + x - x e^{- y}) - \frac{\partial}{\partial y} (\frac{3}{2} y^{2} + e^{- y} + \sin x)} d x d y \\ = \iint_{T} {(x + 1 - e^{- y}) - (3 y - e^{- y})} d x d y \\ = \iint_{T} {x - 3 y + 1} d x d y \end{aligned}

Now

\begin{aligned} \iint_{T} d x d y & = Area (T) = \frac{1}{2} (4) (1) = 2 \\ \iint_{T} y d x d y & = 0 since y is odd under y \to - y \\ \iint_{T} x d x d y & = \int_{0}^{1} d x \int_{- 2 x}^{2 x} d y x = \int_{0}^{1} 4 x^{2} d x = \frac{4}{3} \end{aligned}

\int_{C} F \cdot d r = \frac{4}{3} - 3 \times 0 + 2 = \frac{10}{3}

4.3.13. (✳).

Solution.

Set

F = \frac{- y}{x^{2} + y^{2}} \hat{ı ı} + \frac{x}{x^{2} + y^{2}} \hat{ȷ ȷ}

(a) Green’s theorem must be applied to a curve that is closed, so that it is the boundary of a region in

R^{2} .

The given curve

C

is not closed. But it is part of the boundary of

R = {(x, y) | - 2 \leq x \leq 2, \frac{x^{2}}{4} + 1 \leq y \leq 2}

Here is a sketch of

R .

The boundary of

R

consists of two parts —

C

on the bottom and the line segment

L

from

(2, 2)

(- 2, 2)

on the top. Note that

F

is well-defined on all of

R

and that

\begin{aligned} \frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1} & = \frac{\partial}{\partial x} \frac{x}{x^{2} + y^{2}} + \frac{\partial}{\partial y} \frac{y}{x^{2} + y^{2}} \\ = \frac{(x^{2} + y^{2}) - x (2 x)}{{(x^{2} + y^{2})}^{2}} + \frac{(x^{2} + y^{2}) - y (2 y)}{{(x^{2} + y^{2})}^{2}} \\ = 0 \end{aligned}

on all of

R .

So, by Green’s theorem (Theorem 4.3.2),

\begin{aligned} \int_{C} \frac{- y}{x^{2} + y^{2}} d x + \frac{x}{x^{2} + y^{2}} d y & = \iint_{R} (\frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1}) d x d y - \int_{L} F \cdot d r \\ = \int_{- L} F \cdot d r = \int_{- 2}^{2} F_{1} d x \\ = - \int_{- 2}^{2} \frac{2}{x^{2} + 4} d x since y = 2 on L \\ = - \int_{- 1}^{1} \frac{4}{4 u^{2} + 4} d x with x = 2 u, d x = 2 d u \\ = - \arctan u |_{- 1}^{1} = - \frac{π}{2} \end{aligned}

In the second line, we used the notation

- L

for the line segment from

(- 2, 2)

(2, 2) .

(b) This question looks a lot like that of part (a). But there is a critical difference. Again

C

is not closed and again it is part of the boundary of a simple region in the

x y

-plane, namely

R = {(x, y) | - 2 \leq x \leq 2, x^{2} - 2 \leq y \leq 2}

This

R

is sketched below.

We cannot continue as in part (a), using this

R,

because

\frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1}

is not zero througout

R .

In fact, it is not even defined throughout

R

— it is not defined at

(0, 0),

which is a point of

R .

We can work around this obstruction by

choosing a number $ρ > 0$ that is small enough that the circle $C_{ρ}$ parametrized by
$r (θ) = ρ \cos θ \hat{ı ı} + ρ \sin θ \hat{ȷ ȷ} 0 \leq θ \leq 2 π$
is completely contained inside $R$ (ror example, $ρ = 1$ is fine)
and then removing from $R$ the interior of $C_{ρ} .$

This produces the “deformed washer”

W = {(x, y) | - 2 \leq x \leq 2, x^{2} - 2 \leq y \leq 2, x^{2} + y^{2} \geq ρ^{2}}

that is sketched below.

The boundary of

W

consists the three parts — the curve of interest

C

on the bottom, the line segment

L

from

(2, 2)

(- 2, 2)

on the top, and the circle

- C_{ρ}

(that is

C_{ρ}

but oriented clockwise, rather than counter-clockwise) around the hole in the middle. Now

\frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1}

is well-defined and zero throughout

W .

So, by Green’s theorem (Theorem 4.3.2),

\begin{aligned} \int_{C} \frac{- y}{x^{2} + y^{2}} d x + \frac{x}{x^{2} + y^{2}} d y \\ = \iint_{W} (\frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1}) d x d y - \int_{L} F \cdot d r - \int_{- C_{ρ}} F \cdot d r \\ = \int_{- L} F \cdot d r + \int_{C_{ρ}} F \cdot d r \end{aligned}

We have already found, in part (a), that

\int_{- L} F \cdot d r = - \frac{π}{2} .

So it remains only to use

\begin{aligned} r (θ) & = ρ \cos θ \hat{ı ı} + ρ \sin θ \hat{ȷ ȷ} \\ r^{'} (θ) & = - ρ \sin θ \hat{ı ı} + ρ \cos θ \hat{ȷ ȷ} \end{aligned}

to evaluate

\begin{aligned} \int_{C_{ρ}} F \cdot d r & = \int_{0}^{2 π} F (r (θ)) \cdot r^{'} (θ) d θ \\ = \int_{0}^{2 π} (\overset{F (r (θ))}{\overset{⏞}{- \frac{1}{ρ} \sin θ \hat{ı ı} + \frac{1}{ρ} \cos θ \hat{ȷ ȷ}}}) \cdot (\overset{r^{'} (θ)}{\overset{⏞}{- ρ \sin θ \hat{ı ı} + ρ \cos θ \hat{ȷ ȷ}}}) d θ \\ = \int_{0}^{2 π} d θ \\ = 2 π \end{aligned}

All together

\int_{C} \frac{- y}{x^{2} + y^{2}} d x + \frac{x}{x^{2} + y^{2}} d y = \int_{- L} F \cdot d r + \int_{C_{ρ}} F \cdot d r = - \frac{π}{2} + 2 π = \frac{3 π}{2}

F

is not conservative. We found, in parts (a) and (b), two different values for the integrals along two paths, both of which start at

(- 2, 2)

and end at

(2, 2) .

F

does not have the “path independence” property of Theorem 2.4.7.c and cannot be conservative.

4.3.14. (✳).

Solution.

The given integral is of the form

\int_{C} F \cdot d r

with

F = (2 x e^{y} + \sqrt{2} + x^{2}) \hat{ı ı} + x^{2} (2 + e^{y}) \hat{ȷ ȷ}

If we were to try to evaluate this integral directly, then on the

y = x^{2} - 4 x + 3

part of

C,

the integrand would contain

x^{2} e^{y} = x^{2} e^{x^{2} - 4 x + 3} .

That looks hard to integrate, so let’s try Green’s theorem. The parabolas

y = x^{2} - 4 x + 3

and

y = 3 - x^{2} + 2 x

intersect at

(x, y)

with

\begin{aligned} x^{2} - 4 x + 3 = 3 - x^{2} + 2 x & ⟺ 2 x^{2} - 6 x = 0 ⟺ 2 x (x - 3) = 0 \\ ⟺ x = 0 or x = 3 \end{aligned}

The curve

C

is the boundary of

R = {(x, y) | 0 \leq x \leq 3, x^{2} - 4 x + 3 \leq y \leq 3 - x^{2} + 2 x}

It is sketched below.

By Green’s theorem (Theorem 4.3.2),

\begin{aligned} \int_{C} (2 x e^{y} + \sqrt{2} + x^{2}) d x + x^{2} (2 + e^{y}) d y \\ = \iint_{R} [\frac{\partial}{\partial x} (x^{2} (2 + e^{y})) - \frac{\partial}{\partial y} (2 x e^{y} + \sqrt{2} + x^{2})] d x d y \\ = \iint_{R} (2 x (2 + e^{y}) - 2 x e^{y}) d x d y \\ = 4 \int_{0}^{3} d x \int_{x^{2} - 4 x + 3}^{3 - x^{2} + 2 x} d y x \\ = 4 \int_{0}^{3} d x x [(3 - x^{2} + 2 x) - (x^{2} - 4 x + 3)] \\ = 4 \int_{0}^{3} d x (6 x^{2} - 2 x^{3}) = 4 (2 \times 3^{3} - \frac{1}{2} 3^{4}) = 54 \end{aligned}

4.3.15. (✳).

Solution.

(a) Denote by

R_{2} = {(x, y) | (x - 2)^{2} + y^{2} \leq 1}

the interior of the circle

C_{2} .

Note that

(0, 0)

is not in

R_{2} .

Consequently,

Q_{x} - P_{y} = 0

everywhere in

R_{2}

and, by Green’s theorem (Theorem 4.3.2),

I_{2} = \int_{C_{2}} F \cdot d r = \iint_{R_{2}} (Q_{x} - P_{y}) d x d y = 0

(b) We cannot blindly apply Green’s theorem to

I_{3} = \int_{C_{3}} F \cdot d r

because

(0, 0)

is in the interior of

C_{3},

so that

Q_{x} - P_{y}

is not identically zero in the interior of

C_{3}

— it is not even defined throughout the interior of

C_{3} .

We can work around this obstruction by considering the interior of

C_{3}

with the interior of

C_{1}

removed. That is, by considering

R_{3} = {(x, y) | x^{2} + (y - 2)^{2} \leq 9, x^{2} + y^{2} \geq 1}

It is sketched on the right above. The boundary of

R_{3}

consists of two parts

the circle $C_{3},$ oriented counterclockwise, and
the circle $- C_{1} .$ That is, the circle $C_{1}$ but oriented clockwise, rather than counterclockwise.

Then

Q_{x} - P_{y}

is well-defined and zero throughout

R_{3}

and, by Green’s theorem,

\begin{aligned} 0 & = \iint_{R_{3}} (Q_{x} - P_{y}) d x d y = \int_{C_{3}} F \cdot d r + \int_{- C_{1}} F \cdot d r \\ = \int_{C_{3}} F \cdot d r - \int_{C_{1}} F \cdot d r \\ = \int_{C_{3}} F \cdot d r - π \end{aligned}

\int_{C_{3}} F \cdot d r = π .

I_{4} = \int_{C_{4}} F \cdot d r

because

(0, 0)

is in the interior of

C_{4} .

This time we cannot remove the interior of

C_{1}

from the interior of

C_{4},

because

C_{1}

is not contained in the interior of

C_{4} .

Instead we pick a number

ρ > 0

which is small enough that the positively oriented circle

C_{ρ} = {(x, y) | x^{2} + y^{2} = ρ^{2}}

is completely inside

C_{4} .

Then we can define

R_{4} = {(x, y) | (x - 2)^{2} + (y - 2)^{2} \leq 9, x^{2} + y^{2} \geq ρ^{2}}

It is sketched on the left below. We can now argue as in part (b). The boundary of

R_{4}

consists of two parts

the circle $C_{4},$ oriented counterclockwise, and
the circle $- C_{ρ} .$ That is, the circle $C_{ρ}$ but oriented clockwise, rather than counterclockwise.

Then

Q_{x} - P_{y}

is well-defined and zero throughout

R_{4}

and, by Green’s theorem,

\begin{aligned} 0 & = \iint_{R_{4}} (Q_{x} - P_{y}) d x d y \\ = \int_{C_{4}} F \cdot d r + \int_{- C_{ρ}} F \cdot d r \\ = \int_{C_{4}} F \cdot d r - \int_{C_{ρ}} F \cdot d r \end{aligned}

\int_{C_{4}} F \cdot d r = \int_{C_{ρ}} F \cdot d r .

To complete our computation, we have to determine

\int_{C_{ρ}} F \cdot d r .

We can do so by repeating the same “removing a small disk containing

(0, 0)

” argument for the third time. Set

R_{5} = {(x, y) | x^{2} + y^{2} \leq 1, x^{2} + y^{2} \geq ρ^{2}}

Then the boundary of

R_{5}

consists of

C_{1}

and

- C_{ρ},

and, as

Q_{x} - P_{y}

is well-defined and zero throughout

R_{5},

\begin{aligned} 0 & = \iint_{R_{5}} (Q_{x} - P_{y}) d x d y \\ = \int_{C_{1}} F \cdot d r + \int_{- C_{ρ}} F \cdot d r \\ = π - \int_{C_{ρ}} F \cdot d r \end{aligned}

\int_{C_{4}} F \cdot d r = \int_{C_{ρ}} F \cdot d r = π .

4.3.16. (✳).

Solution.

(a) If

(x, y) \neq (0, 0),

we have

\begin{aligned} Q_{x} - P_{y} & = \frac{\partial}{\partial x} (\frac{y - x}{x^{2} + y^{2}}) - \frac{\partial}{\partial y} (\frac{x + y}{x^{2} + y^{2}}) \\ = \frac{- (x^{2} + y^{2}) - (y - x) (2 x)}{{(x^{2} + y^{2})}^{2}} - \frac{(x^{2} + y^{2}) - (x + y) (2 y)}{{(x^{2} + y^{2})}^{2}} \\ = 0 \end{aligned}

(b) Parametrize

C_{R}

r (θ) = R \cos θ \hat{ı ı} + R \sin θ \hat{ȷ ȷ} 0 \leq θ \leq 2 π

\begin{aligned} \int_{C_{R}} F \cdot d r & = \int_{0}^{2 π} \overset{F (r (θ))}{\overset{⏞}{\frac{1}{R} {(\cos θ + \sin θ) \hat{ı ı} + (\sin θ - \cos θ) \hat{ȷ ȷ}}}} \\ \cdot (\overset{r^{'} (θ)}{\overset{⏞}{- R \sin θ \hat{ı ı} + R \cos θ \hat{ȷ ȷ}}}) d θ \\ = \int_{0}^{2 π} (- 1) d θ \\ = - 2 π \end{aligned}

F

were conservative, the line integral

\int_{C} F \cdot d r

would be

0

for any closed curve

C,

by Theorem 2.4.7.b. So

F

is not conservative. Note that

F

is not defined at

(x, y) = (0, 0)

and so fails the screening test

\nabla \nabla \times F = 0

(x, y) = (0, 0) .

(d) Denote by

R

the interior of the triangle

C .

It is the grey region in the figure

Note that

(0, 0)

is not in

R .

Q_{x} - P_{y}

is defined and zero throughout

R .

So, by Green’s theorem (Theorem 4.3.2),

\begin{aligned} \int_{C} F \cdot d r & = \iint_{R} (Q_{x} - P_{y}) d x d y = 0 \end{aligned}

(e) Note that

(0, 0)

is in the interior of triangle

C

specified for this part. So

Q_{x} - P_{y}

is not defined in that interior and we cannot apply Green’s theorem precisely as we did in part (d). We can work around this obstruction by

picking a number $r > 0$ that is small enough that the circle $C_{r},$ of radius $r$ centred on $(0, 0),$ is completely contained in the interior of the triangle $C .$
Then we work with the region $R$ defined by removing the interior of the circle $C_{r}$ from the interior of the triangle $C .$ It is the grey region sketched below.

The boundary of

R

consists of two parts

the triangle $C,$ oriented counterclockwise, and
the circle $- C_{r} .$ That is, the circle $C_{r},$ but oriented clockwise, rather than counterclockwise.

Then

Q_{x} - P_{y}

is well-defined and zero throughout

R

and, by Green’s theorem,

\begin{aligned} 0 & = \iint_{R} (Q_{x} - P_{y}) d x d y \\ = \int_{C} F \cdot d r + \int_{- C_{r}} F \cdot d r \\ = \int_{C} F \cdot d r - \int_{C_{r}} F \cdot d r \end{aligned}

\int_{C} F \cdot d r = \int_{C_{r}} F \cdot d r .

By part (b), with

R = r,

\int_{C_{r}} F \cdot d r = - 2 π,

\int_{C} F \cdot d r = - 2 π

4.3.17. (✳).

Solution.

(a) The given integral is of the form

\int_{C} F_{1} (x, y) d x + F_{2} (x, y) d y

with

F_{1} (x, y) = \sqrt{1 + x^{3}} F_{2} (x, y) = 2 x y^{2} + y^{2} \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} = 2 y^{2}

C

\partial R

with

R = {(x, y) | x^{2} + y^{2} \leq 1}

Green’s theorem (Theorem 4.3.2) gives

\begin{aligned} \int_{C} \sqrt{1 + x^{3}} d x + (2 x y^{2} + y^{2}) d y & = \int_{C} F_{1} (x, y) d x + F_{2} (x, y) d y \\ = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d x d y \\ = 2 \iint_{R} y^{2} d x d y \end{aligned}

Switching to polar coordinates

\begin{aligned} \int_{C} \sqrt{1 + x^{3}} d x + (2 x y^{2} + y^{2}) d y & = 2 \int_{0}^{2 π} d θ \int_{0}^{1} d r r (r \sin θ)^{2} \\ = 2 [\int_{0}^{2 π} d θ \sin^{2} θ] [\int_{0}^{1} d r r^{3}] \\ = 2 π \frac{1}{4} = \frac{π}{2} \end{aligned}

To do the

θ

integral, we have used

\begin{matrix} \int_{0}^{2 π} \sin^{2} θ d θ = \int_{0}^{2 π} (\frac{1 - \cos (2 θ)}{2}) d θ = [\frac{θ - \sin (2 θ) / 2}{2}]_{0}^{2 π} = π \end{matrix}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \sin^{2} θ d θ,

see Example 2.4.4.

(b) It is again natural to use Green’s theorem. But Green’s theorem must be applied to a curve that is closed, so that it is the boundary of a region in

R^{2} .

The given curve

C

is not closed. But it is part of the boundary of

R = {(x, y) | x^{2} + y^{2} \leq 1, x \geq 0}

Here is a sketch of

R .

The boundary of

R

consists of two parts —

C

on the right and the line segment

L

from

(0, 1)

(0, - 1)

on the left. Note that

F = F_{1} \hat{ı ı} + F_{2} \hat{ȷ ȷ}

is well-defined on all of

R

and that we still have, from part (a),

\begin{aligned} \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} & = 2 y^{2} \end{aligned}

on all of

R .

So, by Green’s theorem (Theorem 4.3.2),

\begin{aligned} \int_{C} \sqrt{1 + x^{3}} d x + (2 x y^{2} + y^{2}) d y & = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{2}}{\partial y}) d x d y - \int_{L} F \cdot d r \\ = \iint_{\binom{x^{2} + y^{2} \leq 1}{x \geq 0}} 2 y^{2} d x d y + \int_{- L} F_{2} d y \\ = \iint_{x^{2} + y^{2} \leq 1} y^{2} d x d y + \int_{- 1}^{1} y^{2} d y \\ by symmetry for the first integral and \\ since x = 0 and d x = 0 in the \\ second integral \\ = \frac{π}{4} + \frac{2}{3} \end{aligned}

In the second line, we used the notation

- L

for the line segment from

(0, - 1)

(0, 1) .

4.3.18. (✳).

Solution.

First, here is a sketch of the curve

C .

We’ll evaluate this integral in three different ways.

Direct evaluation: To evaluate the integral directly, we’ll parametrize $C$ using $y$ as the parameter. That is, we’ll make $y (t) = t :$

$\begin{aligned} r (t) & = x (t) \hat{ı ı} + y (t) \hat{ȷ ȷ} = \cos t \hat{ı ı} + t \hat{ȷ ȷ} - \frac{π}{2} \leq t \leq \frac{π}{2} \\ r^{'} (t) & = x^{'} (t) \hat{ı ı} + y^{'} (t) \hat{ȷ ȷ} = - \sin t \hat{ı ı} + \hat{ȷ ȷ} \end{aligned}$

So the integral is

$\begin{aligned} \int_{C} (x^{2} + y e^{x}) d x + (x \cos y + e^{x}) d y \\ = \int_{- π / 2}^{π / 2} {[x (t)^{2} + y (t) e^{x (t)}] \frac{d x}{d t} (t) \\ + [x (t) \cos (y (t)) + e^{x (t)}] \frac{d y}{d t} (t)} d t \\ = \int_{- π / 2}^{π / 2} {- [\cos^{2} t + t e^{\cos t}] \sin t + [\cos^{2} t + e^{\cos t}]} d t \\ = \int_{- π / 2}^{π / 2} {- \cos^{2} t \sin t + \cos^{2} t - t e^{\cos t} \sin t + e^{\cos t}} d t \\ = \int_{- π / 2}^{π / 2} {\cos^{2} t + \frac{d}{d t} [\frac{\cos^{3} t}{3} + t e^{\cos t}]} d t \\ = \int_{- π / 2}^{π / 2} {\frac{\cos (2 t) + 1}{2} + \frac{d}{d t} [\frac{\cos^{3} t}{3} + t e^{\cos t}]} d t \\ = [\frac{\sin (2 t)}{4} + \frac{t}{2} + \frac{\cos^{3} t}{3} + t e^{\cos t}]_{- π / 2}^{π / 2} \\ = \frac{3 π}{2} \end{aligned}$

For an efficient, sneaky, way to evaluate $\int_{- π / 2}^{π / 2} \cos^{2} t d t$ see Example 2.4.4.
Green’s (or Stokes’) theorem: The curve $C$ is not closed so we cannot apply Green’s theorem directly. However the boundary of the region

$R = {(x, y) | 0 \leq x \leq \cos y, - \frac{π}{2} \leq y \leq \frac{π}{2}}$

(sketched below) consists of two parts, one of which is $C .$ The other is the line $L$ from $(0, \frac{π}{2})$ to $(0, - \frac{π}{2}) .$

So Green’s theorem gives

$\begin{aligned} \int_{C} (x^{2} + y e^{x}) d x + (x \cos y + e^{x}) d y \\ = \iint_{R} {\frac{\partial}{\partial x} (x \cos y + e^{x}) - \frac{\partial}{\partial y} (x^{2} + y e^{x})} d x d y \\ - \int_{L} (x^{2} + y e^{x}) d x + (x \cos y + e^{x}) d y \\ = \iint_{R} \cos y d x d y - \int_{L} d y since x = 0 and d x = 0 on L \\ = \int_{- π / 2}^{π / 2} d y \int_{0}^{\cos y} d x \cos y - \int_{π / 2}^{- π / 2} d y \\ = \int_{- π / 2}^{π / 2} d y \cos^{2} y + π \\ = \int_{- π / 2}^{π / 2} \frac{\cos (2 y) + 1}{2} d y + π \\ = [\frac{\sin (2 y)}{4} + \frac{y}{2}]_{- π / 2}^{π / 2} + π \\ = \frac{3 π}{2} \end{aligned}$
(Sort of) conservative fields: The given integral is $\int_{C} F \cdot d r$ with $F = (x^{2} + y e^{x}) \hat{ı ı} + (x \cos y + e^{x}) \hat{ȷ ȷ} .$ The curl of this field is

$\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{2} + y e^{x} & x \cos y + e^{x} & 0 \end{array}] = \cos y \hat{k} \end{aligned}$

So $F$ violates our screening test and consequently is not conservative. But it violates the screening test only because of the term $x \cos y \hat{ȷ ȷ} .$ This suggests that we split up

$F = G + H with G = (x^{2} + y e^{x}) \hat{ı ı} + e^{x} \hat{ȷ ȷ}, H = x \cos y \hat{ȷ ȷ}$

Then $G$ is conservative with potential $g = \frac{x^{3}}{3} + y e^{x}$ and $H$ is pretty simple, so that it is not hard to evaluate $\int_{C} H \cdot d r$ directly. Using the parametrization $r (t) = \cos t \hat{ı ı} + t \hat{ȷ ȷ},$ $- \frac{π}{2} \leq t \leq \frac{π}{2}$ as above,

$\begin{aligned} \int_{C} F \cdot d r & = \int_{C} G \cdot d r + \int_{C} H \cdot d r \\ = \int_{C} \nabla \nabla g \cdot d r + \int_{C} H \cdot d r \\ = g (r (\frac{π}{2})) - g (r (- \frac{π}{2})) + \int_{- π / 2}^{π / 2} x (t) \cos (y (t)) \frac{d y}{d t} (t) d t \\ = g (0, \frac{π}{2}) - g (0, - \frac{π}{2}) + \int_{- π / 2}^{π / 2} \cos^{2} t d t \\ = \frac{π}{2} - (- \frac{π}{2}) + \int_{- π / 2}^{π / 2} \frac{\cos (2 t) + 1}{2} d t \\ = π + [\frac{\sin (2 t)}{4} + \frac{t}{2}]_{- π / 2}^{π / 2} \\ = \frac{3 π}{2} \end{aligned}$

4.3.19. (✳).

Solution.

Call the region enclosed by the curve

R .

By Green’s theorem, Theorem 4.3.2,

\begin{aligned} \frac{1}{2} \oint_{C} x d y - y d x & = \frac{1}{2} \iint_{R} (\frac{\partial}{\partial x} x - \frac{\partial}{\partial y} (- y)) d x d y = \frac{1}{2} \iint_{R} 2 d x d y \\ = A \end{aligned}

as desired. The curve

x^{2 / 3} + y^{2 / 3} = 1

may be parametrized in the counterclockwise orientation by

x (θ) = \cos^{3} θ,

y (θ) = \sin^{3} θ,

0 \leq θ \leq 2 π .

Then

\begin{aligned} A & = \frac{1}{2} \oint_{C} x d y - y d x \\ = \frac{1}{2} \int_{0}^{2 π} (x (θ) y^{'} (θ) - y (θ) x^{'} (θ)) d θ \\ = \frac{1}{2} \int_{0}^{2 π} (3 \cos^{4} θ \sin^{2} θ + 3 \sin^{4} θ \cos^{2} θ) d θ \\ = \frac{3}{2} \int_{0}^{2 π} \sin^{2} θ \cos^{2} θ d θ = \frac{3}{8} \int_{0}^{2 π} \sin^{2} (2 θ) d θ \\ = \frac{3}{16} \int_{0}^{2 π} (1 - \cos (4 θ)) d θ = \frac{3}{16} [θ - \frac{1}{4} \sin (4 θ)]_{0}^{2 π} = \frac{3 π}{8} \end{aligned}

4.3.20. (✳).

Solution.

If we use

D

to denote the disk inside the circle

C

then we want

\begin{aligned} \oint_{C} F \cdot d r - A \oint_{C} G \cdot d r & = \oint_{C} (F - A G) \cdot d r \\ = \iint_{D} [\frac{\partial}{\partial x} (F - A G)_{2} - \frac{\partial}{\partial y} (F - A G)_{1}] d x d y \end{aligned}

to vanish for all disks

D .

We used Green’s theorem, which is Theorem 4.3.2, in the last step. This is the case if and only if

\begin{aligned} \frac{\partial}{\partial x} (F - A G)_{2} = \frac{\partial}{\partial y} (F - A G)_{1} \\ ⟺ & \frac{\partial}{\partial x} [(x + y) - A (2 x - 3 y)] = \frac{\partial}{\partial y} [(x + 3 y) - A (x + y)] \\ ⟺ & 1 - 2 A = 3 - A \\ ⟺ & A = - 2 \end{aligned}

4.3.21. (✳).

Solution.

(a) Parametrize the circle

r (θ) = (\cos θ, \sin θ) .

Then

\begin{aligned} F (r (θ)) & = \sin^{3} θ \hat{ı ı} - \cos θ \sin^{2} θ \hat{ȷ ȷ} \\ \frac{d r}{d θ} (θ) & = - \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ} \\ F (r (θ)) \cdot \frac{d r}{d θ} (θ) & = - \sin^{4} θ - \cos^{2} θ \sin^{2} θ = - \sin^{2} θ \\ \oint_{C} F \cdot d r & = \int_{0}^{2 π} F (r (θ)) \cdot \frac{d r}{d θ} (θ) d θ = - \int_{0}^{2 π} \sin^{2} θ d θ \\ = - \int_{0}^{2 π} \frac{1 - \cos (2 θ)}{2} d θ \\ = - {[\frac{θ}{2} - \frac{\sin (2 θ)}{4}]}_{0}^{2 π} = - π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \sin^{2} θ d θ

see Example 2.4.4.

(b) Denote by

W

the washer shaped region between the circle

x^{2} + y^{2} = 1

and the ellipse

\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 .

It is sketched below. By Green’s theorem

\begin{matrix} \oint_{C_{0}} F \cdot d r - \oint_{C} F \cdot d r = \iint_{W} [\frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1}] d x d y \end{matrix}

For the specified

F

\begin{aligned} \frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1} = - \frac{\partial}{\partial x} \frac{x y^{2}}{{(x^{2} + y^{2})}^{2}} - \frac{\partial}{\partial y} \frac{y^{3}}{{(x^{2} + y^{2})}^{2}} \\ = - \frac{y^{2}}{{(x^{2} + y^{2})}^{2}} + 2 \frac{x y^{2} (2 x)}{{(x^{2} + y^{2})}^{3}} - \frac{3 y^{2}}{{(x^{2} + y^{2})}^{2}} + 2 \frac{y^{3} (2 y)}{{(x^{2} + y^{2})}^{3}} \\ = \frac{- y^{2} (x^{2} + y^{2}) + 4 x^{2} y^{2} - 3 y^{2} (x^{2} + y^{2}) + 4 y^{4}}{{(x^{2} + y^{2})}^{3}} \\ = 0 \end{aligned}

Consequently

\oint_{C_{0}} F \cdot d r - \oint_{C} F \cdot d r = 0 ⟹ \oint_{C_{0}} F \cdot d r = \oint_{C} F \cdot d r = - π

4.3.22. (✳).

Solution.

Observe that

\begin{aligned} \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} & = \frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}}) - \frac{\partial}{\partial y} (\frac{- y}{x^{2} + y^{2}}) \\ = \frac{(x^{2} + y^{2}) - x (2 x)}{{(x^{2} + y^{2})}^{2}} + \frac{(x^{2} + y^{2}) - y (2 y)}{{(x^{2} + y^{2})}^{2}} = 0 \end{aligned}

except at

(0, 0),

where

F

is not defined. Hence by Green’s theorem (Theorem 4.3.2),

\oint_{C} F \cdot d r = 0

for any closed curve that does not contain

(0, 0)

in its interior. In particular,

\oint_{C_{1}} F \cdot d r = 0 .

On the other hand,

(0, 0)

is contained in the interior of

C_{2},

so we cannot use Green’s theorem to conclude that

\oint_{C_{2}} F \cdot d r = 0 .

Let

C_{3}

be the circle of radius one centred on

(0, 0)

and denote by

W

the washer shaped region between the circle

C_{2}

and the circle

C_{3} .

It is sketched below.

By Green’s theorem (Theorem 4.3.2),

\begin{matrix} \oint_{C_{2}} F \cdot d r - \oint_{C_{3}} F \cdot d r = \iint_{W} [\frac{\partial}{\partial x} F_{2} - \frac{\partial}{\partial y} F_{1}] d x d y = 0 \end{matrix}

\oint_{C_{2}} F \cdot d r = \oint_{C_{3}} F \cdot d r .

Parameterize

C_{3}

x = \cos θ,

y = \sin θ .

Then

\begin{aligned} r (θ) & = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} 0 \leq θ \leq 2 π \\ r^{'} (θ) & = - \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ} \\ F (r (θ)) & = - \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ} \\ F (r (θ)) \cdot r^{'} (θ) & = 1 \end{aligned}

so that

\oint_{C_{2}} F \cdot d r = \oint_{C_{3}} F \cdot d r = \int_{0}^{2 π} d θ 1 = 2 π

4.3.23. (✳).

Solution.

(a) Let

C_{1}

be the line segment from

(0, 1)

(0, 0),

C_{2}

be the line segment from

(0, 0)

(1, 0)

and

C_{3}

be the curve

y = 1 - x^{2}

from

(1, 0)

(0, 1) .

Then

\begin{aligned} \int_{C_{1}} x d s & = \int_{C_{1}} 0 d s = 0 \\ \int_{C_{2}} x d s & = \int_{0}^{1} x d x = \frac{1}{2} \end{aligned}

To evaluate the integral along

C_{3},

we use the parametrization

\begin{aligned} (x (t), y (t)) & = (1 - t, 1 - (1 - t)^{2}) 0 \leq t \leq 1 \\ (x^{'} (t), y^{'} (t)) & = (- 1, 2 (1 - t)) \\ \frac{d s}{d t} (t) & = \sqrt{x^{'} (t)^{2} + y^{'} (t)^{2}} = \sqrt{1 + 4 (1 - t)^{2}} \end{aligned}

\begin{aligned} \int_{C_{3}} x d s & = \int_{0}^{1} x (t) \frac{d s}{d t} (t) d t = \int_{0}^{1} (1 - t) \sqrt{1 + 4 (1 - t)^{2}} d t \\ = [- \frac{1}{12} {(1 + 4 (1 - t)^{2})}^{3 / 2}]_{0}^{1} = \frac{1}{12} [- 1 + 5^{3 / 2}] \end{aligned}

All together

\int_{C} x d s = \frac{1}{2} + \frac{1}{12} [5^{3 / 2} - 1] \approx 1.3484

(b) By either Stokes’ theorem or Green’s theorem

\begin{aligned} \int_{C} F \cdot d r & = \iint_{R} [\frac{\partial}{\partial x} (x^{2} + \cos (y^{2})) - \frac{\partial}{\partial y} (\sin (x^{2}) - x y)] d x d y \\ = \iint_{R} 3 x d x d y \\ = 3 \int_{0}^{1} d x \int_{0}^{1 - x^{2}} d y x \\ = 3 \int_{0}^{1} d x (1 - x^{2}) x \\ = 3 [\frac{1}{2} - \frac{1}{4}] = \frac{3}{4} \end{aligned}

4.3.24. (✳).

Solution.

(a) If

(x, y, z)

is on the curve, it must obey both

z = x + y

and

z = x^{2} + y^{2}

and hence it must also obey

x^{2} + y^{2} = x + y

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{1}{2} .

That’s a circle. We can parametrize the curve by

\begin{aligned} x (θ) & = \frac{1}{2} + \frac{1}{\sqrt{2}} \cos θ \\ y (θ) & = \frac{1}{2} + \frac{1}{\sqrt{2}} \sin θ \\ z (θ) & = x + y = 1 + \frac{1}{\sqrt{2}} [\cos θ + \sin θ] \end{aligned}

with

0 \leq θ < 2 π .

θ

runs from

0

2 π,

(x (θ), y (θ))

runs once around the circle without crossing itself so that

(x (θ), y (θ), z (θ))

runs once around the curve without crossing itself. As

(x (2 π), y (2 π), z (2 π)) = (x (0), y (0), z (0)),

C

is a simple closed curve.

(b) (i) The vector field

F = x^{2} \hat{ı ı} + y^{2} \hat{ȷ ȷ} + 3 e^{z} \hat{k}

is conservative (with potential

\frac{1}{3} x^{3} + \frac{1}{3} y^{3} + 3 e^{z}

). So

\oint_{C} F \cdot d r = 0 .

(b) (ii) Note that the question did not specify the orientation of

C .

It should have. We’ll stick with the most commonly used orientation — counterclockwise when viewed from high on the

z

-axis. The vector field

G = 3 e^{z} \hat{k}

is conservative (with potential

3 e^{z}

). So

\oint_{C} G \cdot d r = 0

and, using the parametrization

\begin{aligned} r (θ) & = [\frac{1}{2} + \frac{1}{\sqrt{2}} \cos θ] \hat{ı ı} + [\frac{1}{2} + \frac{1}{\sqrt{2}} \sin θ] \hat{ȷ ȷ} + [1 + \frac{1}{\sqrt{2}} \sin θ + \frac{1}{\sqrt{2}} \cos θ] \hat{k} \\ r^{'} (θ) & = - \frac{1}{\sqrt{2}} \sin θ \hat{ı ı} + \frac{1}{\sqrt{2}} \cos θ \hat{ȷ ȷ} + [\frac{1}{\sqrt{2}} \cos θ - \frac{1}{\sqrt{2}} \sin θ] \hat{k} \end{aligned}

of part (a), we have

\begin{aligned} \oint_{C} F \cdot d r = \oint_{C} (F - G) \cdot d r \\ = \int_{0}^{2 π} [y (θ)^{2} x^{'} (θ) + x (θ)^{2} y^{'} (θ)] d θ \\ = \int_{0}^{2 π} {- [\frac{1}{2} + \frac{1}{\sqrt{2}} \sin θ]^{2} \frac{1}{\sqrt{2}} \sin θ + [\frac{1}{2} + \frac{1}{\sqrt{2}} \cos θ]^{2} \frac{1}{\sqrt{2}} \cos θ} d θ \end{aligned}

Because the integral of any odd power of

\sin θ

\cos θ

over

0 \leq θ \leq 2 π

is zero (see Example 4.4.6 in the text),

\begin{aligned} \oint_{c} F \cdot d r & = \int_{0}^{2 π} {- \frac{1}{2} \sin^{2} θ + \frac{1}{2} \cos^{2} θ} d θ \\ = 0 \end{aligned}

since (see Example 2.4.4 in the text)

\int_{0}^{2 π} \cos^{2} θ d θ = \int_{0}^{2 π} \sin^{2} θ d θ = π

4.3.25.

Solution.

By Green’s Theorem

\begin{aligned} \oint_{C} (y^{3} - y) d x - 2 x^{3} d y & = \iint_{R} [\frac{\partial}{\partial x} (- 2 x^{3}) - \frac{\partial}{\partial y} (y^{3} - y)] d x d y \\ = \iint_{R} [1 - 6 x^{2} - 3 y^{2}] d x d y \end{aligned}

where

R

is the region in the

x y

-plane whose boundary is

C .

Observe that the integrand

1 - 6 x^{2} - 3 y^{2}

is positive in the elliptical region

6 x^{2} + 3 y^{2} \leq 1

and negative outside of it. To maximize the integral

\iint_{R} [1 - 6 x^{2} - 3 y^{2}] d x d y

we should choose

R

to contain all points

(x, y)

with the integrand

1 - 6 x^{2} - 3 y^{2} \geq 0

and to exclude all points

(x, y)

with the integrand

1 - 6 x^{2} - 3 y^{2} < 0 .

So we choose

R = {(x, y) | 6 x^{2} + 3 y^{2} \leq 1}

The corresponding

C

6 x^{2} + 3 y^{2} = 1 .

4.4 Stokes’ Theorem
4.4.3 Exercises

4.4.3.1.

Solution.

One approach is to first consider

The correct normal to this surface is sketched in

It is correct because

if you walk along $\partial S$ in the direction of the arrow on $\partial S,$
with the vector from your feet to your head having direction $\hat{n}$
then $S$ is on your left hand side.

Now pretend that the surface

S

is made of rubber and that

\hat{n}

is glued to

S .

We can push on this

S

to deform it to the

S

of part (a) or to the

S

of part (b). This gives the solutions to parts (a) and (b).

(a)

(b)

To deal with part (c), we can first rotate the flat disk that we considered above to get

We can push on this

S

to deform it to the

S

of part (c). This gives the solution to part (c).

(c)

4.4.3.2.

Solution.

Think of the

x y

-plane as being the plane

z = 0

R^{3} .

We are going to apply Stokes’ theorem (Theorem 4.4.1) with

S

being the given region

R

in the

x y

-plane and with

F (x, y, z) = F_{1} (x, y) \hat{ı ı} + F_{2} (x, y) \hat{ȷ ȷ} .

Then

the unit normal vector to $S$ specified in Stokes theorem is $\hat{k}$ (if you walk along $\partial S = C$ in the direction of the arrow on $C$ with the vector from your feet to your head having direction $\hat{k}$ then $S = R$ is on your left hand side) and
$d S = d x d y$ and
the curl of $F$ is
$\nabla \nabla \times F = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_{1} (x, y) & F_{2} (x, y) & 0 \end{matrix}] = (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k}$

So Stokes’ theorem gives

\begin{aligned} \oint_{C} [F_{1} (x, y) d x + F_{2} (x, y) d y] & = \oint_{\partial S} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d x d y \end{aligned}

4.4.3.3.

Solution.

We are to show that

\oint_{C} [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] \cdot d r = 0 .

Suppose that

C = \partial S .

Then, by Stokes’ theorem

\oint_{C} [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] \cdot d r = \iint_{S} \nabla \nabla \times [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] \cdot \hat{n} d S

We will show below that

\nabla \nabla \times [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] = 0 .

This will imply that

\oint_{C} [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] \cdot d r = 0 .

One way to see that

\nabla \nabla \times [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] = 0

\begin{aligned} \nabla \nabla \times [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] & = \nabla \nabla \times [\nabla \nabla (ϕ ψ)] & (by part (c) of Theorem 4.1.3) \\ = 0 & (by part (b) of Theorem 4.1.7) \end{aligned}

Another way to see that

\nabla \nabla \times [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] = 0

\begin{aligned} \nabla \nabla \times [ϕ \nabla \nabla ψ + ψ \nabla \nabla ϕ] & = \nabla \nabla ϕ \times \nabla \nabla ψ + ϕ \nabla \nabla \times (\nabla \nabla ψ) + \nabla \nabla ψ \times \nabla \nabla ϕ + ψ \nabla \nabla \times (\nabla \nabla ϕ) \\ = \nabla \nabla ϕ \times \nabla \nabla ψ + \nabla \nabla ψ \times \nabla \nabla ϕ \\ since ϕ \nabla \nabla \times (\nabla \nabla ψ) = ψ \nabla \nabla \times (\nabla \nabla ϕ) = 0 \\ = 0 \end{aligned}

4.4.3.4.

Solution.

(a) Observe that

x (t) = \cos t

and

y (t) = \sin t

obey

x (t)^{2} + y (t)^{2} = 1 .

Then

z (t) = y (t)^{2} = \sin^{2} t .

So we may parametrize the curve by

r (t) = (\cos t, \sin t, \sin^{2} t)

with

0 \leq t \leq 2 π .

Then

\begin{aligned} r^{'} (t) & = (- \sin t, \cos t, 2 \sin t \cos t) \\ F (r (t)) & = (\cos^{2} t - \sin t, \sin^{2} t + \cos t, 1) \\ F (r (t)) \cdot r^{'} (t) & = - \sin t \cos^{2} t + \sin^{2} t + \sin^{2} t \cos t + \cos^{2} t + 2 \sin t \cos t \\ = 1 + \frac{1}{3} \frac{d}{d t} [\cos^{3} t + \sin^{3} t] + \sin (2 t) \\ \oint_{C} F \cdot d r & = \int_{0}^{2 π} {1 + \frac{1}{3} \frac{d}{d t} [\cos^{3} t + \sin^{3} t] + \sin (2 t)} d t \\ = {[t + \frac{1}{3} [\cos^{3} t + \sin^{3} t] - \frac{1}{2} \cos (2 t)]}_{0}^{2 π} \\ = 2 π \end{aligned}

(b) Let

S

be the surface

z = f (x, y)

with

f (x, y) = y^{2}

and

x^{2} + y^{2} \leq 1 .

Since

C

is oriented counter clockwise when viewed from high on the

z

-axis, Stokes’ theorem requires that we use the normal

\hat{n}

S

with positive

z

component. Hence

\begin{aligned} \hat{n} d S & = [- \frac{\partial f}{\partial x} \hat{ı ı} - \frac{\partial f}{\partial y} \hat{ȷ ȷ} + \hat{k}] d x d y = [- 2 y \hat{ȷ ȷ} + \hat{k}] d x d y \\ \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{2} - y & y^{2} + x & 1 \end{array}] = 2 \hat{k} \\ \nabla \nabla \times F \cdot \hat{n} d S & = 2 d x d y \\ \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = 2 \iint_{x^{2} + y^{2} \leq 1} d x d y \\ = 2 π \end{aligned}

4.4.3.5.

Solution.

We apply Stokes’ theorem. First,

\nabla \nabla \times F = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y e^{x} & x + e^{x} & z^{2} \end{matrix}] = (1 + e^{x} - e^{x}) \hat{k} = \hat{k}

Note that

r (t) = x (t) \hat{ı ı} + y (t) \hat{ȷ ȷ} + z (t) \hat{k}

obeys

x (t) + y (t) + z (t) = 3,

for every

t,

and that

x (t) \hat{ı ı} + y (t) \hat{ȷ ȷ} = (1 + \cos t) \hat{ı ı} + (1 + \sin t) \hat{ȷ ȷ}

runs counterclockwise around the circle of radius 1 centered on

(1, 1) .

So we choose

S

to be the part of the plane

G (x, y, z) = x + y + z = 3

with

(x - 1)^{2} + (y - 1)^{2} \leq 1 .

Then, by Stokes’ Theorem,

\oint_{C} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{S} \hat{k} \cdot \hat{n} d S

with

\hat{n} d S = \pm \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} d x d y = \pm (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d x d y

(1 + \cos t) \hat{ı ı} + (1 + \sin t) \hat{ȷ ȷ}

runs counterclockwise around the circle

(x - 1)^{2} + (y - 1)^{2} \leq 1,

Stokes’ theorem specifies the plus sign and

\oint_{C} F \cdot d r = \iint_{(x - 1)^{2} + (y - 1)^{2} \leq 1} d x d y = π

4.4.3.6. (✳).

Solution.

The boundary of

S

\partial S = {(x, y, z) | z = 0, x^{2} + y^{2} = 4}

and can be parametrized

r (θ) = 2 \cos θ \hat{ı ı} + 2 \sin θ \hat{ȷ ȷ} 0 \leq θ \leq 2 π

So, by Stokes’ theorem (Theorem 4.4.1)

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \oint_{\partial S} F \cdot d r \\ = \int_{0}^{2 π} (\overset{F (r (θ))}{\overset{⏞}{- 2 \sin θ \hat{ı ı} + 2 \cos θ \hat{ȷ ȷ} - 2 \cos θ \hat{k}}}) \cdot (\overset{r^{'} (θ)}{\overset{⏞}{- 2 \sin θ \hat{ı ı} + 2 \cos θ \hat{ȷ ȷ}}}) d θ \\ = 4 \int_{0}^{2 π} d θ \\ = 8 π \end{aligned}

4.4.3.7. (✳).

Solution 1.

The boundary of

S

is the circle

x^{2} + y^{2} = 4,

z = 0 .

Let

C

be this circle, oriented by the parametrization

x (t) = 2 \cos t,

y (t) = 2 \sin t,

z (t) = 0 .

By Stokes’ theorem

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \int_{C} F \cdot d r = \int_{0}^{2 π} F (2 \cos t, 2 \sin t, 0) \cdot \frac{d r}{d t} (t) d t \\ = \int_{0}^{2 π} [0 \hat{ı ı} + 2 \cos t (3 + 2 \sin t) \hat{ȷ ȷ} + 2 \sin t \hat{k}] \cdot [- 2 \sin t \hat{ı ı} + 2 \cos t \hat{ȷ ȷ}] d t \\ = \int_{0}^{2 π} [12 \cos^{2} t + 8 \cos^{2} t \sin t] d t \\ = \int_{0}^{2 π} [6 + 6 \cos (2 t) + 8 \cos^{2} t \sin t] d t \\ = [6 t + 3 \sin (2 t) - \frac{8}{3} \cos^{3} t]_{0}^{2 π} = 12 π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} t d t,

see Example 2.4.4.

Solution 2.

$S$ be the surface specified in the question, with upward pointing normal, and
$D$ be the disk ${(x, y, z) | x^{2} + y^{2} \leq 4, z = 0},$ with normal $\hat{n} = \hat{k},$ and
$C$ be the circle ${(x, y, z) | x^{2} + y^{2} = 4, z = 0},$ oriented as in the figure below.

Note that

C

is the boundary curve for both

S

and

D .

So, by Stokes’ theorem, twice

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S & = \int_{C} F \cdot d r = \iint_{D} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{D} \nabla \nabla \times F \cdot \hat{k} d S \end{aligned}

Now the

\hat{k}

component of

\nabla \nabla \times F

\begin{aligned} \nabla \nabla \times F \cdot \hat{k} & = \frac{\partial}{\partial x} [x (3 + y)] - \frac{\partial}{\partial y} [x \ln (1 + z)] = 3 + y \end{aligned}

y

is odd under

y \to - y

and

D

is invariant under

y \to - y,

we have

\iint_{D} y d S = 0

and

\begin{matrix} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{D} (3 + y) d S = 3 \iint_{D} d S = 3 Area (S) = 3 π 2^{2} = 12 π \end{matrix}

4.4.3.8. (✳).

Solution.

Let

S

be the portion of the paraboloid

z = f (x, y) = 4 - x^{2} - y^{2}

with

x^{2} + (y - 1)^{2} \leq 1

and let

\hat{n}

be the upward normal to

S .

For this surface

\hat{n} d S = (- f_{x} (x, y) \hat{ı ı} - f_{y} (x, y) \hat{ȷ ȷ} + \hat{k}) d x d y = (2 x \hat{ı ı} + 2 y \hat{ȷ ȷ} + \hat{k}) d x d y

by (3.3.2). As

(x, y, z)

runs over

S,

(x, y)

runs over the circular disk

D = {(x, y) | x^{2} + (y - 1)^{2} \leq 1}

For the given vector field

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x z & x & y z \end{array}] \\ = z \hat{ı ı} + x \hat{ȷ ȷ} + \hat{k} \end{aligned}

so that, by Stokes’ theorem (Theorem 4.4.1),

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{D} [2 x \overset{z = f (x, y)}{\overset{⏞}{(4 - x^{2} - y^{2})}} + 2 x y + 1] d x d y \end{aligned}

By oddness under

x \to - x,

all terms integrate to zero except for the last. So

\oint_{C} F \cdot d r = \iint_{D} d x d y = Area (D) = π

4.4.3.9.

Solution.

The surface

\begin{aligned} S & = {(x, y, z) | - 1 \leq x \leq 1, - 1 \leq y \leq 1, z \geq 0, z = (1 - x^{2}) (1 - y^{2})} \\ = {(x, y, z) | - 1 \leq x \leq 1, - 1 \leq y \leq 1, z = (1 - x^{2}) (1 - y^{2})} \end{aligned}

Note that when

x = 1

x = - 1

y = 1

y = - 1,

we have

z = (1 - x^{2}) (1 - y^{2}) = 0 .

So the boundary of

S,

call it

C,

is the boundary of the square

- 1 \leq x, y \leq 1,

z = 0,

oriented counterclockwise. Here is a sketch of

C .

Apply Stokes’ theorem. Observing that

z = 0

C

so that

F = - y \hat{ı ı} + x^{3} \hat{ȷ ȷ},

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \oint_{C} F \cdot d r = \oint_{C} [- y \hat{ı ı} + x^{3} \hat{ȷ ȷ}] \cdot d r \\ = \underset{y = - 1 s i d e}{\underset{⏟}{\int_{- 1}^{1} - (- 1) d x}} + \underset{x = 1 s i d e}{\underset{⏟}{\int_{- 1}^{1} (1)^{3} d y}} + \underset{y = 1 s i d e}{\underset{⏟}{\int_{1}^{- 1} - (1) d x}} + \underset{x = - 1 s i d e}{\underset{⏟}{\int_{1}^{- 1} (- 1)^{3} d y}} \\ = 8 \end{aligned}

4.4.3.10.

Solution.

We shall apply Stokes’ Theorem. The curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{x^{2}} - y z & \sin y - y z & x z + 2 y \end{array} | \\ = (2 + y) \hat{ı ı} - (z + y) \hat{ȷ ȷ} + (0 + z) \hat{k} \end{aligned}

The curve

C

is a triangle. All three vertices of the triangle obey

x + y + z = 1 .

So the triangle is the boundary of the surface

S = {(x, y, z) | x \geq 0, y \geq 0, z = 1 - x - y \geq 0} .

The equation of the surface is

z = f (x, y) = 1 - x - y .

So, by (3.3.2),

\begin{aligned} \hat{n} d S & = (- f_{x} \hat{ı ı} - f_{y} \hat{ȷ ȷ} + \hat{k}) d x d y \\ = (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d x d y \end{aligned}

Here

\hat{n}

is the upward pointing unit normal. The set of points

(x, y)

for which there is a corresponding

(x, y, z)

S

T = {(x, y) | x \geq 0, y \geq 0, x + y \leq 1},

which is a triangle of area

\frac{1}{2} .

Since

\begin{aligned} \nabla \nabla \times F \cdot \hat{n} d S & = [(2 + y) \hat{ı ı} - (z + y) \hat{ȷ ȷ} + (0 + z) \hat{k}] \cdot (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d x d y \\ = 2 d x d y \end{aligned}

we have

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{T} 2 d x d y = 2 Area (T) = 1 \end{aligned}

4.4.3.11. (✳).

Solution.

Stokes’ theorem, which is Theorem 4.4.1, says that

\oint_{C} F \cdot d r = \iint_{S} \nabla \times F \cdot \hat{n} d S

for any surface

S

whose boundary is

C .

For the given vector field

\begin{aligned} \nabla \times F (x, y, z) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ - z & x & y \end{array}] \\ = \hat{ı ı} - \hat{ȷ ȷ} + \hat{k} \end{aligned}

Choose

\begin{aligned} S & = {(x, y, z) | z = y, \frac{x^{2}}{4} + \frac{y^{2}}{2} + \frac{z^{2}}{2} \leq 1} \\ = {(x, y, z) | z = y, \frac{x^{2}}{4} + y^{2} \leq 1} \end{aligned}

to be the part of the plane

z = y

bounded by the ellipsoid.

S

is part of the plane

z = f (x, y) = y,

(3.3.2), gives that

\begin{aligned} \hat{n} d S & = \pm (- f_{x}, - f_{y}, 1) d x d y \\ = \pm (0, - 1, 1) d x d y \end{aligned}

C

has the standard orientation (counter-clockwise when viewed from high on the

z

-axis), we want

\hat{n}

to have a positive

z

-component. So

\hat{n} d S = (0, - 1, 1) d x d y .

From the second form of

S

given above, we see that as

(x, y, z)

runs over

S,

(x, y)

runs over

D = {(x, y) | \frac{x^{2}}{4} + y^{2} \leq 1}

Consequently, Stokes’ theorem gives that

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{D} \overset{\nabla \times F}{\overset{⏞}{(1, - 1, 1)}} \cdot \overset{\hat{n} d S}{\overset{⏞}{(0, - 1, 1) d x d y}} \\ = 2 \iint_{D} d x d y = 2 Area (D) \end{aligned}

The ellipse

D,

that is

\frac{x^{2}}{4} + y^{2} \leq 1,

has semi-axes

a = 2

and

b = 1

and hence area

π a b = 2 π .

Finally

\oint_{C} F \cdot d r = 2 Area (D) = 4 π

4.4.3.12. (✳).

Solution.

Note that the curve of part (a) is a simple closed curve that lies in the plane

x + y + z = 2

and is oriented in a counterclockwise direction as observed from the positive

x

-axis. The curve of part (a) encloses a triangle.

Two of the sides of the triangle are

(0, 2, 0) - (2, 0, 0) = (- 2, 2, 0)

and

(0, 0, 2) - (0, 2, 0) = (0, - 2, 2)

so the area of the triangle is

\frac{1}{2} | (- 2, 2, 0) \times (0, - 2, 2) | = \frac{1}{2} det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - 2 & 2 & 0 \\ 0 & - 2 & 2 \end{matrix}] = \frac{1}{2} | (4, 4, 4) | = 2 \sqrt{3}

So let’s do part (b) first.

(b) We are not told explicitly what

C_{2}

is, so we certainly can’t do a direct evaluation. Instead, let’s use Stokes’ theorem (Theorem 4.4.1). The curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^{2} & x^{2} & y^{2} \end{array}] \\ = 2 y \hat{ı ı} + 2 z \hat{ȷ ȷ} + 2 x \hat{k} \end{aligned}

The upward pointing unit normal to

E

\hat{n} = \frac{\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}}{\sqrt{3}} .

So, by Stokes’ theorem,

\begin{aligned} I_{2} & = \int_{C_{2}} F \cdot d r = \iint_{R} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{R} (2 y \hat{ı ı} + 2 z \hat{ȷ ȷ} + 2 x \hat{k}) \cdot \frac{\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}}{\sqrt{3}} d S \\ = \frac{2}{\sqrt{3}} \iint_{R} (\overset{= 2 o n R}{\overset{⏞}{y + z + x}}) d S = \frac{4}{\sqrt{3}} Area (R) = 4 \sqrt{3} \end{aligned}

(a) Denote by

T

the triangle enclosed by

C_{1} .

By the computation that we have just done in part (b)

\begin{aligned} I_{1} & = \int_{C_{1}} F \cdot d r = \frac{4}{\sqrt{3}} Area (T) = 8 \end{aligned}

4.4.3.13. (✳).

Solution.

(a) Observe that

the curve $C_{1}$ is one quarter of a circle in the $x y$ -plane, centred on the origin, of radius $2,$ starting at $(2, 0, 0)$ and ending at $(0, 2, 0)$ and
the curve $C_{2}$ is one quarter of a circle in the $y z$ -plane, centred on the origin, of radius $2,$ starting at $(0, 2, 0)$ and ending at $(0, 0, 2)$ and
the curve $C_{3}$ is one quarter of a circle in the $x z$ -plane, centred on the origin, of radius $2,$ starting at $(0, 0, 2)$ and ending at $(2, 0, 0) .$

Here is a sketch.

(b)

C

lies completely on the sphere

x^{2} + y^{2} + z^{2} = 4 .

So it is natural to choose

S = {(x, y, z) | x^{2} + y^{2} + z^{2} = 4, x \geq 0, y \geq 0, z \geq 0}

and to parametrize

S

using spherical coordinates

\begin{aligned} r (θ, φ) = 2 \cos θ \sin φ \hat{ı ı} + 2 \sin θ \sin φ \hat{ȷ ȷ} + 2 \cos φ \hat{k} \\ 0 \leq θ \leq \frac{π}{2}, 0 \leq φ \leq \frac{π}{2} \end{aligned}

Since

\begin{aligned} \frac{\partial r}{\partial θ} & = - 2 \sin θ \sin φ \hat{ı ı} + 2 \cos θ \sin φ \hat{ȷ ȷ} \\ \frac{\partial r}{\partial φ} & = 2 \cos θ \cos φ \hat{ı ı} + 2 \sin θ \cos φ \hat{ȷ ȷ} - 2 \sin φ \hat{k} \end{aligned}

so that

\begin{aligned} \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ - 2 \sin θ \sin φ & 2 \cos θ \sin φ & 0 \\ 2 \cos θ \cos φ & 2 \sin θ \cos φ & - 2 \sin φ \end{array}] \\ = - 4 \cos θ \sin^{2} φ \hat{ı ı} - 4 \sin θ \sin^{2} φ \hat{ȷ ȷ} - 4 \sin φ \cos φ \hat{k} \end{aligned}

(3.3.1) gives

\begin{aligned} \hat{n} d S & = \pm \frac{\partial r}{\partial θ} \times \frac{\partial r}{\partial φ} d θ d φ \\ = \mp 4 (\cos θ \sin φ \hat{ı ı} + \sin θ \sin φ \hat{ȷ ȷ} + \cos φ \hat{k}) \sin φ d θ d φ \end{aligned}

We want

\hat{n}

to point outward, for compatibility with the orientation of

C .

So we choose the

+

sign.

\begin{aligned} \hat{n} d S & = 4 (\cos θ \sin φ \hat{ı ı} + \sin θ \sin φ \hat{ȷ ȷ} + \cos φ \hat{k}) \sin φ d θ d φ \\ = 2 r (θ, φ) \sin φ d θ d φ \end{aligned}

F

looks too complicated for a direct evaluation of the line integral. So, in preparation for an application of Stokes’ theorem, we compute

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y + \sin (x^{2}) & z - 3 x + \ln (1 + y^{2}) & y + e^{z^{2}} \end{array}] \\ = - 4 \hat{k} \end{aligned}

So, by Stokes’ theorem (Theorem 4.4.1),

\begin{aligned} \int_{C} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \int_{0}^{π / 2} d φ \int_{0}^{π / 2} d θ (- 4 \hat{k}) \cdot (\cos θ \sin φ \hat{ı ı} + \sin θ \sin φ \hat{ȷ ȷ} + \cos φ \hat{k}) 4 \sin φ \\ = - 16 \int_{0}^{π / 2} d φ \int_{0}^{π / 2} d θ \cos φ \sin φ = 16 \frac{π}{2} \frac{\cos^{2} φ}{2} |_{0}^{π / 2} \\ = - 4 π \end{aligned}

4.4.3.14. (✳).

Solution.

(a) The boundary,

\partial S_{1},

S_{1}

as specified in Stokes’ theorem (Theorem 4.4.1) is the circle

\sqrt{x^{2} + y^{2}} = 4,

z = 4

oriented clockwise when viewed from high on the

z

-axis. That is, we can parametrize

\partial S_{1}

\begin{matrix} r (t) = 4 \cos t \hat{ı ı} - 4 \sin t \hat{ȷ ȷ} + 4 \hat{k}, 0 \leq t \leq 2 π \end{matrix}

\begin{aligned} F (r (t)) \cdot d r & = (16 \sin t, 16 \cos t, - 16 \sin t \cos t \cos (- 16 \sin t)) \\ \cdot (- 4 \sin t, - 4 \cos t, 0) d t \\ = - 64 d t \end{aligned}

and, by Stokes’ theorem,

\begin{aligned} \iint_{S_{1}} \nabla \nabla \times F \cdot \hat{n} d S & = \oint_{\partial S_{1}} F (r (t)) \cdot d r = - 64 \int_{0}^{2 π} d t = - 128 π \end{aligned}

(b) The boundary,

\partial S_{2},

S_{2}

consists of two parts, a circle in the plane

z = 4

and a circle in the plane

z = 1 .

We’ll call the first part

\partial S_{2 a} .

It is the same as

\partial S_{1} .

We’ll call the second part

\partial S_{2 b} .

It is the circle

\sqrt{x^{2} + y^{2}} = 1,

z = 1

oriented counterclockwise when viewed from high on the

z

-axis. We can parametrize it

\begin{matrix} r (t) = \cos t \hat{ı ı} + \sin t \hat{ȷ ȷ} + \hat{k}, 0 \leq t \leq 2 π \end{matrix}

So, on

\partial S_{2 b},

\begin{aligned} F (r (t)) \cdot d r & = (- \sin t, \cos t, \sin t \cos t \cos (\sin t)) \cdot (- \sin t, \cos t, 0) d t \\ = d t \end{aligned}

and, by Stokes’ theorem,

\begin{aligned} \iint_{S_{2}} \nabla \nabla \times F \cdot \hat{n} d S & = \oint_{\partial S_{2 a}} F (r (t)) \cdot d r + \oint_{\partial S_{2 b}} F (r (t)) \cdot d r \\ = - 128 π + \int_{0}^{2 π} d t \\ = - 126 π \end{aligned}

4.4.3.15. (✳).

Solution.

Denote by

S = {(x, y, z) | z = x + 4, x^{2} + y^{2} \leq 4}

the part of the plane

z = x + 4

that is contained in the cylinder

x^{2} + y^{2} = 4 .

Orient

S

by the downward pointing normal

\hat{n} = \frac{1}{\sqrt{2}} (1, 0, - 1) .

Then

C

is the boundary of

S .

The part of

C

and

S

that are in the first octant are sketched below.

We may parametrize

S

r (x, y) = (x, y, x + 4) with x^{2} + y^{2} \leq 4

So,

\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] = (- 1, 0, 1)

and, by (3.3.1),

\begin{matrix} \hat{n} d S = - \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} d x d y = (1, 0, - 1) d x d y \end{matrix}

We have chosen to “

-

” sign in

\hat{n} d S = \pm \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} d x d y

to give the downward pointing normal. As the curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{3} + 2 y & \sin (y) + z & x + \sin (z^{2}) \end{array}] \\ = - \hat{ı ı} - \hat{ȷ ȷ} - 2 \hat{k} \end{aligned}

Stokes’ theorem (Theorem 4.4.1) gives

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{S} (- 1, - 1, - 2) \cdot (1, 0, - 1) d x d y \\ = \iint_{S} d x d y = 4 π \end{aligned}

4.4.3.16. (✳).

Solution.

(a) Note that all three vertices,

(2, 0, 0),

(0, 2, 0)

and

0, 0, 2),

lie in the plane

x + y + z = 2 .

So the entire path lies in that plane too.

In part (b) we will need to evaluate a line integral that clearly cannot be computed directly — we will need to use Stokes’ theorem. So let’s use Stokes’s theorem in part (a) too. First, we find

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^{2} & x^{2} & y^{2} \end{array}] = 2 y \hat{ı ı} + 2 z \hat{ȷ ȷ} + 2 x \hat{ı ı} \end{aligned}

Let

S

be the triangular surface that is contained in the plane

x + y + z = 2

and is bounded by

L_{1},

L_{2}

and

L_{3} .

Orient

S

by the normal vector

\hat{n} = \frac{1}{\sqrt{3}} (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) .

Then,

\nabla \nabla \times F \cdot \hat{n} = \frac{1}{\sqrt{3}} (2 y \hat{ı ı} + 2 z \hat{ȷ ȷ} + 2 x \hat{ı ı}) \cdot (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) = \frac{2}{\sqrt{3}} (x + y + z)

and, by Stokes’ theorem,

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \frac{2}{\sqrt{3}} \iint_{S} (x + y + z) d S = \frac{4}{\sqrt{3}} \iint_{S} d S \\ = \frac{4}{\sqrt{3}} Area (S) \end{aligned}

The triangle

S

is half of the prallelogram with sides

(0, 2, 0) - (2, 0, 0) = (- 2, 2, 0)

and

(0, 0, 2) - (2, 0, 0) = (- 2, 0, 2) .

The area of the parallelogram is

\begin{matrix} | (- 2, 2, 0) \times (- 2, 0, 2) | = | (4, 4, 4) | = 4 \sqrt{3} \end{matrix}

\begin{matrix} \oint_{C} F \cdot d r = \frac{4}{\sqrt{3}} 2 \sqrt{3} = 8 \end{matrix}

(b) Let

\tilde{S}

be the specified surface. Then, as in part (a),

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{\tilde{S}} \nabla \nabla \times F \cdot \hat{n} d S = \frac{2}{\sqrt{3}} \iint_{\tilde{S}} (x + y + z) d S = \frac{4}{\sqrt{3}} \iint_{\tilde{S}} d S \\ = \frac{4}{\sqrt{3}} Area (\tilde{S}) \\ = 4 \sqrt{3} \end{aligned}

4.4.3.17. (✳).

Solution.

Let’s try Stokes’ theorem with

F = (z + \frac{1}{1 + z}) \hat{ı ı} + x z \hat{ȷ ȷ} + (3 x y - \frac{x}{(z + 1)^{2}}) \hat{k}

The curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z + \frac{1}{1 + z} & x z & 3 x y - \frac{x}{(z + 1)^{2}} \end{array}] \\ = (3 x - x) \hat{ı ı} - (3 y - \frac{1}{(z + 1)^{2}} - 1 + \frac{1}{(1 + z)^{2}}) \hat{ȷ ȷ} + z \hat{k} \\ = 2 x \hat{ı ı} + (1 - 3 y) \hat{ȷ ȷ} + z \hat{k} \end{aligned}

Write

S = {(x, y, z) | z = f (x, y) = 1 - x^{2} y, x^{2} + y^{2} \leq 1}

For

S,

with the upward pointing normal, by (3.3.2),

\begin{aligned} \hat{n} d S & = (- f_{x}, - f_{y}, 1) d x d y \\ = (2 x y, x^{2}, 1) d x d y \end{aligned}

so that

\begin{aligned} \nabla \nabla \times F \cdot \hat{n} d S & = {4 x^{2} y + (x^{2} - 3 x^{2} y) + \overset{z}{\overset{⏞}{(1 - x^{2} y)}}} d x d y \end{aligned}

and, by Stokes’ theorem,

\begin{aligned} \int_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{x^{2} + y^{2} \leq 1} {4 x^{2} y + x^{2} - 3 x^{2} y + 1 - x^{2} y} d x d y \end{aligned}

\begin{aligned} \int_{C} F \cdot d r & = \iint_{x^{2} + y^{2} \leq 1} {x^{2} + 1} d x d y = π + \iint_{x^{2} + y^{2} \leq 1} x^{2} d x d y \end{aligned}

To evaluate the final remaining integral, let’s switch to polar coordinates.

\begin{aligned} \iint_{x^{2} + y^{2} \leq 1} x^{2} d x d y & = \int_{0}^{1} d r r \int_{0}^{2 π} d θ (r \cos θ)^{2} \\ = \int_{0}^{1} d r r^{3} \int_{0}^{2 π} d θ \cos^{2} θ \end{aligned}

Since

\begin{aligned} \int_{0}^{2 π} \cos^{2} θ d θ & = \int_{0}^{2 π} \frac{1 + \cos (2 θ)}{2} d θ = {[\frac{θ}{2} + \frac{\sin (2 θ)}{4}]}_{0}^{2 π} = π \end{aligned}

we finally have

\int_{0}^{1} d r r^{3} \int_{0}^{2 π} d θ \cos^{2} θ = \frac{π}{4}

and

\int_{C} F \cdot d r = π + \frac{π}{4} = \frac{5 π}{4}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} θ d θ,

see Example 2.4.4.

4.4.3.18. (✳).

Solution.

We are to evaluate a line integral around a curve

C .

We are told that

C

is the boundary of a surface

S

that is contained in the plane

x + y + z = 1,

but we are not told precisely what

C

is. So we are going to have to use Stokes’ theorem. The curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^{2} & x^{2} & y^{2} \end{array}] = 2 y \hat{ı ı} + 2 z \hat{ȷ ȷ} + 2 x \hat{k} \end{aligned}

and, by (3.3.3) with

G (x, y, z) = x + y + z,

\begin{aligned} d S & = | \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} | d x d y = \sqrt{3} d x d y \\ \hat{n} d S & = \pm \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} d x d y = \pm (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d x d y = \pm \frac{1}{\sqrt{3}} (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d S \end{aligned}

Because

C

is oriented in a clockwise direction as

observed from the positive

z

-axis looking down at the plane,

\hat{n}

is to point downwards, so that

\hat{n} d S = - \frac{1}{\sqrt{3}} (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d S

S

we have

x + y + z = 1,

so that Stokes’ theorem gives

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} (\nabla \nabla \times F) \cdot \hat{n} d S \\ = \iint_{S} 2 (y \hat{ı ı} + z \hat{ȷ ȷ} + x \hat{k}) \cdot (- \frac{1}{\sqrt{3}}) (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d S \\ = - \frac{2}{\sqrt{3}} \iint_{S} (y + z + x) d S = - \frac{2}{\sqrt{3}} \iint_{S} d S \\ = - \frac{10}{\sqrt{3}} \end{aligned}

since

S

has area

5 .

4.4.3.19. (✳).

Solution.

We are to evaluate the line integral of a complicated vector field around a relatively complicated closed curve. That certainly suggests that we should not try to evaluate the integral directly. To see if Stokes’ theorem looks promising, let’s compute the curl

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ - y + e^{x} \sin x & y^{4} & \sqrt{z} \tan z \end{array}] = \hat{k} \end{aligned}

That’s suggestive. Next we need to find a surface whose boundary is

C .

First, here is a sketch of

C .

We can choose the surface

S

to be the union of two flat parts:

the quadralateral $Q$ in the $y z$ -plane with vertices $(0, 0, 0),$ $(0, 1, 1),$ $(0, 1, 2)$ and $(0, 2, 0)$ and
the triangle $T$ in the $x y$ -plane with vertices $(0, 0, 0),$ $(0, 2, 0)$ and $(2, 2, 0) .$

The normal to

Q

- \hat{ı ı}

and the normal to

T

- \hat{k} .

Then Stokes’ theorem gives

\begin{aligned} \int_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{Q} \hat{k} \cdot (- \hat{ı ı}) d S + \iint_{T} \hat{k} \cdot (- \hat{k}) d S \\ = - \iint_{T} d S \\ = - Area (T) \\ = - \frac{1}{2} \overset{base}{\overset{⏞}{2}} \overset{height}{\overset{⏞}{2}} \\ = - 2 \end{aligned}

4.4.3.20. (✳).

Solution.

The integral looks messy. Let’s compute the curl of

F = (z + \sin z) \hat{ı ı} + (x^{3} - x^{2} y) \hat{ȷ ȷ} + (x \cos z - y) \hat{k}

to help gauge if Stokes’ theorem would be easier.

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z + \sin z & x^{3} - x^{2} y & x \cos z - y \end{array}] \\ = - \hat{ı ı} + \hat{ȷ ȷ} + (3 x^{2} - 2 x y) \hat{k} \end{aligned}

That’s a lot simpler than

F .

For the surface

z = f (x, y) = x y^{2},

with downward pointing normal (since

C

is traversed clockwise)

\hat{n} d S = - (- f_{x}, - f_{y}, 1) d x d y = (y^{2}, 2 x y, - 1) d x d y

by (3.3.2), So, writing

\begin{aligned} S & = {(x, y, z) | z = x y^{2}, x^{2} + y^{2} \leq 1} \\ D & = {(x, y) | x^{2} + y^{2} \leq 1} \end{aligned}

Stoke’s theorem gives

\begin{aligned} \int_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{D} {- y^{2} + 2 x y - 3 x^{2} + 2 x y} d x d y \\ = - \iint_{x^{2} + y^{2} \leq 1} {3 x^{2} + y^{2} - 4 x y} d x d y \end{aligned}

To evaluate this integral, switch to polar coordinates.

\begin{aligned} \int_{C} F \cdot d r & = - \int_{0}^{1} d r r \int_{0}^{2 π} d θ {3 r^{2} \cos^{2} θ + r^{2} \sin^{2} θ - 4 r^{2} \sin θ \cos θ} \\ = - 4 π \int_{0}^{1} d r r^{3} = - π \end{aligned}

since

\int_{0}^{2 π} \sin θ \cos θ d θ = \frac{1}{2} \int_{0}^{2 π} \sin (2 θ) d θ = 0

and

\int_{0}^{2 π} \sin^{2} θ d θ = \int_{0}^{2 π} \cos^{2} θ d θ = π .

(See Example 2.4.4.)

4.4.3.21. (✳).

Solution.

Here is a sketch of the part of

S

in the first octant.

The boundary,

\partial S,

S

is the circle

x^{2} + y^{2} = 1,

z = 1,

oriented counterclockwise when viewed from above. It is parametrized by

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + \hat{k} 0 \leq θ \leq 2 π

So Stokes’ theorem gives

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \oint_{\partial S} F \cdot d r \\ = \int_{0}^{2 π} (\overset{F (r (t))}{\overset{⏞}{- \sin^{2} θ \hat{ı ı} + \cos^{3} θ \hat{ȷ ȷ} + (mess) \hat{k}}}) \cdot (\overset{r^{'} (t)}{\overset{⏞}{- \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ}}}) d θ \\ = \int_{0}^{2 π} (\sin^{3} θ + \cos^{4} θ) d θ \end{aligned}

The integral of any odd power of

\sin θ

\cos θ

over

0 \leq θ \leq 2 π

is zero. (See Example 4.4.6.) In particular,

\int_{0}^{2 π} \sin^{3} θ d θ = 0 .

To integrate

\cos^{4} θ

we use the trig identity

\begin{aligned} \cos^{2} θ & = \frac{\cos (2 θ) + 1}{2} \\ ⟹ \cos^{4} θ & = \frac{\cos^{2} (2 θ) + 2 \cos (2 θ) + 1}{4} \\ = \frac{1}{4} \frac{\cos (4 θ) + 1}{2} + \frac{\cos (2 θ)}{2} + \frac{1}{4} \\ = \frac{3}{8} + \frac{\cos (4 θ)}{8} + \frac{\cos (2 θ)}{2} \end{aligned}

Finally

\iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \int_{0}^{2 π} (\frac{3}{8} + \frac{\cos (4 θ)}{8} + \frac{\cos (2 θ)}{2}) d θ = \frac{3 π}{4}

4.4.3.22. (✳).

Solution.

We are to evaluate the line integral of a complicated vector field around a relatively complicated closed curve. That certainly suggests that we should not try to evaluate the integral directly. As we are to use Stokes’ theorem, let’s compute the curl

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x \sin y & - y \sin x & (x - y) z^{2} \end{array}] \\ = - z^{2} \hat{ı ı} - z^{2} \hat{ȷ ȷ} - (y \cos x + x \cos y) \hat{k} \end{aligned}

Next we need to find a surface whose boundary is

C .

First, here is a sketch of

C .

We can choose the surface

S

to be the union of two flat parts:

the rectangle $S_{x}$ in the $x z$ -plane with vertices $(0, 0, 0),$ $(\frac{π}{2}, 0, 0),$ $(\frac{π}{2}, 0, 1)$ and $(0, 0, 1)$ and
the rectangle $S_{y}$ in the $y z$ -plane with vertices $(0, 0, 0),$ $(0, 0, 1),$ $(0, \frac{π}{2}, 1)$ and $(0, \frac{π}{2}, 0)$

The normal to

S_{x}

- \hat{ȷ ȷ}

and the normal to

S_{y}

- \hat{ı ı} .

Then Stokes’ theorem gives

\begin{aligned} \int_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{S_{x}} \nabla \nabla \times F \cdot (- \hat{ȷ ȷ}) d S + \iint_{S_{y}} \nabla \nabla \times F \cdot (- \hat{ı ı}) d S \\ = \int_{0}^{\frac{π}{2}} d x \int_{0}^{1} d z z^{2} + \int_{0}^{\frac{π}{2}} d y \int_{0}^{1} d z z^{2} \\ = \int_{0}^{\frac{π}{2}} d x \frac{1}{3} + \int_{0}^{\frac{π}{2}} d y \frac{1}{3} \\ = \frac{π}{3} \end{aligned}

4.4.3.23. (✳).

Solution.

(a) Here is a sketch.

(b) We are to evaluate the line integral of a complicated vector field around a relatively complicated closed curve. That certainly suggests that we should not try to evaluate the integral directly. Let’s try Stokes’ theorem. First, we compute the curl

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{2 z}{1 + y} + \sin (x^{2}) & \frac{3 z}{1 + x} + \sin (y^{2}) & 5 (x + 1) (y + 2) \end{array}] \\ = (5 (x + 1) - \frac{3}{1 + x}) \hat{ı ı} - (5 (y + 2) - \frac{2}{1 + y}) \hat{ȷ ȷ} \\ + (- \frac{3 z}{(1 + x)^{2}} + \frac{2 z}{(1 + y)^{2}}) \hat{k} \end{aligned}

Next we need to find a surface

S

whose boundary is

C .

We can choose the surface

S

to be the union of two flat parts:

the triangle $S_{x}$ in the $x z$ -plane with vertices $(0, 0, 0),$ $(2, 0, 0),$ and $(0, 0, 2)$ and
the triangle $S_{y}$ in the $y z$ -plane with vertices $(0, 0, 0),$ $(0, 0, 2),$ and $(0, 3, 0)$

Note that

The normal to $S_{x}$ specified by Stokes’ theorem is $- \hat{ȷ ȷ} .$ On $S_{x}$ we have $y = 0,$ so that $\nabla \nabla \times F \cdot \hat{ȷ ȷ}$ simplifies to $- (5 (0 + 2) - \frac{2}{1 + 0}) = - 8 .$
The normal to $S_{y}$ specified by Stokes’ theorem is $- \hat{ı ı} .$ On $S_{y}$ we have $x = 0,$ so that $\nabla \nabla \times F \cdot \hat{ı ı}$ simplifies to $(5 (0 + 1) - \frac{3}{1 + 0}) = 2 .$

So Stokes’ theorem gives

\begin{aligned} \int_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{S_{x}} \overset{8}{\overset{⏞}{\nabla \nabla \times F \cdot (- \hat{ȷ ȷ})}} d S + \iint_{S_{y}} \overset{- 2}{\overset{⏞}{\nabla \nabla \times F \cdot (- \hat{ı ı})}} d S \\ = 8 Area (S_{x}) - 2 Area (S_{y}) = 8 \frac{1}{2} (2) (2) - 2 \frac{1}{2} (3) (2) \\ = 10 \end{aligned}

4.4.3.24. (✳).

Solution.

The boundary,

\partial S,

S

is the circle

x^{2} + y^{2} = 1

oriented counter clockwise as usual. It may be parametrized by

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ},

0 \leq θ \leq 2 π .

By Stokes’ theorem

\begin{aligned} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S & = \oint_{\partial S} F \cdot d r = \int_{0}^{2 π} F (r (θ)) \cdot \frac{d r}{d θ} (θ) d θ \\ = \int_{0}^{2 π} (\sin θ, 0, 3 \cos θ) \cdot (- \sin θ, \cos θ, 0) d θ \\ = - \int_{0}^{2 π} \sin^{2} θ d θ = - \int_{0}^{2 π} d θ \frac{1 - \cos (2 θ)}{2} \\ = - {[\frac{θ}{2} - \frac{\sin (2 θ)}{4}]}_{0}^{2 π} \\ = - π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} d θ \sin^{2} θ,

see Example 2.4.4.

4.4.3.25. (✳).

Solution.

The given surface is an ellipsoid centred at

(x, y, z) = (0, 0, 1) .

It caps a curve

C

in the plane

z = 0,

given by

x^{2} + y^{2} = 4 .

This is a circle of radius

2

centred at the origin, oriented counterclockwise when viewed from the positive

z

-axis.

Method I — double Stokes’: Let

D

denote the plane disk

x^{2} + y^{2} \leq 4,

z = 0 .

Using Stokes’ theorem twice gives

\begin{aligned} \iint_{S} G \cdot \hat{n} d S & = \iint_{S} \nabla \times F \cdot \hat{n} d S = \oint_{C} F \cdot d r = \iint_{D} \nabla \times F \cdot \hat{n} d S \\ = \iint_{D} G \cdot \hat{n} d S \end{aligned}

Now in

D

we have

\hat{n} = \hat{k}

and

z = 0,

so on this surface,

\begin{aligned} G \cdot \hat{n} = (\nabla \times F) \cdot \hat{k} & = det {[\begin{array}{c} 0 & 0 & 1 \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (x z - y^{3} \cos z) & x^{3} e^{z} & x y z e^{x^{2} + y^{2} + z^{2}} \end{array}]}_{z = 0} \\ = [3 x^{2} e^{z} + 3 y^{2} \cos z]_{z = 0} = 3 (x^{2} + y^{2}) \end{aligned}

Hence, using polar coordinates,

\begin{aligned} \iint_{S} G \cdot \hat{n} d S & = \iint_{D} 3 (x^{2} + y^{2}) d x d y = 3 \int_{θ = 0}^{2 π} \int_{r = 0}^{2} (r^{2}) r d r d θ \\ = 3 (2 π) (4) = 24 π \end{aligned}

Method II — single Stokes’: By Stokes’ theorem

\iint_{S} G \cdot \hat{n} d S = \iint_{S} \nabla \times F \cdot \hat{n} d S = \oint_{C} F \cdot d r

Parametrize the circle

C

using

r (θ) = 2 \cos θ \hat{ı ı} + 2 \sin θ \hat{ȷ ȷ}, 0 \leq θ \leq 2 π

to obtain

d r = \frac{d r}{d θ} d θ = (- 2 \sin θ \hat{ı ı} + 2 \cos θ \hat{ȷ ȷ}) d θ .

Then since

z = 0

C,

\begin{aligned} \iint_{S} G \cdot \hat{n} d S & = \int_{0}^{2 π} (\overset{F (r (θ))}{\overset{⏞}{- (2 \sin θ)^{3} \hat{ı ı} + (2 \cos θ)^{3} \hat{ȷ ȷ}}}) \cdot (\overset{r^{'} (θ)}{\overset{⏞}{- 2 \sin θ \hat{ı ı} + 2 \cos θ \hat{ȷ ȷ}}}) d θ \\ = 16 \int_{0}^{2 π} (\sin^{4} θ + \cos^{4} θ) d θ \end{aligned}

By the double angle trig identities

\begin{matrix} \cos^{2} θ = \frac{1 + \cos (2 θ)}{2} \sin^{2} θ = \frac{1 - \cos (2 θ)}{2} \end{matrix}

we have

\begin{aligned} \sin^{4} θ + \cos^{4} θ & = \frac{[1 - \cos (2 θ)]^{2}}{4} + \frac{[1 + \cos (2 θ)]^{2}}{4} \\ = \frac{1}{2} + \frac{\cos^{2} (2 θ)}{2} = \frac{1}{2} + \frac{1 + \cos (4 θ)}{4} \end{aligned}

\iint_{S} G \cdot \hat{n} d S = 16 \int_{0}^{2 π} (\frac{3}{4} + \frac{1}{4} \cos (4 θ)) d θ = 16 \times \frac{3}{4} \times (2 π) = 24 π

4.4.3.26. (✳).

Solution.

Note that

\nabla \nabla \times F = det | \begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^{2} & x^{2} & y^{2} \end{matrix} | = 2 y \hat{ı ı} + 2 z \hat{ȷ ȷ} + 2 x \hat{k}

Let

D

be the disk in the plane

x + y + z = 3

whose boundary is

C

and let

\hat{n} = \frac{1}{\sqrt{3}} (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k})

be the upward unit normal to

D .

If the circle is oriented counterclockwise, when viewed from above, then, by Stokes’ theorem (Theorem 4.4.1),

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{D} \nabla \nabla \times F \cdot \hat{n} d S \\ = \frac{1}{\sqrt{3}} \iint_{D} (2 y \hat{ı ı} + 2 z \hat{ȷ ȷ} + 2 x \hat{k}) \cdot (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) d S \\ = \frac{1}{\sqrt{3}} \iint_{D} 2 \overset{= 3 on D}{\overset{⏞}{(x + y + z)}} d S \\ = 2 \sqrt{3} \iint_{D} d S = 2 \sqrt{3} π R^{2} \end{aligned}

4.4.3.27.

Solution 1.

Let

S^{'}

be the bottom surface of the cube, oriented with normal

\hat{k} .

Then, by Stokes’ theorem, since

\partial S = \partial S^{'},

\iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \oint_{\partial S} F \cdot d r = \oint_{\partial S^{'}} F \cdot d r = \iint_{S^{'}} \nabla \nabla \times F \cdot \hat{n} d S

Since

\begin{array}{r} \nabla \nabla \times F = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x y z & x y^{2} & x^{2} y z \end{array}] = (\dots, \dots, y^{2} - x z) \end{array}

and

\hat{n} = \hat{k}

S^{'}

and

z = - 1

S^{'}

\begin{aligned} \iint_{S^{'}} \nabla \nabla \times F \cdot \hat{n} d S & = \int_{- 1}^{1} d x \int_{- 1}^{1} d y (\dots, \dots, y^{2} - x z) \cdot \hat{k} |_{z = - 1} \\ = \int_{- 1}^{1} d x \int_{- 1}^{1} d y (y^{2} + x) \\ = \int_{- 1}^{1} d x \int_{- 1}^{1} d y y^{2} = 2 \times 2 \int_{0}^{1} d y y^{2} \\ = \frac{4}{3} \end{aligned}

Solution 2.

The boundary of

S

is the square

C,

with sides

C_{1},

\dots,

C_{4},

in the sketch

By Stokes’ theorem,

\iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \oint_{C} F \cdot d r

Parametrize

C_{1}

x .

That is,

r (x) = x \hat{ı ı} - \hat{ȷ ȷ} - \hat{k},

- 1 \leq x \leq 1 .

Since

r^{'} (x) = \hat{ı ı},

and

y = z = - 1

C_{1},

\begin{aligned} \int_{C_{1}} F \cdot d r & = \int_{- 1}^{1} F (r (x)) \cdot r^{'} (x) d x = \int_{- 1}^{1} F (r (x)) \cdot \hat{ı ı} d x \\ = \int_{- 1}^{1} \overset{x y z}{\overset{⏞}{x (- 1) (- 1)}} d x \\ = 0 (since x is odd) \end{aligned}

Parametrize

C_{2}

y .

That is,

r (y) = \hat{ı ı} + y \hat{ȷ ȷ} - \hat{k},

- 1 \leq y \leq 1 .

Since

r^{'} (y) = \hat{ȷ ȷ},

and

x = 1

C_{2},

\begin{matrix} \int_{C_{2}} F \cdot d r = \int_{- 1}^{1} F (r (y)) \cdot \hat{ȷ ȷ} d y = \int_{- 1}^{1} \overset{x y^{2}}{\overset{⏞}{y^{2}}} d y = {[\frac{y^{3}}{3}]}_{- 1}^{1} = \frac{2}{3} \end{matrix}

Parametrize

C_{3}

x .

That is,

r (x) = x \hat{ı ı} + \hat{ȷ ȷ} - \hat{k}

with

x

running from

1

- 1 .

(If you’re nervous about this, parametrize by

t = - x .

That is

r (t) = - t \hat{ı ı} + \hat{ȷ ȷ} - \hat{k},

- 1 \leq t \leq 1 .

) Since

r^{'} (x) = \hat{ı ı},

and

y = 1,

z = - 1

C_{3},

\begin{matrix} \int_{C_{3}} F \cdot d r = \int_{1}^{- 1} F (r (x)) \cdot \hat{ı ı} d x = \int_{1}^{- 1} \overset{x y z}{\overset{⏞}{x (1) (- 1)}} d x = 0 (since x is odd) \end{matrix}

Parametrize

C_{4}

y .

That is,

r (y) = - \hat{ı ı} + y \hat{ȷ ȷ} - \hat{k},

with

y

running from

1

- 1 .

Since

r^{'} (y) = \hat{ȷ ȷ},

and

x = - 1

C_{4},

\begin{matrix} \int_{C_{4}} F \cdot d r = \int_{1}^{- 1} F (r (y)) \cdot \hat{ȷ ȷ} d y = \int_{1}^{- 1} \overset{x y^{2}}{\overset{⏞}{(- 1) y^{2}}} d y = - {[\frac{y^{3}}{3}]}_{1}^{- 1} = \frac{2}{3} \end{matrix}

All together

\iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \int_{C_{1}} F \cdot d r + \int_{C_{2}} F \cdot d r + \int_{C_{3}} F \cdot d r + \int_{C_{4}} F \cdot d r = \frac{4}{3}

4.4.3.28.

Solution.

Let’s try Stokes’ Theorem. Call

F = y \hat{ı ı} - x \hat{ȷ ȷ} + x y \hat{k} .

Then

\begin{array}{r} \nabla \nabla \times F = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & - x & x y \end{array}] = x \hat{ı ı} - y \hat{ȷ ȷ} - 2 \hat{k} \end{array}

Now compute

\hat{n} d S

in the

(u, v)

-parametrization.

\begin{aligned} r (u, v) & = (u \cos v, u \sin v, v) \\ \frac{\partial r}{\partial u} (u, v) & = (\cos v, \sin v, 0) \\ \frac{\partial r}{\partial v} (u, v) & = (- u \sin v, u \cos v, 1) \\ \frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v} & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \cos v & \sin v & 0 \\ - u \sin v & u \cos v & 1 \end{array}] = (\sin v, - \cos v, u) \\ \hat{n} d S & = (\sin v, - \cos v, u) d u d v \end{aligned}

Since

u \geq 0,

we do indeed have the upward pointing normal. So, Stokes’ theorem tells us

\begin{aligned} \int_{C} y d x - x d y + x y d z = \int_{C} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \int_{0}^{1} d u \int_{0}^{2 π} d v (u \cos v, - u \sin v, - 2) \cdot (\sin v, - \cos v, u) \\ = \int_{0}^{1} d u \int_{0}^{2 π} d v (2 u \sin v \cos v - 2 u) \\ = [\int_{0}^{1} d u u] [\int_{0}^{2 π} d v (\sin 2 v - 2)] \\ = \frac{1}{2} (- 4 π) = - 2 π \end{aligned}

4.4.3.29.

Solution.

Given the form of

F,

direct evaluation looks hard. So let’s try Stokes’ theorem, using as

S

the part of the plane

G (x, y, z) = x + 2 y - z = 7

that is inside

x^{2} - 2 x + 4 y^{2} = 15 .

Then

\hat{n} d S = \pm \frac{\nabla \nabla G}{\nabla \nabla G \cdot \hat{k}} d x d y = \pm (\hat{ı ı} + 2 \hat{ȷ ȷ} - \hat{k}) d x d y

C

is oriented counterclockwise when viewed from high, Stokes’ theorem specifies the upward pointing normal so that

\hat{n} d S = - (\hat{ı ı} + 2 \hat{ȷ ȷ} - \hat{k}) d x d y .

From the observations that

\nabla \nabla \times F = det [\begin{matrix} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{x^{2}} + y z & \cos (y^{2}) - x^{2} & \sin (z^{2}) + x y \end{matrix}] = x \hat{ı ı} - (z + 2 x) \hat{k}

and that we can rewrite

x^{2} - 2 x + 4 y^{2} = 15

(x - 1)^{2} + 4 y^{2} = 16,

we have

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{(x - 1)^{2} + 4 y^{2} \leq 16} [x \hat{ı ı} - (z + 2 x) \hat{k}] |_{z = - 7 + x + 2 y} \cdot (- 1, - 2, 1) d x d y \\ = \iint_{(x - 1)^{2} + 4 y^{2} \leq 16} [- x - (- 7 + x + 2 y + 2 x)] d x d y \\ = \iint_{(x - 1)^{2} + 4 y^{2} \leq 16} [7 - 4 x - 2 y] d x d y \end{aligned}

To evaluate the integrals of

x

and

y

we use that, for any region

R

in the

x y

--plane,

\bar{x} = \frac{\iint_{R} x d x d y}{Area (R)} \bar{y} = \frac{\iint_{R} y d x d y}{Area (R)}

Our ellipse is

\frac{(x - 1)^{2}}{4^{2}} + \frac{y^{2}}{2^{2}} = 1

and so has area

π a b = π \times 4 \times 2 = 8 π

and centroid

(\bar{x}, \bar{y}) = (1, 0) .

So, using

R = {(x, y) | \frac{(x - 1)^{2}}{4^{2}} + \frac{y^{2}}{2^{2}} \leq 1},

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{R} [7 - 4 x - 2 y] d x d y \\ = Area (R) {7 - 4 \bar{x} - 2 \bar{y}} \\ = 8 π [7 - 4 \times 1 - 2 \times 0] = 24 π \end{aligned}

4.4.3.30. (✳).

Solution.

(a) The curl is

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2 + x^{2} + z & 0 & 3 + x^{2} z \end{array}] = (1 - 2 x z) \hat{ȷ ȷ} \end{aligned}

(b) We are going to use Stokes’ theorem. The specified curve

C

is not closed and so is not the boundary of a surface. So we extend

C

to a closed curve

\tilde{C}

by appending to

C

the line segment

L

from

(2, 0, 0)

(0, 0, 0) .

In the figure below,

C

is the red curve and

\tilde{C}

C

plus the blue line segment.

The closed curve

\tilde{C}

is boundary of the surface

S

that is the union of

the triangle $T_{1}$ in the $y z$ -plane with vertices $(0, 0, 0),$ $(0, 0, 3)$ and $(0, 1, 0)$ and with normal vector $- \hat{ı ı}$ and
the triangle $T_{2}$ in the $x y$ -plane with vertices $(0, 0, 0),$ $(0, 1, 0)$ and $(2, 0, 0)$ and with normal vector $- \hat{k} .$

So, by Stokes’ theorem

\begin{aligned} \int_{C} F \cdot d r + \int_{L} F \cdot d r = \int_{\partial S} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{T_{1}} \nabla \nabla \times F \cdot (- \hat{ı ı}) d S + \iint_{T_{2}} \nabla \nabla \times F \cdot (- \hat{k}) d S \\ = \iint_{T_{1}} (1 - 2 x z) \hat{ȷ ȷ} \cdot (- \hat{ı ı}) d S + \iint_{T_{2}} (1 - 2 x z) \hat{ȷ ȷ} \cdot (- \hat{k}) d S \\ = 0 \end{aligned}

Consequently the integral of interest

\begin{aligned} \int_{C} F \cdot d r & = - \int_{L} F \cdot d r = - \int_{2}^{0} (2 + x^{2}) d x since d y = d z = z = 0 on L \\ = \int_{0}^{2} (2 + x^{2}) d x = {[2 x + \frac{x^{3}}{3}]}_{0}^{2} = \frac{20}{3} \end{aligned}

4.4.3.31. (✳).

Solution.

(a) by direct evaluation: The curl of

G

\begin{aligned} \nabla \nabla \times G & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & - z & y \end{array}] = 2 \hat{ı ı} \end{aligned}

The part of

S

in the first octant is sketched in the figure on the left below.

S

consists of two parts — the cylindrical surface

S_{1} = {(x, y, z) | y^{2} + z^{2} = 9, 0 \leq x \leq 5}

and the disc

S_{2} = {(x, y, z) | x = 0, y^{2} + z^{2} \leq 9}

The normal

\hat{n}

S_{1}

always points radially outward from the cylinder and so always has

\hat{ı ı}

component zero. The normal of

S_{2}

- \hat{ı ı} .

So the flux is

\begin{aligned} \iint_{S} \nabla \nabla \times G \cdot \hat{n} d S & = \iint_{S_{1}} 2 \hat{ı ı} \cdot \hat{n} d S + \iint_{S_{2}} 2 \hat{ı ı} \cdot (- \hat{ı ı}) d S \\ = - 2 \iint_{S_{2}} d S \\ = - 2 (π 3^{2}) = - 18 π \end{aligned}

(a) using Stokes’ theorem: Let’s use Stokes’ theorem. The boundary

\partial S

S

is the cirlce

y^{2} + z^{2} = 9,

x = 5,

oriented clockwise when viewed from far down the

x

-axis. We’ll parametrize it by

r (θ) = (5, 3 \cos θ, - 3 \sin θ) .

Then Stokes’ theorem gives

\begin{aligned} \iint_{S} \nabla \nabla \times G \cdot \hat{n} d S & = \oint_{\partial S} G \cdot d r \\ = \int_{0}^{2 π} (5, 3 \sin θ, 3 \cos θ) \cdot (0, - 3 \sin θ, - 3 \cos θ) d θ \\ = \int_{0}^{2 π} (- 9 \sin^{2} θ - 9 \cos^{2} θ) d θ \\ = - 18 π \end{aligned}

(b) This time we’ll use the divergence theorem. The surface

S

is not closed. So we’ll use the auxilary surface formed by “topping

S

off” with the cap

T = {(5, y, z) | y^{2} + z^{2} \leq 9} .

If we give

T

the normal vector

\hat{ı ı},

this auxiliary surface, the union of

S

and

T,

is the boundary of

V = {(x, y, z) | y^{2} + z^{2} \leq 9, 0 \leq x \leq 5} .

So the divergence theorem gives

\begin{aligned} \iint_{S} F \cdot \hat{n} d S + \iint_{T} F \cdot \hat{n} d S & = \iint_{\partial V} F \cdot \hat{n} d S \\ = ∭_{V} \nabla \nabla \cdot F d V \\ = 0 \end{aligned}

since

\nabla \nabla \cdot F = 0 .

Thus the flux of interest is

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = - \iint_{T} F \cdot \hat{n} d S = - \iint_{T} F \cdot \hat{ı ı} d S = - \iint_{T} (2 + z) d S \\ = - 2 \iint_{T} d S since \iint_{T} z d S = 0, because z is odd \\ = - 18 π since T has area 9 π \end{aligned}

4.4.3.32. (✳).

Solution.

(a) Since

$\frac{y}{x}$ is defined when $x \neq 0$ and
$x^{1 + x^{2}} = e^{(1 + x^{2}) \ln x}$ is defined when $\ln x$ is defined, which is when $x > 0$ (assuming that we are not allowed to use complex numbers) and
$y^{1 + y^{2}} = e^{(1 + y^{2}) \ln y}$ is defined when $\ln y$ is defined, which is when $y > 0$ and
$\cos^{5} (\ln z)$ is defined when $\ln z$ is defined, which when $z > 0$

the domain of

F

D = {(x, y, z) | x > 0, y > 0, z > 0}

(b) The domain

D

is both connected (any two points in

D

can be joined by a curve that lies completely in

D

) and simply connected (any simple closed curve in

D

can be shrunk to a point continuously in

D

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{y}{x} + x^{1 + x^{2}} & x^{2} - y^{1 + y^{2}} & \cos^{5} (\ln z) \end{array}] = (2 x - \frac{1}{x}) \hat{k} \end{aligned}

(d) The integrand for direct evaluation looks very complicated. On the other hand

\nabla \nabla \times F

is quite simple. So let’s try Stokes’ thoerem. Denote

S = {(x, y, z) | 2 \leq x \leq 4, 2 \leq y \leq 4, z = 2}

The boundary of

S

C .

Because of the clockwise orientation of

C,

we assign the normal vector

- \hat{k}

S .

See the sketch below

Then, by Stokes’ theorem,

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{S} \nabla \nabla \times F \cdot (- \hat{k}) d S \\ = - \iint_{S} (2 x - \frac{1}{x}) d S \\ = - \int_{2}^{4} d x \int_{2}^{4} d y (2 x - \frac{1}{x}) = - \int_{2}^{4} d x 2 (2 x - \frac{1}{x}) \\ = - 2 [x^{2} - \ln x]_{2}^{4} = - 2 [12 - \ln 2] \\ = 2 \ln 2 - 24 \end{aligned}

(e) Since

\nabla \nabla \times F

is not

0,

F

cannot be conservative.

4.4.3.33. (✳).

Solution.

(a) By the vector identity

\nabla \nabla \cdot (\nabla \nabla \times G) = 0

(Theorem 4.1.7.a). So we must have

\begin{aligned} 0 & = \nabla \nabla \cdot F = \frac{\partial}{\partial x} (x z) + \frac{\partial}{\partial y} (a x e^{y} z + b y z) + \frac{\partial}{\partial z} (y^{2} - x e^{y} z^{2}) \\ = z + (a x e^{y} z + b z) + (- 2 x e^{y} z) \\ = (1 + b) z + (a - 2) x e^{y} z \end{aligned}

So we need

a = 2

and

b = - 1 .

(b) Note that the boundary,

\partial S,

is the circle

x^{2} + y^{2} = 1,

z = 0,

oriented counter-clockwise. Also note that, if we knew what

G

was, we would be able to use Stokes’ theorem to give

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \iint_{S} (\nabla \nabla \times G) \cdot \hat{n} d S = \oint_{\partial S} G \cdot d r \end{aligned}

So let’s find a vector potential

G .

That is, let’s try and find a vector field

G = G_{1} \hat{ı ı} + G_{2} \hat{ȷ ȷ} + G_{3} \hat{k}

that obeys

\nabla \nabla \times G = F,

or equivalently,

\begin{aligned} \frac{\partial G_{3}}{\partial y} - \frac{\partial G_{2}}{\partial z} & = F_{1} = x z \\ - \frac{\partial G_{3}}{\partial x} + \frac{\partial G_{1}}{\partial z} & = F_{2} = 2 x e^{y} z - y z \\ \frac{\partial G_{2}}{\partial x} - \frac{\partial G_{1}}{\partial y} & = F_{3} = y^{2} - x e^{y} z^{2} \end{aligned}

Let’s also require that

G_{3} = 0 .

(If this is mysterious to you, review §4.1.2.) Then the equations above simplify to

\begin{aligned} - \frac{\partial G_{2}}{\partial z} & = x z \\ \frac{\partial G_{1}}{\partial z} & = 2 x e^{y} z - y z \\ \frac{\partial G_{2}}{\partial x} - \frac{\partial G_{1}}{\partial y} & = y^{2} - x e^{y} z^{2} \end{aligned}

Now the first equation contains only a single unknown, namely

G_{2}

and we can find all

G_{2}

’s that obey the first equation simply by integrating with respect to

z :

G_{2} = - \frac{x z^{2}}{2} + N (x, y)

Note that, because

\frac{\partial}{\partial z}

treats

x

and

y

as constants, the constant of integration

N

is allowed to depend on

x

and

y .

Similarly, the second equation contains only a single unknown,

G_{1},

and is easily solved by integrating with respect to

z .

The second equation is satisfied if and only if

G_{1} = x e^{y} z^{2} - \frac{1}{2} y z^{2} + M (x, y)

for some function

M .

Finally, the third equation is also satisfied if and only if

M (x, y)

and

N (x, y)

obey

\frac{\partial}{\partial x} (- \frac{x z^{2}}{2} + N (x, y)) - \frac{\partial}{\partial y} (x e^{y} z^{2} - \frac{y z^{2}}{2} + M (x, y)) = y^{2} - x e^{y} z^{2}

which simplifies to

\frac{\partial N}{\partial x} (x, y) - \frac{\partial M}{\partial y} (x, y) = y^{2}

This is one linear equation in two unknowns,

M

and

N .

Typically, we can easily solve one linear equation in one unknown.

So we are free to eliminate one of the unknowns by setting, for example,

M = 0,

and then choosing any

N

that obeys

\frac{\partial N}{\partial x} (x, y) = y^{2}

Integrating with respect to

x

gives, as one possible choice,

N (x, y) = x y^{2} .

So we have found a vector potential. Namely

G = (x e^{y} z^{2} - \frac{1}{2} y z^{2}) \hat{ı ı} + (x y^{2} - \frac{x z^{2}}{2}) \hat{ȷ ȷ}

We can now evaluate the flux. Parametrize

\partial S

\begin{aligned} r (θ) & = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} \\ r^{'} (θ) & = - \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ} \end{aligned}

with

0 \leq θ \leq 2 π .

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = \oint_{\partial S} G \cdot d r \\ = \int_{0}^{2 π} (\overset{G (r (θ))}{\overset{⏞}{\cos θ \sin^{2} θ \hat{ȷ ȷ}}}) \cdot \overset{r^{'} (θ)}{\overset{⏞}{(- \sin θ \hat{ı ı} + \cos θ \hat{ȷ ȷ})}} d θ \\ = \int_{0}^{2 π} \sin^{2} θ \cos^{2} θ d θ \\ = \int_{0}^{2 π} \frac{1 - \cos (2 θ)}{2} \frac{1 + \cos (2 θ)}{2} d θ \\ = \frac{1}{4} \int_{0}^{2 π} {1 - \cos^{2} (2 θ)} d θ \\ = \frac{1}{4} \int_{0}^{2 π} {1 - \frac{1 + \cos (4 θ)}{2}} d θ \\ = \frac{1}{4} \frac{1}{2} 2 π since \int_{0}^{2 π} \cos (4 θ) d θ = 0 \\ = \frac{π}{4} \end{aligned}

4.4.3.34. (✳).

Solution.

Considering that there are ten line segments in

C,

it is probably not very efficient to use direct evaluation. Two other possible methods come to mind. If

F

is conservative, we can use

F

’s potential. Even if

F

is not conservative, it may be possible to efficiently use Stokes’ (or Green’s) theorem. So let’s compute

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & 2 x - 10 & 0 \end{array}] = \hat{k} \end{aligned}

\nabla \nabla \times F \neq 0,

the vector field

F

is not conservative. As

\nabla \nabla \times F \neq 0

is very simple, it looks like Stokes’ theorem could provide an efficient way to compute the integral. The left figure below contains a sketch of

C .

The curve

C

is not closed, and so is not the boundary of a surface, so we cannot apply Stokes’ theorem directly. But we can easily come up with a surface whose boundary contains

C .

Let

R

be the shaded region in the figure on the right above. The boundary

\partial R

R

consists of two parts —

C

and the line segment

L .

The normal of

R

for

- \hat{k}

(since

\partial R

is oriented clockwise). So Stokes’ theorem gives

\begin{aligned} \int_{C} F \cdot d r + \int_{L} F \cdot d r & = \iint_{R} \nabla \nabla \times F \cdot (- \hat{k}) d S = \iint_{R} (\hat{k}) \cdot (- \hat{k}) d S \\ = - Area (R) \end{aligned}

R

is the union of

5

triangles, each of height

1

and base

1 .

Area (R) = 5 \times \frac{1}{2} \times 1 \times 1 = \frac{5}{2}

If we denote by

- L

the line segment from

(0, 0)

(5, 5),

we can parametrize

- L

r (t) = t (5, 5),

0 \leq t \leq 1

and

\begin{aligned} \int_{- L} F \cdot d r & = \int_{0}^{1} \overset{F (r (t))}{\overset{⏞}{(5 t \hat{ı ı} + (10 t - 10) \hat{ȷ ȷ})}} \cdot \overset{r^{'} (t)}{\overset{⏞}{(5 \hat{ı ı} + 5 \hat{ȷ ȷ})}} d t = \int_{0}^{1} 5 (15 t - 10) d t \\ = 25 (\frac{3}{2} - 2) = - \frac{25}{2} \end{aligned}

All together

\begin{aligned} \int_{C} F \cdot d r & = - Area (R) - \int_{L} F \cdot d r = - Area (R) + \int_{- L} F \cdot d r = - \frac{5}{2} - \frac{25}{2} \\ = - 15 \end{aligned}

4.4.3.35. (✳).

Solution.

If we parametrize the curve as

x = 2 \cos θ y = 2 \sin θ z = x^{2} = 4 \cos^{2} θ 0 \leq θ \leq 2 π

then the term

\sin x (θ)^{2} x^{'} (θ)

in the integral will be

\sin (4 \cos^{2} θ) (- 2 \sin θ) .

That looks hard to integrate. So let’s try Stokes’ theorem. The curl of

F

\begin{aligned} \nabla \nabla \times F & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \sin x^{2} & x z & z^{2} \end{array}] = - x \hat{ı ı} + z \hat{k} \end{aligned}

The curve

C

is the boundary of the surface

S = {(x, y, z) | x^{2} + y^{2} \leq 4, z = x^{2}}

with upward pointing normal. For the surface

z = f (x, y) = x^{2},

(3.3.2) gives

\begin{aligned} \hat{n} d S & = \pm [- f_{x} (x, y) \hat{ı ı} - f_{y} (x, y) \hat{ȷ ȷ} + \hat{k}] d x d y \\ = \pm [- 2 x \hat{ı ı} + \hat{k}] d x d y \end{aligned}

Since we want the upward pointing normal

\begin{aligned} \hat{n} d S & = [- 2 x \hat{ı ı} + \hat{k}] d x d y \end{aligned}

So by Stokes’ theorem (Theorem 4.4.1)

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{x^{2} + y^{2} \leq 4} (- x \hat{ı ı} + \overset{z}{\overset{⏞}{x^{2}}} \hat{k}) \cdot [- 2 x \hat{ı ı} + \hat{k}] d x d y \\ = 3 \iint_{x^{2} + y^{2} \leq 4} x^{2} d x d y \end{aligned}

Switching to polar coordinates

\begin{aligned} \oint_{C} F \cdot d r & = 3 \int_{0}^{2} d r r \int_{0}^{2 π} d θ r^{2} \cos^{2} θ \\ = 3 [\int_{0}^{2} r^{3} d r] [\int_{0}^{2 π} \cos^{2} θ d θ] \\ = 3 \frac{2^{4}}{4} [\int_{0}^{2 π} \frac{\cos (2 θ) + 1}{2} d θ] \\ = 12 π \end{aligned}

For an efficient, sneaky, way to evaluate

\int_{0}^{2 π} \cos^{2} t d t

see Example 2.4.4.

4.4.3.36. (✳).

Solution.

By Stokes’ Theorem,

\oint_{C} E \cdot d r = \iint_{S} (\nabla \nabla \times E) \cdot \hat{n} d S

so Faraday’s law becomes

\iint_{S} (\nabla \nabla \times E + \frac{1}{c} \frac{\partial H}{\partial t}) \cdot \hat{n} d S = 0

This is true for all surfaces

S .

So the integrand, assuming that it is continuous, must be zero.

To see this, let

G = (\nabla \nabla \times E + \frac{1}{c} \frac{\partial H}{\partial t}) .

Suppose that

G (x_{0}) \neq 0 .

Pick a unit vector

\hat{n}

in the direction of

G (x_{0}) .

Let

S

be a very small flat disk centered on

x_{0}

with normal

\hat{n}

(the vector we picked). Then

G (x_{0}) \cdot \hat{n} > 0

and, by continuity,

G (x) \cdot \hat{n} > 0

for all

x

S,

if we have picked

S

small enough. Then

\iint_{S} (\nabla \nabla \times E + \frac{1}{c} \frac{\partial H}{\partial t}) \cdot \hat{n} d S > 0,

which is a contradiction. So

G = 0

everywhere and we conclude that

\nabla \nabla \times E + \frac{1}{c} \frac{\partial H}{\partial t} = 0

4.4.3.37. (✳).

Solution.

The curl of the specified vector field is

\begin{aligned} \nabla \times F & = \nabla \times (z \hat{ı ı} + x \hat{ȷ ȷ} + y^{3} z^{3} \hat{k}) \\ = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z & x & y^{3} z^{3} \end{array}] \\ = 3 y^{2} z^{3} \hat{ı ı} + \hat{ȷ ȷ} + \hat{k} \end{aligned}

For every

t,

we have

x (t) = z (t)

and

x (t)^{2} + y (t)^{2} + z (t)^{2} = 2 .

So the specified curve is the intersection of the plane

x = z

and the sphere

x^{2} + y^{2} + z^{2} = 2 .

This curve is the boundary of the circular disk

D = {(x, y, z) | x = z, x^{2} + y^{2} + z^{2} \leq 2}

The curve is oriented so that

(x (t), y (t)) = (\cos t, \sqrt{2} \sin t)

runs in the standard (counterclockwise) direction. So the unit normal to

D

used in Stokes’ theorem has positive

\hat{k}

component. Since the plane

x - z = 0

has unit normal

\pm \frac{1}{\sqrt{2}} (1, 0, - 1),

the unit normal used in Stokes’ theorem is

\hat{n} = \frac{1}{\sqrt{2}} (- 1, 0, 1) .

By Stokes’ theorem

\begin{aligned} \oint_{C} F \cdot d r & = \iint_{D} \nabla \times F \cdot \hat{n} d S = \frac{1}{\sqrt{2}} \iint_{D} (3 y^{2} z^{3}, 1, 1) \cdot (- 1, 0, 1) d S \\ = \frac{1}{\sqrt{2}} \iint_{D} (1 - 3 y^{2} z^{3}) d S \end{aligned}

The disk

D

is invariant under the reflection

(x, y, z) \to (- x, y, - z) .

Since

y^{2} z^{3}

is odd under this reflection,

\iint_{D} y^{2} z^{3} d s = 0

and

\oint F \cdot d r = \frac{1}{\sqrt{2}} \iint_{D} d S = \frac{1}{\sqrt{2}} A r e a (D)

Because the centre of the ball

x^{2} + y^{2} + z^{2} \leq 2

(namely

(0, 0, 0)

) is contained in the plane

x = z,

the radius of the disk

D

is the same as the radius of the sphere

x^{2} + y^{2} + z^{2} = 2 .

D

has radius

\sqrt{2}

and

\oint F \cdot d r = \frac{1}{\sqrt{2}} A r e a (D) = \frac{1}{\sqrt{2}} π {(\sqrt{2})}^{2} = \sqrt{2} π

4.4.3.38. (✳).

Solution.

The curl of the vector field

F = z \hat{ı ı} + x \hat{ȷ ȷ} - y \hat{k}

\nabla \nabla \times F = - \hat{ı ı} + \hat{ȷ ȷ} + \hat{k}

The unit normal to the plane

x + y + z = 1,

with positive

\hat{k}

component as required by Stokes’ theorem in this case, is

\hat{n} = \frac{1}{\sqrt{3}} (1, 1, 1) .

If we denote by

D

the circular disk

x + y + z = 1,

x^{2} + y^{2} + z^{2} \leq 1,

then Stokes’ theorem (Theorem 4.4.1) says

\begin{aligned} \oint_{C} z d x + x d y - y d z = \oint_{C} F \cdot d r & = \iint_{D} \nabla \nabla \times F \cdot \hat{n} d S \\ = \iint_{D} (- 1, 1, 1) \cdot \frac{1}{\sqrt{3}} (1, 1, 1) d S \\ = \frac{1}{\sqrt{3}} A r e a (D) \end{aligned}

A reasonable guess for the centre of the disk is

\frac{1}{3} (1, 1, 1) .

(This guess is just based on symmetry.) To check this we just need to observe that it is indeed on the plane

x + y + z = 1

and that the distance from

\frac{1}{3} (1, 1, 1)

to any point

(x, y, z)

obeying

x + y + z = 1

and

x^{2} + y^{2} + z^{2} = 1,

namely

\begin{aligned} \sqrt{{(x - \frac{1}{3})}^{2} + {(y - \frac{1}{3})}^{2} + {(z - \frac{1}{3})}^{2}} \\ = \sqrt{x^{2} + y^{2} + z^{2} - \frac{2}{3} (x + y + z) + \frac{3}{9}} = \sqrt{1 - \frac{2}{3} + \frac{1}{3}} \\ = \sqrt{\frac{2}{3}} \end{aligned}

is the same. This also tells us that

D

has radius

\sqrt{\frac{2}{3}}

and hence area

\frac{2}{3} π .

So the specified line integral is

\frac{2 π}{3 \sqrt{3}} .

4.4.3.39. (✳).

Solution.

(a) We parametrize

S

in cylindrical coordinates:

r (r, θ) = r \cos θ \hat{ı ı} + r \sin θ \hat{ȷ ȷ} + r \hat{k} with 0 \leq r \leq 1, 0 \leq θ \leq π

(b) We compute

\begin{aligned} \frac{\partial r}{\partial r} & = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + \hat{k} \\ \frac{\partial r}{\partial θ} & = - r \sin θ \hat{ı ı} + r \cos θ \hat{ȷ ȷ} \\ \hat{n} d S & = \pm \frac{\partial r}{\partial r} \times \frac{\partial r}{\partial θ} d r d θ = \pm (- r \cos θ \hat{ı ı} - r \sin θ \hat{ȷ ȷ} + r \hat{k}) d r d θ \end{aligned}

To calculate the downward flux, we use the minus sign. We find

\begin{aligned} \iint_{S} v \cdot \hat{n} d S & = \int_{0}^{π} d θ \int_{0}^{1} d r (r \cos θ, r \sin θ, - 2 r) \cdot (r \cos θ, r \sin θ, - r) \\ = \int_{0}^{π} d θ \int_{0}^{1} d r 3 r^{2} = π r^{3} |_{r = 0}^{1} = π \end{aligned}

P

be the path along line segments from

(1, 0, 1)

(0, 0, 0)

and from

(0, 0, 0)

(- 1, 0, 1) .

Here is a sketch.

P

is in blue.

Then

\int_{C} F \cdot d r + \int_{P} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S

by Stokes’ Theorem. Along

P,

the vector field

F

is orthogonal to the curve so that

\int_{P} F \cdot d r = 0 .

Note that

\nabla \times F

is the vector field

v

from part (b). Thus

\int_{C} F \cdot d r = \iint_{S} v \cdot \hat{n} d S = π

L

be the line segment from

(1, 0, 1)

(- 1, 0, 1)

and let

\begin{matrix} R = {(x, y, z)} x^{2} + y^{2} \leq 1, y \geq 0, z = 1 \end{matrix}

Here is a sketch.

L

is in blue and

R

is shaded.

Then

\int_{C} F \cdot d r + \int_{L} F \cdot d r = \iint_{R} \nabla \nabla \times F \cdot (- \hat{k}) d S

by Stokes’ Theorem. Along

L,

the vector field

F = \hat{ȷ ȷ}

is orthogonal to the curve (which has direction

- \hat{ı ı}

so that

\int_{L} F \cdot d r = 0 .

Note that

\nabla \times F

is the vector field

v

from part (b). Thus

\int_{C} F \cdot d r = - \iint_{R} v \cdot \hat{k} d S = \iint_{R} 2 z d S = 2 \iint_{R} d S = 2 Area (R) = π

4.4.3.40.

Solution.

Let

S^{'}

be the portion of

x + y + z = 1

that is inside the sphere

x^{2} + y^{2} + z^{2} = 1 .

Then

\partial S = \partial S^{'},

so, by Stokes’ theorem, (with

\hat{n}

always the upward pointing normal)

\iint_{S^{'}} (\nabla \nabla \times F) \cdot \hat{n} d S = \oint_{\partial S^{'}} F \cdot d r = \oint_{\partial S} F \cdot d r = \iint_{S} (\nabla \nabla \times F) \cdot \hat{n} d S

\begin{array}{r} \nabla \nabla \times F = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y - z & z - x & x - y \end{array}] = - 2 (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k}) \end{array}

and, on

S^{'},

\hat{n} = \frac{1}{\sqrt{3}} (\hat{ı ı} + \hat{ȷ ȷ} + \hat{k})

\iint_{S^{'}} (\nabla \nabla \times F) \cdot \hat{n} d S = \iint_{S^{'}} (- 2 \sqrt{3}) d S = - 2 \sqrt{3} \times A r e a (S^{'})

S^{'}

is the intersection of a plane with a sphere and so is a circular disk. It’s center

(x_{c}, y_{c}, z_{c})

has to obey

x_{c} + y_{c} + z_{c} = 1 .

By symmetry,

x_{c} = y_{c} = z_{c},

x_{c} = y_{c} = z_{c} = \frac{1}{3} .

Any point,

(x, y, z),

which satisfies both

x + y + z = 1

and

x^{2} + y^{2} + z^{2} = 1,

obeys

\begin{aligned} {(x - \frac{1}{3})}^{2} + {(y - \frac{1}{3})}^{2} + {(z - \frac{1}{3})}^{2} \\ = x^{2} + y^{2} + z^{2} - \frac{2}{3} (x + y + z) + 3 \frac{1}{9} \\ = 1 - \frac{2}{3} + \frac{1}{3} = \frac{2}{3} \end{aligned}

That is, any point on the boundary of

S^{'}

is a distance

\sqrt{\frac{2}{3}}

from

(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) .

So the radius of

S^{'}

\sqrt{\frac{2}{3}},

the area of

S^{'}

\frac{2}{3} π

and

\iint_{S^{'}} (\nabla \nabla \times F) \cdot \hat{n} d S = - 2 \sqrt{3} \times A r e a (S^{'}) = - \frac{4}{\sqrt{3}} π

5 True/False and Other Short Questions
5.2 Exercises

5.2.1. (✳).

Solution.

(a) True. For any constant vector

a = (a_{1}, a_{2}, a_{3}),

\begin{aligned} a \times r & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ x & y & z \end{array}] = (a_{2} z - a_{3} y) \hat{ı ı} - (a_{1} z - a_{3} x) \hat{ȷ ȷ} + (a_{1} y - a_{2} x) \hat{ȷ ȷ} \end{aligned}

This vector field does indeed have divergence

0 .

(b) True. This is our conservative field screening condition Theorem 4.1.7.b.

(d) False. The trap here is that

F

need not be defined at the origin. We saw, in Example 3.4.2, that the point source

F_{S} = \frac{m r}{| r |^{3}}

had flux

4 π m

through every sphere centred on the origin. We also saw, in Example 4.2.7, that the divergence

\nabla \nabla \cdot F_{S} = 0

everywhere except at the origin (where it is not defined). So if we choose

m

to be a very big negative number (say

- 10^{100}

) and add in a very small vector field with positive divergence (say

10^{- 100} (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k})

), we will get the vector field

F = - 10^{100} \frac{r}{| r |^{3}} + 10^{- 100} (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k})

which has divergence

\nabla \nabla \cdot F = 3 \times 10^{- 100} > 0

everywhere except at the origin. The flux of this field through the specified sphere will be

- 4 π \times 10^{100}

plus a very small positive number.

(e) True. The statement that “the flux out of one hemisphere is equal to the flux into the opposite hemisphere” is equivalent to the statement that “the flux out of the sphere is equal to zero”. Since

\nabla \nabla \cdot F = 0

everywhere, that is true by the divergence theorem.

(f) That depends.

κ = 0,

then

\frac{d \hat{T}}{d s} = 0,

so that

\frac{d r}{d s} = \hat{T}

is a constant. So

r (s) = s \hat{T} + r (0)

is part of a straight line.

κ > 0,

then, because the curve is in a plane, the torsion

τ

is zero and the Frenet-Serret formulae reduce to

\frac{d \hat{T}}{d s} = κ \hat{N} \frac{d \hat{N}}{d s} = - κ \hat{T}

Now consider the centre of curvature

c (s) = r (s) + \frac{1}{κ} \hat{N} (s) .

Since

\frac{d c}{d s} = \frac{d r}{d s} + \frac{1}{κ} \frac{d \hat{N}}{d s} = \hat{T} (s) + \frac{1}{κ} (- κ \hat{T} (s)) = 0

c (s)

is a constant and

| r (s) - c | = \frac{1}{κ}

which says that the curve is part of the circle of radius

\frac{1}{κ}

centred on

c .

(g) False. We saw in Examples 2.3.14 and 4.3.8 that the given vector field is not conservative.

(h) False. For example, if

P = - y,

then

\oint_{C} F \cdot d r = - \oint_{C} y d x

is the area inside

C .

See Corollary 4.3.5.

(i) False.

\frac{d v}{d t} = a

is a constant, then

v (t) = a t + v_{0} .

Integrating a second time,

r (t) = \frac{1}{2} a t^{2} + v_{0} t + r_{0} .

This is not a spiral, whether or not the speed is constant. (In fact, for the speed

| v (t) | = | a t + v_{0} |

to be constant,

a

has to be

0,

so that

r (t) = v_{0} t + r_{0}

is a straight line.)

Another way to come to the same conclusion uses

a (t) = \frac{d^{2} s}{d t^{2}} (t) \hat{T} (t) + κ (t) (\frac{d s}{d t} (t))^{2} \hat{N} (t)

As the speed

\frac{d s}{d t}

is a constant, it reduces to

a (t) = κ (t) (\frac{d s}{d t} (t))^{2} \hat{N} (t)

a (t)

is a constant, its direction,

\hat{N} (t),

is also a constant. The normal vector to a spiral is not constant.

5.2.2. (✳).

Solution.

(a) False. For any constant vector

a = (a_{1}, a_{2}, a_{3}),

\begin{aligned} a \times r & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ x & y & z \end{array}] = (a_{2} z - a_{3} y) \hat{ı ı} - (a_{1} z - a_{3} x) \hat{ȷ ȷ} + (a_{1} y - a_{2} x) \hat{ȷ ȷ} \end{aligned}

\begin{aligned} \nabla \nabla \times (a \times r) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ a_{2} z - a_{3} y & - a_{1} z + a_{3} x & a_{1} y - a_{2} x \end{array}] \\ = 2 a_{1} \hat{ı ı} + 2 a_{2} \hat{ȷ ȷ} + 2 a_{3} \hat{k} \end{aligned}

is nonzero, unless the constant vector

a = 0 .

(b) False. For example, if

f (x) = x^{2},

then

\begin{matrix} \nabla \nabla \cdot (\nabla \nabla f) = \nabla \nabla \cdot (\nabla \nabla x^{2}) = \nabla \nabla \cdot (2 x \hat{ı ı}) = 2 \end{matrix}

F = x^{2} \hat{ı ı},

then

\begin{matrix} \nabla \nabla (\nabla \nabla \cdot F) = \nabla \nabla (\nabla \nabla \cdot (x^{2} \hat{ı ı})) = \nabla \nabla (2 x) = 2 \hat{ı ı} \end{matrix}

(d) False. The trap here is that

F

need not be defined at the origin. We saw, in Example 3.4.2, that the point source

F = \frac{m r}{| r |^{3}}

had flux

4 π m

through every sphere centred on the origin. We also saw, in Example 4.2.7, that the divergence

\nabla \nabla \cdot F = 0

everywhere except at the origin (where it is not defined).

(e) True. Any simple, smooth, closed curve in

R^{3}

that avoids the origin is the boundary of a surface

S

that also avoids the origin. Then, by Stokes’ theorem,

\oint_{C} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = 0

(f) True. Let

S = {r | | r - c | = R}

be a sphere. Denote by

V = {r | | r - c | \leq R}

the ball whose boundary is

S .

Let

H

be one hemisphere of

S

with outward pointing normal and let

H^{'}

be the other hemisphere of

S

with inward point normal. Then the boundary of

V,

with outward pointing normal, can be viewed as consisting of two parts, namely

H

and

- H^{'},

where by

- H^{'}

we mean

H^{'}

but with outward pointing normal. Then, by the divergence theorem

\begin{aligned} \iint_{H} F \cdot \hat{n} d S - \iint_{H^{'}} F \cdot \hat{n} d S & = \iint_{\partial V} F \cdot \hat{n} d S \\ = ∭_{V} \nabla \nabla \cdot F d V > 0 \end{aligned}

which implies that

\iint_{H} F \cdot \hat{n} d S > \iint_{H^{'}} F \cdot \hat{n} d S .

(g) False. The trap here is that the curve is in

R^{3},

not

R^{2} .

As we saw in Example 1.4.4, a helix has constant curvature, but does not lie in a plane and so is not part of a circle.

(h) False. Even if we restrict

F

to the

x y

-plane (i.e. to

z = 0

), this vector field is not conservative. We saw that in Examples 2.3.14 and 4.3.8.

(i) False. For example, the vector field

F = x \hat{k}

is always parallel to the

z

-axis. So its flow lines are also all parallel to the

z

-axis. But if the closed curve

C

consists of the line segments

$L_{1}$ from $(0, 0, 0)$ to $(1, 0, 0),$ followed by
$L_{2}$ from $(1, 0, 0)$ to $(1, 0, 1),$ followed by
$L_{3}$ from $(1, 0, 1)$ to $(0, 0, 1),$ followed by
$L_{4}$ from $(0, 0, 1)$ back to $(0, 0, 0) .$

then

$\int_{L_{1}} F \cdot d r = \int_{0}^{1} (x \hat{k}) \cdot \hat{ı ı} d x = 0$ since $\hat{k} ⊥ \hat{ı ı},$ $d r = \hat{ı ı} d x$ on $L_{1}$ and
$\int_{L_{2}} F \cdot d r = \int_{0}^{1} (1 \hat{k}) \cdot \hat{k} d z = 1$ since $x = 1$ and $d r = \hat{k} d z$ on $L_{2}$ and
$\int_{L_{3}} F \cdot d r = - \int_{0}^{1} (x \hat{k}) \cdot \hat{ı ı} d x = 0$ since $\hat{k} ⊥ \hat{ı ı}$ and
$\int_{L_{4}} F \cdot d r = - \int_{0}^{1} (0 \hat{k}) \cdot \hat{k} d z = 0$ since $x = 0$ on $L_{4} .$

All together

\int_{C} F \cdot d r = \int_{L_{1}} F \cdot d r + \int_{L_{2}} F \cdot d r + \int_{L_{3}} F \cdot d r + \int_{L_{4}} F \cdot d r = 1

(j) True. If the speed

| v |

is constant then

\begin{matrix} 0 = \frac{d}{d t} | v |^{2} = \frac{d}{d t} (v \cdot v) = 2 v \cdot a \end{matrix}

5.2.3. (✳).

Solution.

(a) False.

r^{″} (t)

is the full acceleration. So

| r^{″} (t) |

is the magnitude of the full acceleration, not just the tangential component of acceleration. For example, if

r (t) = \cos t \hat{ı ı} + \sin t \hat{ȷ ȷ}

(i.e. the particle is just going around in circles), the acceleration

r^{″} (t) = - \cos t \hat{ı ı} - \sin t \hat{ȷ ȷ}

is perpendicular to the direction of motion. So the tangential component of acceleration is zero, while

| r^{″} (t) | = 1 .

(b)

\hat{T} (t)

is the tangent vector to the curve at

r (t) .

\hat{N} (t)

and

\hat{B} (t)

are both perpendicular to

\hat{T} (t)

(and to each other) and so span the plane normal to the curve at

r (t) .

(d) False. The statement

\nabla \nabla \times (\nabla \nabla \cdot F) = 0

is just plain gibberish, because

\nabla \nabla \cdot F

is a scalar valued function and there is no such thing as the curl of a scalar valued function.

(e) False. For example if

F = \hat{ı ı},

then, by the divergence theorem,

\iint_{S} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V = 0

Here

V = {x, y, z | x^{2} + y^{2} + z^{2} \leq 1}

is the inside of the sphere.

(f) True. If

S

is the boundary of the solid region

E,

then we can orient

S

by always choosing the normal vector that points into

E .

5.2.4. (✳).

Solution.

(a) The helix is approximately a bunch of circles stacked one on top of each other. The radius of the circles increase as

z

increases. So the curvature decreases as

z

increases.

(b) Here are two arguments both of which conclude that

f (x)

D .

If $C$ were the graph $y = f (x),$ then $f^{'} (x)$ would have two points of discontinuity. The curvature $κ (x)$ would not the defined at those two points. The function whose graph is $D$ is defined everywhere and so cannot be the curvature of the function whose graph is $C .$
The function whose graph is $D$ has two inflection points. So its curvature is zero at two points. The function whose graph is $C$ is indeed zero at two points (that in fact correspond to the inflection points of $D$ ). So $D$ is the graph of $f (x)$ and $C$ is the graph of $κ (x) .$

y,

x^{2} + z^{2} = 1

is a circle of radius

1 .

So we can parametrize it by

x (θ) = \cos θ,

z (θ) = \sin θ,

0 \leq θ < 2 π .

The

y

-coordinate of any point on the intersection is determined by

y = x z .

So we can use

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{k} + \sin θ \cos θ \hat{ȷ ȷ} 0 \leq θ < 2 π

(d) We are told that the helical ramp starts starts with the

y

-axis when

z = 0 .

In the cases of parametrisations (a) and (c), $z = 0$ forces $u = 0$ and $u = 0$ forces $x = y = 0 .$ That is only the origin, not the $y$ -axis. So we can rule out (a) and(c).
In the case of parametrisation (b), $z = 0$ forces $v = 0$ and $v = 0$ forces $y = 0$ and $x = u .$ As $u$ varies that sweeps out the $x$ -axis, not the $y$ -axis. So we can rule out (b).
In the case of parametrisation (d), $z = 0$ forces $v = 0$ and $v = 0$ forces $x = 0$ and $y = u .$ As $u$ varies that sweeps out the $y$ -axis, which is what we want.

Furthermore

we are told that $z = v$ runs from $0$ to $5$ and that
$x^{2} + y^{2} = u^{2} \geq 4$

So we want parametrisation (d) with domain

| u | \geq 2,

0 \leq v \leq 5 .

(e) Straight lines have curvature

0 .

So one acceptable parametrized curve is

r (t) = t \hat{ı ı},

0 \leq t \leq 1 .

(f) The cube

S

has six sides. So the outward flux through

\partial S

6

and, by the divergence theorem,

\begin{aligned} 6 & = \iint_{\partial S} F \cdot \hat{n} d S = ∭_{S} \nabla \nabla \cdot F d V = ∭_{S} C d V = C \end{aligned}

since

S

has volume one. So

C = 6 .

(g) For the vector field

F

to be conservative, we need

\begin{aligned} \frac{\partial F_{1}}{\partial y} & = \frac{\partial F_{2}}{\partial x} \\ ⟺ & \frac{\partial}{\partial y} (a x + b y) & = \frac{\partial}{\partial x} (c x + d y) \\ ⟺ & b & = c \end{aligned}

When

b = c,

an allowed potential is

\frac{a}{2} x^{2} + b x y + \frac{d}{2} y^{2} .

The specified set is

\begin{matrix} {(a, b, c, d) | a, b, c, d all real and b = c} \end{matrix}

(h) By the definition of arclength parametrisation, the arclength along the curve between

r (0)

and

r (s)

s .

In particular, the arclength between

r (0)

and

r (3)

3

and the arclength between

r (0)

and

r (5),

which is the same as the arclength between

r (0)

and

r (3)

plus the arclength between

r (3)

and

r (5),

5 .

So the arclength between

r (3)

and

r (5)

5 - 3 = 2 .

(i) In this solution, we’ll use, for example

- T

to refer to the curve

T,

but with the arrow pointing in the opposite direction to that of the arrow on

T .

In parts (2), (3) and (4) we will choose

F

to be the vector field

G (x, y) = - \frac{y}{x^{2} + y^{2}} \hat{ı ı} + \frac{x}{x^{2} + y^{2}} \hat{ȷ ȷ}

We saw, in Example 2.3.14, that

\nabla \nabla \times G = 0

except at the origin where it is not defined. We also saw, in Example 4.3.8, that

\oint_{C} G \cdot d r = 2 π

for any counterclockwise oriented circle centred on the origin.

(1): Let $R_{1}$ be the region between $S$ and $T .$ It is the shaded region in the figure on the left below. Note that $R_{1}$ is contained in the domain of $F,$ so that $\nabla \nabla \times F = 0$ on all of $R_{1} .$ The boundary of $R_{1}$ is $S - T,$ meaning that the boundary consists of two parts, with one part being $S$ and the other part being $- T .$ So, by Stokes’ theorem

$\int_{S} F \cdot d r - \int_{T} F \cdot d r = \int_{\partial R_{1}} F \cdot d r = \iint_{R_{1}} \nabla \nabla \times F \cdot \hat{k} d S = 0$

and (1) is true.
(2): False. Choose a coordinate system so that $Q$ is at the origin and choose $F = G .$ We saw, in Examples 2.3.14 and 4.3.8, that the curl of $G$ vanished everywhere except at the origin, where it was not defined, but that $\int_{R} G \cdot d r \neq 0 .$
(3),(4): False. Here is a counterxample that shows that both (3) and (4) are false. Choose a coordinate system so that $Q$ is at the origin and choose $F = G .$ By Stokes’ theorem

$\int_{S} G \cdot d r = \int_{T} G \cdot d r = 0$

because $\nabla \nabla \times G = 0$ everywhere inside $S,$ including at $P .$ So now both parts (3) and (4) reduce to the claim that $\int_{U} G \cdot d r = \int_{R} G \cdot d r .$

We saw, in Example 4.3.8, that $\int_{R} G \cdot d r = 2 π .$

To finish off the counterexample, we’ll now show that $\int_{U} G \cdot d r = - 2 π .$ Let $R_{2}$ be the region between $U$ and $R .$ It is the shaded region in the figure on the right above. Note that $\nabla \nabla \times G = 0$ on all of $R_{2} .$ including at $P .$ The boundary of $R_{2}$ is $- U - R,$ meaning that the boundary consists of two parts, with one part being $- U$ and the other part being $- R .$ So, by Stokes’ theorem

$- \int_{U} G \cdot d r - \int_{R} G \cdot d r = \int_{\partial R_{2}} G \cdot d r = \iint_{R_{2}} \nabla \nabla \times G \cdot \hat{k} d S = 0$

and $\int_{U} G \cdot d r = - \int_{R} G \cdot d r = - 2 π$
(5): False. For any conservative vector field $F,$ with potential $f,$ $\int_{V} F \cdot d r$ is just the difference of the values of $f$ at the two end points of $V .$ It is easy to choose an $f$ for which those two values are different. For example $f (x, y) = x$ does the job.

(j) Let

S

be any closed surface and denote by

V

the volume that it encloses. Presumably the question assumes that

S

is oriented so that

S = \partial V .

Then by the divergence theorem

\iint_{S} F \cdot \hat{n} d S = \iint_{\partial V} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V

This is exactly the volume of

V

\nabla \nabla \cdot F = 1

everywhere. One vector field

F

with

\nabla \nabla \cdot F = 1

everywhere is

F = x \hat{ı ı} .

(k) Let

C

be the counterclockwise boundary of a small square centred on

P,

like the blue curve in the figure below, but much smaller. Call the square (the inside of

C

)

S .

By Stokes’ theorem

\iint_{S} \nabla \nabla \times F \cdot \hat{k} d S = \oint_{C} F \cdot d r

The contribution to $\oint_{C} F \cdot d r$ coming from the left and right sides of $C$ will be zero, because $F$ is perpendicular to $d r$ there.
The contribution to $\oint_{C} F \cdot d r$ coming from the top of $C$ will be negative, because there $F$ is a positive number times $\hat{ı ı}$ and $d r$ is a negative number times $\hat{ı ı} .$
The contribution to $\oint_{C} F \cdot d r$ coming from the bottom of $C$ will be positive, because there $F$ is a positive number times $\hat{ı ı}$ and $d r$ is a positive number times $\hat{ı ı} .$
The magnitude of the contribution from the top of $C$ will be larger than the magnitude of the contribution from the bottom of $C,$ because $| F |$ is larger on the top than on the bottom.

So, all together,

\oint_{C} F \cdot d r < 0,

and consequently (taking a limit as the square size tends to zero)

\nabla \nabla \times F \cdot \hat{k}

is negative at

P .

5.2.5. (✳).

Solution.

(a) False. We could have, for example,

\nabla \nabla \cdot F

zero at one point and strictly positive elsewhere. One example would be

F = x^{3} \hat{ı ı} + y^{3} \hat{ȷ ȷ} + z^{3} \hat{k},

with

S_{1}

and

S_{2}

being the upward oriented top and bottom hemispheres, respectively, of the unit sphere

x^{2} + y^{2} + z^{2} = 1 .

(b) False. The conditions that (1)

\nabla \nabla \times F

= 0 and (2) the domain of

F

is simply-connected, are sufficient, but not necessary, to imply that

F

is conservative. For example the vector field

F = 0,

with any domain at all, is conservative with potential

0 .

Another example (which does not depend on choosing a domain that is smaller than the largest possible domain) is

F = \nabla \nabla \frac{1}{x^{2} + y^{2}}

with domain

{(x, y, z) | (x, y) \neq (0, 0)} .

That is, the domain is

R^{3}

with the

z

-axis removed.

r (t_{0})

on a parametrized curve

r (t) .

That’s the blue point in the figure below.

The centre of curvature for the curve at

r (t_{0})

c = r (t_{0}) + ρ (t_{0}) \hat{N} (t_{0}) .

It is the red dot in the figure.

The radius of the osculating circle is the distance from its centre, $c,$ to any point of the circle, like $r (t_{0}) .$ That’s $| r (t_{0}) - c | = | ρ (t_{0}) \hat{N} (t_{0}) | = ρ (t_{0}) .$ The curvature of the osculating circle is one over its radius. So its curvature is $\frac{1}{ρ (t_{0})} = κ (t_{0}) .$
The unit normal to the osculating circle at $r (t_{0})$ is a unit vector in the opposite direction to the radius vector from the centre $c$ to $r (t_{0}) .$ The radius vector is $r (t_{0}) - c_{0} = - ρ (t_{0}) \hat{N} (t_{0}),$ so the unit normal is $\hat{N} (t_{0}) .$
The osculating circle lies in the plane that best fits the curve near $r (t_{0}) .$ (See the beginning of §1.4.) So the unit tangents to the osculating circle at $r (t_{0})$ are perpendicular to both $\hat{N} (t_{0})$ and $\hat{B} (t_{0})$ and so are either $\hat{T} (t_{0})$ or $- \hat{T} (t_{0}),$ depending on how we orient the osculating circle.

(d) False. Kepler’s third law is that a planet orbiting a sun has the square of the period proportional to the cube of the major axis of the orbit.

(e) True. That’s part (a) of Theorem 4.1.7.

(f) True. Every domain contains closed surfaces. This has nothing to do with vector fields.

(g) True. We saw this in Example 2.3.4.

(h) False. Let

F

be an everywhere defined conservative vector field with potential

φ .

Then

\nabla \nabla \times F = 0

everywhere. If

P

and

Q

are two points and if

φ (P) - φ (Q) = 3

and if

C

is a curve from

Q

P,

then

\int_{C} F \cdot d r = 3 .

One example would be

φ (x, y, z) = x,

F = \hat{ı ı},

P = (3, 0, 0),

Q = (0, 0, 0) .

(i) False. The normal component of acceleration depends on speed, as well as curvature.

(j) False. The curve

r_{1}

contains only points in the

x y

-plane. Every

r_{2} (t)

with

t \neq 0

has a nonzero

z

-coordinate.

5.2.6. (✳).

Solution.

(a) False. Changing the orientation of a surface does not change

d S

at all. (It changes

\hat{n} d S

by a factor of

(- 1) .

) So

\iint_{S} f d S = + \iint_{- S} f d S

which is not

- \iint_{- S} f d S,

unless the integral is zero.

(b) False. For every vector field with two continuous partial derivatives,

\nabla \nabla \cdot (\nabla \nabla \times F) = 0

(see Theorem 4.1.7.a), so the divergence theorem gives

\begin{aligned} \iint_{S} (\nabla \nabla \times F) \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot (\nabla \nabla \times F) d V = 0 \end{aligned}

whether or not

F

is conservative.

F = f \hat{ı ı} .

Then, by Stokes’ theorem,

\begin{matrix} \int_{C} f d x = \int_{C} F \cdot d r = \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{S} (\frac{\partial f}{\partial z} \hat{ȷ ȷ} - \frac{\partial f}{\partial y} \hat{k}) \cdot \hat{n} d S \end{matrix}

(d) True. The left hand side,

(\nabla \nabla f) \times (\nabla \nabla f),

is zero because

(\nabla \nabla f)

is parallel to itself and the right hand side

\nabla \nabla \times (\nabla \nabla f)

is zero by Theorem 4.1.7.b (the screening test for conservative fields).

(e) True. The curve

r (t) = (2, 0, 1) + t^{3} (4, - 1, - 2)

is a straight line. Straight lines have curvature

0 .

(f) True. In general

| r^{'} (t) | = \frac{d s}{d t} .

Under arc length parametrization

t = s

so that

\frac{d s}{d t} = 1 .

(g) True. If

F

is a constant vector fleld, then, by the divergence thoerem,

\iint_{S} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V = ∭_{V} 0 d V = 0

(h) False. The statement

\nabla \nabla \times F = (x, y, z)

means that

F

is a vector potential for the vector field

G = (x, y, z) .

But

G

fails the screening test

\nabla \nabla \cdot G = 0

for vector potentials.

5.2.7. (✳).

Solution.

(a)

P

is the

x

-component of

F .

As we travel vertically upward through

A,

that

x

-component decreases. Hence

P_{y} < 0

A .

(b)

Q

is the

y

-component of

F .

As we travel horizontally to the right through

A,

that

y

-component increases. Hence

Q_{x} > 0

A .

(c)

\nabla \nabla \times F = (Q_{x} - P_{y}) \hat{k}

and

Q_{x} - P_{y} > 0

A,

so that the curl of

F

A

is in the direction of

+ \hat{k} .

(d) Along the curve

C_{1}

the magnitude of the angle between

F

and

d r

is less than

90^{\circ},

so that

F \cdot d r > 0

and

\int_{C_{1}} F \cdot d r > 0 .

(e) Along the curve

C_{2}

the magnitude of the angle between

F

and

d r

is greater than

90^{\circ},

so that

F \cdot d r < 0

and

\int_{C_{2}} F \cdot d r < 0 .

(f) If

F

were conservative, we would have

\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r .

As these two integrals have opposite signs

F

is not conservative. (Since

F

is not conservative, it is not the gradient of some function. At

A,

P_{x} > 0

and

Q_{y} > 0 .

F

is not divergence free and is not the curl of a vector potential.)

5.2.8. (✳).

Solution.

(a) False. The curve

r_{1}

contains only points with

z \geq 0 .

Every

r_{2} (t)

with

t < 0

has

z < 0 .

(b) True.

r_{2} (t^{2}) = r_{1} (t)

and

t^{2}

runs from

0

1

t

runs from

0

1 .

| r^{'} (t) | = \frac{d s}{d t} .

When

t = s,

\frac{d s}{d t} = 1 .

(d) False. The curve need not even lie in a plane. For example, as we saw in Example 1.4.4, the helix

r (t) = a \cos t \hat{ı ı} + b \sin t \hat{ȷ ȷ} + b t \hat{k}

has constant curvature

κ = \frac{a}{a^{2} + b^{2}}

but is not a circle.

(e) True. If the speed

| v | = \sqrt{v \cdot v}

of a moving object is constant, then

0 = \frac{d}{d t} (v \cdot v) = 2 v \cdot a

(f) False. If the vector field

F (x, y, z) = \frac{- y}{x^{2} + y^{2}} \hat{ı ı} + \frac{x}{x^{2} + y^{2}} \hat{ȷ ȷ} + z \hat{k}

were conservative, its restriction,

\frac{- y}{x^{2} + y^{2}} \hat{ı ı} + \frac{x}{x^{2} + y^{2}} \hat{ȷ ȷ},

to the

x y

-plane would also be conservative. But we saw in Examples 2.3.14 and 4.3.8 that the vector field

\frac{- y}{x^{2} + y^{2}} \hat{ı ı} + \frac{x}{x^{2} + y^{2}} \hat{ȷ ȷ}

is not conservative.

(g) False. The vector field of part (f), with domain

{(x, y, z) | x^{2} + y^{2} > 1},

provides a counterexample.

(h) False. The curve

x^{2} + y^{2} = 2

can not be shrunk to a point continuously in

{(x, y) | x^{2} + y^{2} > 1} .

(i) True. Any curve in

{(x, y) | y > x^{2}}

can be shrunk to a point continuously in

{(x, y) | y > x^{2}} .

(j) True. By the divergence theorem,

\begin{matrix} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = ∭_{E} \nabla \nabla \cdot (\nabla \nabla \times F) d V = 0 \end{matrix}

since

\nabla \nabla \cdot (\nabla \nabla \times F) = 0

by the vector identity of Theorem 4.1.7.a.

5.2.9. (✳).

Solution.

(a) True. In general

| r^{'} (t) | = \frac{d s}{d t} .

When

t = s,

\frac{d s}{d t} = 1 .

(b) False. The curve need not even lie in a plane. For example, as we saw in Example 1.4.4, the helix

r (t) = a \cos t \hat{ı ı} + b \sin t \hat{ȷ ȷ} + b t \hat{k}

has constant curvature

κ = \frac{a}{a^{2} + b^{2}}

but is not a circle.

(d) False. The vector field

F (x, y, z) = \frac{- y}{x^{2} + y^{2}} \hat{ı ı} + \frac{x}{x^{2} + y^{2}} \hat{ȷ ȷ},

with domain

{(x, y, z) | x^{2} + y^{2} > 1}

provides a counterexample.

(e) False. The curve

r_{1}

contains only points with

z \geq 0 .

Every

r_{2} (t)

with

t < 0

has

z < 0 .

(f) True.

r_{2} (t^{2}) = r_{1} (t)

and

t^{2}

runs from

0

1

t

runs from

0

1 .

(g) True.

\nabla \nabla \cdot (\nabla \nabla \times F) = 0

by the vector identity of Theorem 4.1.7.a.

(h) False. A counterexample is

f (x, y, z) = x^{2} .

It has

\nabla \nabla f = 2 x \hat{ı ı}

and hence

\nabla \nabla \cdot (\nabla \nabla f) = 2 .

(i) False. The curve

x^{2} + y^{2} = 2

can not be shrunk to a point continuously in

{(x, y) | x^{2} + y^{2} > 1} .

(j) True. Any curve in

{(x, y) | y > x^{2}}

can be shrunk to a point continuously in

{(x, y) | y > x^{2}} .

5.2.10. (✳).

Solution.

(a) False.

\nabla \nabla f = 0

if and only if

f

is constant. But if

f

is the constant

K,

then

\int_{C} f d s

K

times the length of

C,

which need not be zero.

(b) False. Any curve which lies in a plane has constant binormal. For example, the circle

r (t) = \cos t \hat{ı ı} + \sin t \hat{ȷ ȷ} + 0 \hat{k}

has constant binormal

\hat{B} = \hat{k},

but is not a straight line.

r (t)

has constant speed, the

(\frac{d s}{d t} (t))^{2} = r^{'} (t) \cdot r^{'} (t)

is constant and

\begin{matrix} 0 = \frac{d}{d t} (r^{'} (t) \cdot r^{'} (t)) = 2 r^{'} (t) \cdot r^{″} (t) \end{matrix}

(d) False. For the line integral

\int_{C} (F \times G) \cdot d r

to be independent of the path

C,

the vector field

F \times G

has to be conservative and so has to obey

\nabla \nabla \times (F \times G) = 0 .

But

Not all vector fields are conservative. For example, the vector field $H = x \hat{ȷ ȷ}$ obeys $\nabla \nabla \times H = \hat{k}$ and so is not conservative.
We can make $F \times G$ be any vector field through judicious choices of $F$ and $G .$ For example, if $F = x \hat{k}$ and $G = \hat{ı ı},$ then $F \times G = x \hat{k} \times \hat{ı ı} = x \hat{ȷ ȷ} = H .$

(e) True. The contribution to

\int_{C} f d s

from an “infinitesmal piece of

C

” is the value of

f

on the piece times the length of the piece. That does not depend on the orientation of the piece.

(f) False. The two vectors in the cross product

\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial u}

are identical. So the cross product is

0 .

(g) False. The integral is completely independent of

x (u, v)

and

y (u, v) .

In particular if, for example,

x (u, v) = 157 u,

y (u, v) = 157 v,

z (u, v) = 0

then

\iint_{D} {(1 + (\frac{\partial z}{\partial u})^{2} + (\frac{\partial z}{\partial v})^{2})}^{1 / 2} d u d v

is always exactly the area of

D,

while the area of

S

157^{2}

times the area of

D .

(h) True. If the fluid is incompressible then its flow preserves volumes and consequently

\nabla \nabla \cdot F = 0 .

(i) Not only False, but Ridiculous. The left had side is scalar valued while the right hand side is vector valued.

5.2.11. (✳).

Solution.

(a) True. That

\nabla \nabla \cdot (\nabla \nabla \times F) = 0

is the vector identity of Theorem 4.1.7.a. That identity is the basis of the vector potential screening test.

(b) False. If

F

is not conservative, then

\int_{C} F \cdot d r

will depend on the endpoints of

C .

\nabla \nabla f = 0,

then

\begin{aligned} \frac{\partial f}{\partial x} (x, y, z) & = 0 & ⟹ & f (x, y, z) & = g (y, z) \\ \frac{\partial f}{\partial y} (x, y, z) & = 0 & ⟹ & \frac{\partial g}{\partial y} (y, z) & = 0 & ⟹ & g (y, z) & = h (z) \\ \frac{\partial f}{\partial z} (x, y, z) & = 0 & ⟹ & h^{'} (z) & = 0 & ⟹ & h (z) & = C \end{aligned}

for some functions

g (y, z),

h (z)

and constant

C .

(d) False. The curl

\nabla \nabla \times F

is zero for every conservative vector fields

F .

There are many nonconstant conservative vector fields, like

F (x, y, z) = x \hat{ı ı} .

(e) True. As

S

is closed, it is the boundary of a solid region

V .

Then, by the divergence theorem,

\iint_{S} F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot F d V = 0

(f) True. If

\int_{C} F \cdot d r = 0

for every closed curve

C,

then

F

is conservative by Theorem 2.4.7. Consequently,

\nabla \nabla \times F = 0

by Theorem 2.3.9.

(g) True. If the speed

| v |

is constant then

\begin{matrix} 0 = \frac{d}{d t} | v |^{2} = \frac{d}{d t} (v \cdot v) = 2 v \cdot a \end{matrix}

Since

\hat{T} = \frac{v}{| v |},

\hat{T} \cdot a = 0

too. Here, we have assumed that the constant

| v |

is not zero. If the constant

| v |

is zero, then

\hat{T}

is not defined at all (and

a = 0

(h) False. The trap here is that the curve is in

R^{3},

not

R^{2} .

As we saw in Example 1.4.4, a helix has constant curvature, but does not lie in a plane and so is not part of a circle.

(i) False. The trap here is that we are told nothing about

\nabla \nabla \cdot F .

As an example, let

S_{1}

be the hemisphere

S_{1} = {(x, y, z) | x^{2} + y^{2} + z^{2} = 1, z \geq 0}

with upward pointing normal and

S_{2}

be the disk

S_{2} = {(x, y, 0) | x^{2} + y^{2} \leq 1}

also with upward pointing normal.

Set

V = {(x, y, z) | 0 \leq z \leq \sqrt{x^{2} + y^{2}}, x^{2} + y^{2} \leq 1}

Then the boundary,

\partial V,

V

consists of two parts, namely

S_{1}

(with normal pointing upwards) and

S_{2}

(but with normal pointing downwards). The divergence theorem (Theorem 4.2.2) gives

\begin{aligned} \iint_{S_{1}} F \cdot \hat{n} d S - \iint_{S_{2}} F \cdot \hat{n} d S & = ∭_{V} \nabla \nabla \cdot F d V \end{aligned}

\nabla \nabla \cdot F > 0

(as is the case, for example, if

F = x \hat{ı ı}

) then

\iint_{S_{1}} F \cdot \hat{n} d S - \iint_{S_{2}} F \cdot \hat{n} d S

is definitely nonzero.

(j) True. This is one of Kepler’s laws. See §1.9.

5.2.12. (✳).

Solution.

It’s (b). (a) is gibberish — the left hand side is a scalar while the right hand side is a vector. (c) is also gibberish — the left hand side is a vector while the right hand side is a scalar. (b) is the vector identity of Theorem 4.1.4.c.

5.2.13. (✳).

Solution.

(a) False. For example, if

f (x, y, z) = x^{2},

then

\nabla \nabla f = 2 x \hat{ı ı}

and

\nabla \nabla \cdot \nabla \nabla f = 2 .

(b) Not only false, but ridiculous. The left hand side is a vector while the right hand side is a scalar.

(d) True. That’s the screening test for conservative fields, Theorem 4.1.7.b.

(e) Not only false, but ridiculous. The curl of a scalar function is not defined.

(f) True. That’s the screening test for vector potentials, Theorem 4.1.7.a.

(g) False.

\begin{aligned} \nabla \nabla \cdot \frac{r}{| r |^{2}} = \frac{\partial}{\partial x} \frac{x}{x^{2} + y^{2} + z^{2}} + \frac{\partial}{\partial y} \frac{y}{x^{2} + y^{2} + z^{2}} + \frac{\partial}{\partial z} \frac{z}{x^{2} + y^{2} + z^{2}} \\ = \frac{1}{x^{2} + y^{2} + z^{2}} - \frac{2 x^{2}}{{[x^{2} + y^{2} + z^{2}]}^{2}} + \frac{1}{x^{2} + y^{2} + z^{2}} - \frac{2 y^{2}}{{[x^{2} + y^{2} + z^{2}]}^{2}} \\ + \frac{1}{x^{2} + y^{2} + z^{2}} - \frac{2 z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{2}} \\ = \frac{3 [x^{2} + y^{2} + z^{2}] - 2 x^{2} - 2 y^{2} - 2 z^{2}}{{[x^{2} + y^{2} + z^{2}]}^{2}} \\ = \frac{1}{x^{2} + y^{2} + z^{2}} \end{aligned}

(h) False. For any constant vector

ω ω = (ω_{1}, ω_{2}, ω_{3}),

\begin{aligned} ω ω \times r & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ ω_{1} & ω_{2} & ω_{3} \\ x & y & z \end{array}] \\ = (ω_{2} z - ω_{3} y) \hat{ı ı} - (ω_{1} z - ω_{3} x) \hat{ȷ ȷ} + (ω_{1} y - ω_{2} x) \hat{ȷ ȷ} \end{aligned}

\begin{aligned} \nabla \nabla \times (ω ω \times r) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ ω_{2} z - ω_{3} y & - ω_{1} z + ω_{3} x & ω_{1} y - ω_{2} x \end{array}] \\ = 2 ω_{1} \hat{ı ı} + 2 ω_{2} \hat{ȷ ȷ} + 2 ω_{3} \hat{k} \end{aligned}

is nonzero, unless the constant vector

ω ω = 0 .

(i) True. The given equation is equivalent (by the vector identity Theorem 4.1.4.c) to

∭_{Ω} \nabla \nabla \cdot (f F) d V = \iint_{\partial Ω} f F \cdot \hat{n} d S

which is true by the divergence theorem.

(j) False. One of the variants of the divergence theorem given in Theorem 4.2.9 is

\iint_{\partial Ω} f \hat{n} d S = ∭_{Ω} \nabla \nabla f d V

Note that the sign on the right hand side is “

+

”, not “

-

”. In order for the equation given in part (j) to be true, it would be necessary that

∭_{Ω} \nabla \nabla f d V = 0

for all smooth scalar functions

f .

That’s silly. One counterexample is

f (x) = x Ω = {(x, y, z) | x^{2} + y^{2} + z^{2} \leq 1}

Then

\begin{aligned} \iint_{\partial Ω} f \hat{n} d S & = \iint_{\partial Ω} x (\overset{\hat{n}}{\overset{⏞}{x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}}}) d S = \hat{ı ı} \iint_{\partial Ω} x^{2} d S \\ - ∭_{Ω} \nabla \nabla f d V & = - ∭_{Ω} \hat{ı ı} d V = - \hat{ı ı} ∭_{Ω} d V \end{aligned}

The coefficient of

\hat{ı ı}

is obviously strictly positive in the upper integral and strictly negative in the lower integral.

5.2.14. (✳).

Solution.

(a) True. If the vector field is

F = a \hat{ı ı} + b \hat{ȷ ȷ} + c \hat{k},

then

f (x, y, z) = a x + b y + c z

obeys

F = \nabla \nabla f

and so is a potential for

F .

(b) False. For example the vector field

F = x \hat{ı ı} - y \hat{ȷ ȷ}

obeys

\nabla \nabla \cdot F = 0

but is not a constant vector field.

r (t)

is not indentically

0 .

r (t)

and

\frac{d r}{d t}

are orthogonal at all points of the curve

C,

then

\frac{d}{d t} (r (t) \cdot r (t)) = 2 r (t) \cdot \frac{d r}{d t} (t) = 0

x (t)^{2} + y (t)^{2} + z (t)^{2} = r (t) \cdot r (t)

is a constant. If

r (t)

is not indentically

0,

that constant must be strictly positive. That is

x (t)^{2} + y (t)^{2} + z (t)^{2} = a^{2}

for some constant

a > 0 .

(d) False. The curvature (see §1.5) is

κ (t) = \frac{| \frac{d \hat{T}}{d t} (t) |}{| r^{'} (t) |}

Changing the orientation of the curve amounts to replacing

t

- t .

This changes the signs of

\hat{T}

and

r^{'},

but does not change

κ,

because the absolute values eliminate the signs.

(e) False. For example, the vector field

F = 0,

with domain

{(x, y, z) | x^{2} + y^{2} > 0}

is a conservative vector field (with potential

0

) whose domain is not simply connected. As a less nitpicky example, let

F = \nabla \nabla f

with

f = \frac{1}{x^{2} + y^{2}} .

The biggest possible domain for this vector field is also

{(x, y, z) | x^{2} + y^{2} > 0} .

5.2.15. (✳).

Solution.

(a) We are to compute the divergence of

x^{2} y \hat{ı ı} + e^{y} \sin x \hat{ȷ ȷ} + e^{z x} \hat{k} .

Since

\begin{aligned} \frac{\partial}{\partial x} (x^{2} y) & = 2 x y \\ \frac{\partial}{\partial y} (e^{y} \sin x) & = e^{y} \sin x \\ \frac{\partial}{\partial z} (e^{z x}) & = x e^{x z} \end{aligned}

the specified divergence is

\begin{aligned} \nabla \nabla \cdot (x^{2} y \hat{ı ı} + e^{y} \sin x \hat{ȷ ȷ} + e^{z x} \hat{k}) & = 2 x y + e^{y} \sin x + x e^{x z} \end{aligned}

(b) The specified curl is

\begin{aligned} \nabla \nabla \times (\cos x^{2} \hat{ı ı} - y^{3} z \hat{ȷ ȷ} + x z \hat{k}) & = det [\begin{array}{c} \hat{ı ı} & \hat{ȷ ȷ} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \cos x^{2} & - y^{3} z & x z \end{array}] = y^{3} \hat{ı ı} - z \hat{ȷ ȷ} \end{aligned}

{(x, y, z) | x^{2} + y^{2} > 0} .

We are not told which subset to use, so, by default,

D

is the maximal domain

D = {(x, y, z) | x^{2} + y^{2} > 0} = {(x, y, z) | (x, y) \neq (0, 0)}

This

D

is connected (any two points in

D

can be joined by a curve that lies completely in

D

) but is not simply connected (the simple closed curve

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ},

0 \leq θ \leq 2 π

lies in

D

but cannot be shrunk to a point continuously in

D

). So (I) and (IV) are true. That’s (iii).

(d) False. If the position of the particle at time

t

r (t) = \cos t \hat{ı ı} + \sin t \hat{ȷ ȷ},

then its speed is the constant

1

but its acceleration is

- \cos t \hat{ı ı} - \sin t \hat{ȷ ȷ},

which is nonzero.

5.2.16. (✳).

Solution.

(a) True. By the vector identity of Theorem 4.1.5.c,

\begin{matrix} \nabla \nabla \times (f \nabla \nabla f) = (\nabla \nabla f) \times (\nabla \nabla f) + f \nabla \nabla \times (\nabla \nabla f) = 0 \end{matrix}

The second term vanished because of the screening test vector identity of Theorem 4.1.7.b.

(b) True. That’s the vector identity of Theorem 4.1.4.c.

0

the curve must have unit tangent vector

\hat{T} (s)

obeying

\frac{d \hat{T}}{d s} (s) = 0

(See §1.5.) So

r^{'} (s) = \hat{T} (s)

must be a constant vector. Call it

{\hat{T}}_{0} .

Integrating gives

r (s) = s {\hat{T}}_{0} + r_{0}

for some constant vector

r_{0} .

r (s)

lies on the same straight line for all

s .

(d) False. The trap here is that the curve is in

R^{3},

not

R^{2} .

As we saw in Example 1.4.4, a helix has constant curvature, but does not lie in a plane and so is not part of a circle.

(e) True. The vector field

F = \nabla \nabla f

is conservative. So, by Theorem 2.4.7.b, the work integral

\int_{C} \nabla \nabla f \cdot d r = \int_{C} F \cdot d r = 0

for any closed curve

C,

and, in particular, for any circle

C .

(f) True. The statement that “the flux out of one hemisphere is equal to the flux into the opposite hemisphere” is equivalent to the statement that “the flux out of the sphere is equal to zero”. Since

\nabla \nabla \cdot F = 0

everywhere, that is true by the divergence theorem.

(g) True. Let

S

be the boundary of the solid region

V .

Then, by the divergence theorem (Theorem 4.2.9),

\begin{matrix} \iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot (\nabla \nabla \times F) d V \end{matrix}

But

\nabla \nabla \cdot (\nabla \nabla \times F)

is identically zero, by the screening test vector identity of Theorem 4.1.7.a. So the integral is zero.

5.2.17. (✳).

Solution.

(a) True. Let

F

be the vector field. We are assuming that

\nabla \times F = 0

on all of

R^{3} .

As a result,

F = \nabla ϕ

for some potential function

ϕ .

We are also assuming that

0 = \nabla \cdot F = \nabla \cdot \nabla ϕ = (\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}) ϕ .

This is the definition of “

ϕ

is harmonic”.

(b) False. Let

F

be the vector field. We are assuming that

F = \nabla ϕ

for some potential function

ϕ .

S

is any smooth closed surface, with

S

being the boundary of the solid

V,

then, by the divergence theorem, the outward flux of

F

through

S

\begin{aligned} \iint_{S} F \cdot \hat{n} d S & = ∭_{V} \nabla \cdot F d V = ∭_{V} \nabla \cdot \nabla ϕ d V \\ = ∭_{V} (\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}) ϕ d V \end{aligned}

If, for example,

ϕ = x^{2},

then

(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}) ϕ = 2

and the flux of

F

through

S

is twice the volume of

V,

which is not zero.

5.2.18. (✳).

Solution.

(a) True. The vector field

\nabla \nabla f

is conservative and the work done by a conservative field around any closed curve is zero.

(b) False. By the vector identity Theorem 4.1.7.a, we have

\nabla \nabla \cdot (\nabla \nabla \times F) = 0

for all vector fields

F .

But

\nabla \nabla \cdot (x \hat{ı ı} + y \hat{ȷ ȷ} + z \hat{k}) = 3 .

5.2.19. (✳).

Solution.

(a)

\int_{C} \nabla \nabla f \cdot d r = 0

is the work done along the curve using the conservative force

\nabla \nabla f .

That work is difference between the potential

f

at the final point minus the potential

f

at the initial point. If the final and initial points are both on the level surface

f (x, y, z) = 0,

that difference is zero.

(b) The rate of change of the specified vector is

\frac{d}{d t} v (t) \times r (t) = v^{'} (t) \times r (t) + v (t) \times v (t)

The first term vanishes because

v^{'} (t) = a (t) = f (t) r (t)

is parallel to

r (t) .

The second term vanishes because

v (t) = v (t) .

v \times r

of part (b)

N .

This vector is a constant and is perpendicular to both

v (t)

and

r (t) .

In particular

N \cdot r (t) = 0

Assuming that

N

is nonzero, this is the equation of the plane through the origin with normal vector

N .

(d) Yes, as long as

\hat{T},

\hat{N},

and

\hat{B}

are well-defined, since

\hat{B} = \hat{T} \times \hat{N} .

(e) No. When the maximum speed occurs

\frac{d^{2} s}{d t^{2}} = 0

so that

a = κ (t) (\frac{d s}{d t} (t))^{2} \hat{N} (t) .

If the speed and (constant) curvature are nonzero, the acceleration is nonzero.

5.2.20. (✳).

Solution.

We apply Green’s Theorem:

\int_{C} F_{1} d x + F_{2} d y = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d x d y

(a)

\frac{1}{2} \int_{C} - y d x + x d y = \frac{1}{2} \iint_{R} {1 - (- 1)} d x d y = Area (R)

(b)

\frac{1}{2} \int_{C} - x d x + y d y = \frac{1}{2} \iint_{R} 0 d x d y = 0 \neq Area (R)

(c)

\int_{C} y d x = \iint_{R} {- 1} d x d y = - Area (R) \neq Area (R)

(d)

\int_{C} 3 y d x + 4 x d y = \iint_{R} {4 - 3} d x d y = Area (R)

5.2.21. (✳).

Solution.

(a) True

.

Since

v = | v | = 1

is constant, we have

a = \frac{d v}{d t} \hat{T} + v^{2} κ \hat{N} = 0 \hat{T} + κ \hat{N} .

Thus

1 = | a | = κ | \hat{N} |,

i.e.,

κ = 1 .

(a) (Again.) Since

v \cdot v = | v |^{2} = 1

for all

t,

differentiation gives

v \cdot a = 0,

i.e.,

v ⊥ a

always. It follows that

| v \times a | = | v | | a | \sin θ = 1

always, because the angle

θ

here is always

π / 2 .

Thus, for all

t,

κ = \frac{| v \times a |}{| v |^{3}} = \frac{1}{1} = 1.

(b) True

.

By the divergence theorem, if

V

is the solid bounded by

S,

\iint_{S} \nabla \nabla \times F \cdot \hat{n} d S = ∭_{V} \nabla \nabla \cdot (\nabla \nabla \times F) d V = 0

since

\nabla \nabla \cdot (\nabla \nabla \times F) = 0 .

.

F = 0

and

G

is any nonzero, conservative field, like

G = 2 x \hat{ı ı} = \nabla \nabla (x^{2}),

then

\oint_{C} F \cdot d r = \oint_{C} G \cdot d r = 0

for every closed curve

C .

5.2.22. (✳).

Solution.

(a) Define

Ω Ω (t) = r (t) \times v (t) .

Then by the product rule,

\begin{aligned} \frac{d Ω Ω}{d t} & = \frac{d r}{d t} \times v + r \times \frac{d v}{d t} \\ = v \times v + r \times (f (r, v) r) . \\ = 0 + f (r, v) (r \times r) = 0 . \end{aligned}

It follows that

Ω Ω

is constant.

(b) By the divergence theorem, where

R

is the solid cylinder as described,

\iint_{S} (x \hat{ı ı} - y \hat{ȷ ȷ} + z^{2} \hat{k}) \cdot \hat{n} d S = ∭_{R} (1 - 1 + 2 z) d V = 2 ∭_{R} z d V

The solid

R

clearly has reflection symmetry across the plane

z = 2 .

So the

z

-coordinate of the centre of mass of

R,

i.e. the average value of

z

over

R,

i.e.

\bar{z} = \frac{∭_{R} z d V}{∭_{R} d V} = \frac{∭_{R} z d V}{V o l (R)}

2 .

Hence

\iint_{S} (x \hat{ı ı} - y \hat{ȷ ȷ} + z^{2} \hat{k}) \cdot \hat{n} d S = 2 \bar{z} V o l (R) = 4 V o l (R)

By basic geometry,

V o l (R) = π r^{2} h = π b^{2} 2 .

Hence

\iint_{S} (x \hat{ı ı} - y \hat{ȷ ȷ} + z^{2} \hat{k}) \cdot \hat{n} d S = 8 π b^{2}

\begin{aligned} \oint_{\partial D} F \cdot d r = \iint_{D} G \cdot \hat{n} d S & ⟹ \iint_{D} \nabla \nabla \times F \cdot \hat{n} d S = \iint_{D} G \cdot \hat{n} d S \\ ⟹ \iint_{D} (\nabla \nabla \times F - G) \cdot \hat{n} d S = 0 \end{aligned}

for all disks

D .

Because this is true for all disks

D,

the integrand must be zero. To see this, let

H = \nabla \nabla \times F - G .

Suppose that

H (x_{0}) \neq 0 .

Pick a unit vector

\hat{n}

in the direction of

H (x_{0}) .

Let

D

be a very small flat disk centered on

x_{0}

with normal

\hat{n}

(the vector we picked). Then

H (x_{0}) \cdot \hat{n} > 0

and, by continuity,

H (x) \cdot \hat{n} > 0

for all

x

D,

if we have picked

D

small enough. Then

\iint_{D} (\nabla \nabla \times F - G) \cdot \hat{n} d S > 0,

which is a contradiction. So we conclude that

\nabla \nabla \times F - G = 0

and hence

G = \nabla \nabla \times F .

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Appendix D Solutions to Exercises

1 Curves1.1 Derivatives, Velocity, Etc.

Exercises

1.1.1.

1.1.2.

1.1.3.

1.1.4.

1.1.5.

1.1.6.

1.1.7.

1.1.8.

1.1.9. (✳).

1.1.10.

1.1.11. (✳).

1.1.12.

1.1.13. (✳).

1.1.14.

1.1.15.

1.1.16.

1.1.17. (✳).

1.1.18. (✳).

1.1.19.

1.1.20. (✳).

1.1.21. (✳).

1.1.22. (✳).

1.1.23.

1.1.24.

1.1.25. (✳).

1.1.26. (✳).

1.1.27. (✳).

1.1.28.

1.1.29.

1.1.30.

1.1.31.

1.2 Reparametrization

Exercises

1.2.1.

1.2.2.

1.2.3.

1.2.4. (✳).

1.2.5. (✳).

1.2.6.

1.2.7.

1.3 Curvature

Exercises

1.3.1.

1.3.2.

1.3.3.

1.3.4.

1.3.5.

1.3.6.

1.3.7.

1.3.8.

1.3.9.

1.3.10. (✳).

1.3.11. (✳).

1.3.12.

1.3.13. (✳).

1.4 Curves in Three Dimensions

Exercises

1.4.1.

1.4.2.

1.4.3.

1.4.4.

1.4.5. (✳).

1.4.6. (✳).

1.4.7.

1.4.8.

1.4.9.

1.4.10. (✳).

1.4.11. (✳).

1.4.12. (✳).

1.4.13. (✳).

1.4.14. (✳).

1.4.15. (✳).

1.4.16. (✳).

1.4.17. (✳).

1.4.18. (✳).

1.4.19. (✳).

1.4.20. (✳).

1 Curves
1.1 Derivatives, Velocity, Etc.

1.7 Sliding on a Curve
1.7.4 Exercises

2 Vector Fields
2.1 Definitions and First Examples

2.2 Optional — Field Lines
2.2.2 Exercises