Stokes’ Theorem

Section 4.4 Stokes’ Theorem

Our last variant of the fundamental theorem of calculus is Stokes’

Sir George Gabriel Stokes (1819–1903) was an Irish physicist and mathematician. In addition to Stokes’ theorem, he is known for the Navier-Stokes equations of fluid dynamics and for his work on the wave theory of light. He gave evidence to the Royal Commission on the Use of Iron in Railway Structures after the Dee bridge disaster of 1847.

theorem, which is like Green’s theorem, but in three dimensions. It relates an integral over a finite surface in $\bbbr^3$ with an integral over the curve bounding the surface.

Theorem 4.4.1. Stokes’ Theorem.

Let

$S$ be a piecewise smooth oriented surface (i.e. a unit normal $\hn$ has been chosen at each point of $S$ and this choice depends continuously on the point)
the boundary, $\partial S\text{,}$ of the surface $S$ consist of a finite number of piecewise smooth, simple curves that are oriented consistently with $\hn$ in the sense that
- if you walk along $\partial S$ in the direction of the arrow on $\partial S\text{,}$
- with the vector from your feet to your head having direction $\hn$
- then $S$ is on your left hand side.
$\vF$ be a vector field that has continuous first partial derivatives at every point of $S\text{.}$

Then

\begin{equation*} \oint_{\partial S}\vF\cdot \dee{\vr} =\dblInt_{S}\vnabla\times\vF\cdot\hn\ \dee{S} \end{equation*}

Note that

in Stokes’ theorem, $S$ must be an oriented surface. In particular, $S$ may not be a Möbius strip. (See Example 3.5.3.)
If $S$ is part of the $xy$-plane, then Stokes’ theorem reduces to Green’s theorem. Our proof of Stokes’ theorem will consist of rewriting the integrals so as to allow an application of Green’s theorem.
If $\partial S$ is a simple closed curve and
- when you look at $\partial S$ from high on the $z$-axis, it is oriented counterclockwise (look at the figure in Theorem 4.4.1), then
- $\hn$ is upward pointing, i.e. has positive $z$-component, at least near $\partial S\text{.}$

Proof.

Write $\vF=F_1\,\hi+F_2\,\hj+F_3\,\hk\text{.}$ Both integrals involve $F_1$ terms and $F_2$ terms and $F_3$ terms. We shall show that the $F_1$ terms in the two integrals agree. In other words, we shall assume that $\vF=F_1\hi\text{.}$ The proofs that the $F_2$ and $F_3$ terms also agree are similar. For simplicity, we’ll assume

Otherwise, decompose $S$ into simpler pieces, analogously to what we did in the proof of the divergence theorem.

that the boundary of $S$ consists of just a single curve, and that we can

pick a parametrization of $S$ with
\begin{equation*} S=\Set{\vr(u,v)=\big(x(u,v),y(u,v),z(u,v)\big)}{ (u,v)\text{ in } R\subset\bbbr^2} \end{equation*}
and with $\vr(u,v)$ orientation preserving in the sense that $\hn\,\dee{S} = +\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v}\,\dee{u}\,\dee{v}\text{.}$ Also
pick a parametrization of the curve, $\partial R\text{,}$ bounding $R$ as $\big( u(t), v(t)\big)\text{,}$ $a\le t\le b\text{,}$ in such a way that when you walk along $\partial R$ in the direction of increasing $t\text{,}$ then $R$ is on your left.

Then the curve $\partial S$ bounding $S$ can be parametrized as $\vR(t)=\vr\big(u(t),v(t)\big)\text{,}$ $a\le t\le b\text{.}$

The orientation of $\vR(t)\text{:}$

We’ll now verify that the direction of increasing $t$ for the parametrization $\vR(t)$ of $\partial S$ is the direction of the arrow on $\partial S$ in the figure on the left above.

By continuity, it suffices to check the orientation at a single point.

Find a point $(u_0,v_0)$ on $\partial R$ where the forward pointing tangent vector is a positive multiple of $\,\hi\text{.}$ The horizontal arrow on $\partial R$ in the figure on the left below is at such a point. Suppose that $t=t_0$ at this point — in other words, suppose that $(u_0,v_0)=\big(u(t_0),v(t_0)\big)\text{.}$ Because the forward pointing tangent vector to $\partial R$ at $(u_0,v_0)\text{,}$ namely $\big(u'(t_0),v'(t_0)\big)\text{,}$ is a positive multiple of $\,\hi\text{,}$ we have $u'(t_0) \gt 0$ and $v'(t_0)=0\text{.}$ The tangent vector to $\partial S$ at $\vR(t_0)=\vr\big(u_0,v_0\big)\text{,}$ pointing in the direction of increasing $t\text{,}$ is

\begin{align*} \vR'(t_0)&=\diff{ }{t}\vr\big(u(t),v(t)\big)\big|_{t=t_0} =u'(t_0)\frac{\partial \vr}{\partial u}(u_0,v_0) +v'(t_0) \frac{\partial \vr}{\partial v}(u_0,v_0)\\ &=u'(t_0)\frac{\partial \vr}{\partial u}(u_0,v_0) \end{align*}

and so is a positive multiple of $\frac{\partial \vr}{\partial u}(u_0,v_0)\text{.}$ See the figure on the right below.

If we now walk along a path in the $uv$-plane which starts at $(u_0,v_0)\text{,}$ holds $u$ fixed at $u_0$ and increases $v\text{,}$ we move into the interior of $R$ starting at $(u_0,v_0)\text{.}$ Correspondingly, if we walk along the path, $\vr(u_0,v)\text{,}$ in $\bbbr^3$ with $v$ starting at $v_0$ and increasing, we move into the interior of $S\text{.}$ The forward tangent to this new path, $\frac{\partial \vr}{\partial v}(u_0,v_0)\text{,}$ points from $\vr(u_0,v_0)$ into the interior of $S\text{.}$ It’s the blue arrow in the figure on the right below.

Now imagine that you are walking along $\partial S$ in the direction of increasing $t\text{.}$ At time $t_0$ you are at $\vR(t_0)\text{.}$ You point your right arm straight ahead of you. So it is pointing in the direction $\frac{\partial \vr}{\partial u}(u_0,v_0)\text{.}$ You point your left arm out sideways into the interior of $S\text{.}$ It is pointing in the direction $\frac{\partial \vr}{\partial v}(u_0,v_0)\text{.}$ If the direction of increasing $t$ is the same as the forward direction of the orientation of $\partial S\text{,}$ then the vector from our feet to our head, which is $\frac{\partial \vr}{\partial u}(u_0,v_0) \times \frac{\partial \vr}{\partial v}(u_0,v_0)\text{,}$ should be pointing in the same direction as $\hn\text{.}$ And since $\hn\,\dee{S} = +\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v}\,\dee{u}\,\dee{v}\text{,}$ it is.

Now, with our parametrization and orientation sorted out, we can examine the integrals.

The surface integral:

Since $F=F_1\,\hi\text{,}$ so that

and

\begin{align*} \hn\,\dee{S} &=\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v} \ \dee{u}\,\dee{v} =\det\left[\begin{matrix} \hi & \hj & \hk \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{matrix}\right]\\ &=\left(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v} -\frac{\partial z}{\partial u} \frac{\partial y}{\partial v}\right)\hi + \left(\frac{\partial z}{\partial u} \frac{\partial x}{\partial v} -\frac{\partial x}{\partial u} \frac{\partial z}{\partial v}\right)\hj\\ &\hskip1in +\left(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} -\frac{\partial y}{\partial u} \frac{\partial x}{\partial v}\right) \hk \end{align*}

and

\begin{align*} &\dblInt_{S}\vnabla\times\vF\cdot\hn\,\dee{S} =\dblInt_{R}\left(0,\frac{\partial F_1}{\partial z}, -\frac{\partial F_1}{\partial y}\right) \cdot\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v}\ \dee{u}\,\dee{v}\cr &\hskip.15in=\dblInt_{R}\left\{\frac{\partial F_1}{\partial z} \left(\frac{\partial z}{\partial u} \frac{\partial x}{\partial v} -\frac{\partial x}{\partial u} \frac{\partial z}{\partial v}\right) -\frac{\partial F_1}{\partial y} \left(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} -\frac{\partial y}{\partial u} \frac{\partial x}{\partial v}\right) \right\}\,\dee{u}\,\dee{v} \end{align*}

Now we examine the line integral and show that it equals this one.

The line integral:

\begin{align*} &\oint_{\partial S}\vF\cdot \dee{\vr} =\int_a^b\vF\Big(\vr\big(u(t),v(t)\big)\Big)\cdot\diff{ }{t} \vr\big(u(t),v(t)\big)\ \dee{t}\\ &\hskip0.1in=\int_a^b\vF\Big(\vr\big(u(t),v(t)\big)\Big)\cdot\Big[ \frac{\partial\vr}{\partial u}\big(u(t),v(t)\big)\diff{u}{t}(t) +\frac{\partial\vr}{\partial v}\big(u(t),v(t)\big)\diff{v}{t}(t)\Big]\ \dee{t} \end{align*}

We can write this as the line integral

\begin{align*} &\oint_{\partial R}M(u,v)\ \dee{u}+N(u,v)\ \dee{v}\\ &\hskip1in=\int_a^b \Big[M\big(u(t),v(t)\big)\ \diff{u}{t}(t) +N\big(u(t),v(t))\ \diff{v}{t}(t)\Big]\ \dee{t} \end{align*}

around $\partial R\text{,}$ if we choose

\begin{alignat*}{2} M(u,v)&=\vF\big(\vr(u,v)\big)\cdot\frac{\partial\vr}{\partial u}(u,v) &&=F_1\big({x(u,v),y(u,v),z(u,v)}\big)\frac{\partial x}{\partial u}(u,v)\\ N(u,v)&=\vF\big(\vr(u,v)\big)\cdot\frac{\partial\vr}{\partial v}(u,v) &&=F_1\big({x(u,v),y(u,v),z(u,v)}\big)\frac{\partial x}{\partial v}(u,v) \end{alignat*}

Finally, we show that the surface integral equals the line integral:

By Green’s Theorem, we have

\begin{align*} \oint_{\partial S}\vF\cdot \dee{\vr} &=\oint_{\partial R}M(u,v)\ \dee{u}+N(u,v)\ \dee{v}\cr =\dblInt_{R}\bigg\{&\frac{\partial N}{\partial u}-\frac{\partial M}{\partial v} \bigg\}\ \dee{u} \dee{v}\cr =\dblInt_{R}\bigg\{& \frac{\partial }{\partial u}\big[ F_1\big({x(u,v),y(u,v),z(u,v)}\big) \big]\frac{\partial x}{\partial v} +F_1\frac{\partial^2 x}{\partial u\partial v}\\ -& \frac{\partial }{\partial v}\big[ F_1\big({x(u,v),y(u,v),z(u,v)}\big) \big]\frac{\partial x}{\partial u} -F_1\frac{\partial^2 x}{\partial v\partial u} \bigg\}\ \dee{u} \dee{v}\\ =\dblInt_{R}\bigg\{& {\color{blue}{\Big(\frac{\partial F_1}{\partial x} \frac{\partial x}{\partial u}}} +\frac{\partial F_1}{\partial y} \frac{\partial y}{\partial u} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\partial u} \Big)\frac{\partial x}{\partial v} {\color{red}{ +F_1\frac{\partial^2 x}{\partial u\partial v}}}\cr -&\Big( {\color{blue}{\frac{\partial F_1}{\partial x} \frac{\partial x}{\partial v}}} +\frac{\partial F_1}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\partial v} \Big)\frac{\partial x}{\partial u} {\color{red}{ -F_1\frac{\partial^2 x}{\partial v\partial u}}} \bigg\}\ \dee{u} \dee{v}\cr =\dblInt_{R}\bigg\{&\Big(\frac{\partial F_1}{\partial y} \frac{\partial y}{\partial u} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\partial u} \Big)\frac{\partial x}{\partial v} -\Big(\frac{\partial F_1}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\partial v} \bigg)\frac{\partial x}{\partial u} \Big\}\,\dee{u}\, \dee{v}\\ &\hskip-33pt=\dblInt_{S}\vnabla\times\vF\cdot\hn\,\dee{S} \end{align*}

which is the conclusion that we wanted.

Before we move on to some examples, here are a couple of remarks.

Stokes’ theorem says that $\oint_{C}\vF\cdot \dee{\vr} =\dblInt_{S}\vnabla\times\vF\cdot\hn\ \dee{S}$ for any (suitably oriented) surface whose boundary is $C\text{.}$ So if $S_1$ and $S_2$ are two different (suitably oriented) surfaces having the same boundary curve $C\text{,}$ then

\begin{equation*} \dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{S} =\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S} \end{equation*}

For example, if $C$ is the unit circle

\begin{equation*} C=\Set{(x,y,z)}{x^2+y^2=1,\ z=0} \end{equation*}

oriented counterclockwise when viewed from above, then both

\begin{align*} S_1&= \Set{(x,y,z)}{x^2+y^2\le 1,\ z=0}\\ S_2&= \Set{(x,y,z)}{z\ge0,\ x^2+y^2+z^2=1} \end{align*}

with upward pointing unit normal vectors, have boundary $C\text{.}$ So Stokes’ tells us that $\dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{S} =\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S}\text{.}$

It should not be a surprise that $\dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{S} =\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S}\text{,}$ for the following reason. Let

\begin{equation*} V= \Set{(x,y,z)}{x^2+y^2+z^2\le 1,\ z\ge 0} \\ \end{equation*}

be the solid between $S_1$ and $S_2\text{.}$ The boundary $\partial V$ of $V$ is the union of $S_1$ and $S_2\text{.}$

But beware that the outward pointing normal to $\partial V$ (call it $\hN$) is $+\hn$ on $S_2$ and $-\hn$ on $S_1\text{.}$ So the divergence theorem gives

\begin{align*} &\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S} -\dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{S}\\ &\hskip0.5in= \dblInt_{S_2}\vnabla\times\vF\cdot\hN\ \dee{S} +\dblInt_{S_1}\vnabla\times\vF\cdot\hN\ \dee{S}\\ &\hskip0.5in= \dblInt_{\partial V}\vnabla\times\vF\cdot\hN\ \dee{S}\\ &\hskip0.5in= \tripInt_{V}\vnabla\cdot\big(\vnabla\times\vF\big)\ \dee{V}\\ &\hskip2in\qquad\text{by the divergence theorem}\\ &\hskip0.5in=0 \end{align*}

by the vector identity Theorem 4.1.7.a.
As a second remark, suppose that the vector field $\vF$ obeys $\vnabla\times\vF=\vZero$ everywhere. Then Stokes’ theorem forces $\oint_{C}\vF\cdot \dee{\vr}=0$ are around all closed curves $C\text{,}$ which implies that $\vF$ is conservative, by Theorem 2.4.7. So Stokes’ theorem provides another proof of Theorem 2.4.8.

Here is an easy example which shows that Stokes’ can be very useful when $\vnabla\times\vF$ simplifies.

Example 4.4.2.

Evaluate $\oint_C\vF\cdot\dee{\vr}$ where $\vF= \big[2z+\sin\big(x^{146}\big)\big]\,\hi-5z\,\hj -5y\,\hk$ and the curve $C$ is the circle $x^2+y^2=4\text{,}$ $z=1\text{,}$ oriented counterclockwise when viewed from above.

Solution.

The $x^{146}$ in $\vF$ will probably make a direct evaluation of the integral difficult. So we’ll use Stokes’ theorem. To do so we need a surface $S$ with $\partial S=C\text{.}$ The simplest is just the flat disk

\begin{equation*} S = \Set{(x,y,z)}{ x^2+y^2\le 4,\ \ z=1} \end{equation*}

Since

\begin{align*} \vnabla\times\vF & = \det\left[\begin{matrix} \hi & \hj & \hk \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 2z+\sin\big(x^{146}\big) & -5z & -5y \end{matrix}\right]\\ & =\hi \det\left[\begin{matrix} \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ -5z & -5y \end{matrix}\right] -\hj\det\left[\begin{matrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial z} \\ 2z+\sin\big(x^{146}\big) & -5y \end{matrix}\right]\\ &\hskip1in +\hk\det\left[\begin{matrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \\ 2z+\sin\big(x^{146}\big) & -5z \end{matrix}\right]\\ &=2\hj \end{align*}

and the normal to $S$ is $\hk\text{,}$ Stokes’ theorem gives

\begin{align*} \oint_C\vF\cdot\dee{\vr} & = \dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} = \dblInt_S (2\hj)\cdot\hk\,\dee{S} =0 \end{align*}

Now we’ll repeat the last example with a harder curve.

Example 4.4.3.

Evaluate $\oint_C\vF\cdot\dee{\vr}$ where $\vF= \big[2z+\sin\big(x^{146}\big)\big]\,\hi-5z\,\hj -5y\,\hk$ and the curve $C$ is the intersection of $x^2+y^2+z^2=4$ and $z=y\text{,}$ oriented counterclockwise when viewed from above.

Solution.

The surface $x^2+y^2+z^2=4$ is the sphere of radius $2$ centred on the origin and $z=y$ is a plane which contains the origin. So $C\text{,}$ being the intersection of a sphere with a plane through the centre of the sphere, is a circle, with centre $(0,0,0)$ and radius $2\text{.}$ The part of the circle in the first octant is sketched on the left below.

\begin{equation*} S = \Set{(x,y,z)}{ x^2+y^2+z^2\le 4,\ \ z=y} \end{equation*}

The first octant of $S$ is shown in the figure on the right above. We saw in the last Example 4.4.2 that

\begin{align*} \vnabla\times\vF &=2\hj \end{align*}

So Stokes’ theorem gives

\begin{align*} \oint_C\vF\cdot\dee{\vr} & = \dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} = 2\dblInt_S \hj\cdot\hn\,\dee{S} \end{align*}

We’ll evaluate the integral $2\dblInt_S \hj\cdot\hn\,\dee{S}$ in two ways. The first way is more efficient, but also requires more insight. Since $\vnabla(z-y)=\hk-\hj\text{,}$ the upward unit normal to the plane $z-y=0\text{,}$ and hence to $S\text{,}$ is $\hn = \frac{1}{\sqrt{2}}(\hk-\hj)\text{.}$ Consequently the integrand

\begin{gather*} \hj\cdot\hn=\hj\cdot\Big(\frac{-\hj+\hk}{\sqrt{2}}\Big)=-\frac{1}{\sqrt{2}} \end{gather*}

is a constant and we do not need a formula for $\hn\,dS\text{:}$

\begin{align*} \oint_C\vF\cdot\dee{\vr} & = 2\dblInt_S \hj\cdot\hn\,\dee{S} =-\sqrt{2} \dblInt_S \dee{S} = -\sqrt{2}\text{Area}(S) =-\sqrt{2}\pi\, 2^2\\ &=-4\sqrt{2}\pi \end{align*}

Alternatively, we can evaluate the integral $\dblInt_S \hj\cdot\hn\,\dee{S}$ using our normal protocol. As $S$ is part of the plane $z=f(x,y)=y\text{,}$

\begin{gather*} \hn\,dS = \pm\big(-f_x\,\hi-f_y\,\hj+\hk\big)\,\dee{x}\dee{y} = \pm (-\hj+\hk)\,\dee{x}\dee{y} \end{gather*}

To get the upward pointing normal pointing normal, we take the $+$ sign so that $\hn\,dS= (-\hj+\hk)\,\dee{x}\dee{y}\text{.}$ As $(x,y,z)$ runs over

\begin{align*} S &= \Set{(x,y,z)}{ x^2+y^2+z^2\le 4,\ \ z=y}\\ &= \Set{(x,y,z)}{ x^2+2y^2\le 4,\ \ z=y}\\ &= \Set{(x,y,z)}{ \tfrac{x^2}{4} +\tfrac{y^2}{2}\le 1,\ \ z=y} \end{align*}

$(x,y)$ runs over the elliptical disk $R=\Set{(x,y)}{\frac{x^2}{4} +\frac{y^2}{2}\le 1}\text{.}$ The part of this ellipse in the first octant is the shaded region in the figure below.

This ellipse has semiaxes $a=2$ and $b =\sqrt{2}$ and hence area $\pi a b = 2\sqrt{2} \pi\text{.}$ So

\begin{align*} \oint_C\vF\cdot\dee{\vr} & = 2\dblInt_S \hj\cdot\hn\,\dee{S} =2\dblInt_R \hj\cdot(-\hj+\hk)\,\dee{x}\dee{y} =-2 \dblInt_R \dee{x}\dee{y}\\ &= -2\text{Area}(R)\\ &=-4\sqrt{2}\pi \end{align*}

Example 4.4.4.

Evaluate $\oint_C\vF\cdot\dee{\vr}$ where $\vF= (x+y)\,\hi+2(x-z)\,\hj +(y^2+z)\,\hk$ and $C$ is the oriented curve obtained by going from $(2,0,0)$ to $(0, 3, 0)$ to $(0, 0, 6)$ and back to $(2, 0, 0)$ along straight line segments.

Solution 1.

In this first solution, we’ll evaluate the integral directly. The first line segment ($C_1$ in the figure above) may be parametrized as

\begin{equation*} \vr(t)=(2,0,0)+t\big\{(0, 3, 0)-(2,0,0)\big\} =\big(2-2t\,,\,3t\,,\,0\big)\qquad 0\le t\le 1 \end{equation*}

So the integral along this segment is

\begin{align*} \int_0^1 \vF(\vr(t))\cdot\diff{\vr}{t}\ \dee{t} &=\int_0^1 (2+t\,,\, 2(2-2t)\,,\, (3t)^2)\cdot(-2\,,\, 3\,,\, 0)\ \dee{t}\\ &=\int_0^1 (8-14t)\ \dee{t}\\ &=\Big[8t-7t^2\Big]_0^1=1 \end{align*}

The second line segment ($C_2$ in the figure above) may be parametrized as

\begin{equation*} \vr(t)=(0,3,0)+t\big\{(0, 0, 6)-(0,3,0)\big\} =\big(0\,,\,3-3t\,,\,6t\big)\qquad 0\le t\le 1\text{.} \end{equation*}

So the integral along this segment is

\begin{align*} \int_0^1 \vF(\vr(t))\cdot\diff{\vr}{t}\ \dee{t} &=\int_0^1 \big(3(1-t)\,,\, - 12t\,,\, 9(1-t)^2+ 6t\big)\cdot(0, -3, 6)\ \dee{t}\\ &=\int_0^1 [36t+54(1-t)^2+36t]\ \dee{t}\\ &=\Big[18t^2-18(1-t)^3+18t^2\Big]_0^1\\ &=54 \end{align*}

The final line segment ($C_3$ in the figure above) may be parametrized as

\begin{equation*} \vr(t)=(0,0,6)+t\big\{(2,0,0) -(0,0,6)\big\} = (2t\,,\,0\,,\,6-6t) \qquad 0\le t\le 1 \end{equation*}

So the line integral along this segment is

\begin{align*} \int_0^1 \vF(\vr(t))\cdot\diff{\vr}{t}\ \dee{t} &=\int_0^1 \big(2t\,,\, 4t - 12(1-t)\,,\, 6(1-t)\big)\cdot(2, 0, -6)\ \dee{t}\\ &=\int_0^1 [4t-36(1-t)]\ \dee{t} =\Big[2t^2+18(1-t)^2\Big]_0^1 =-16 \end{align*}

The full line integral is

\begin{equation*} \oint_C\vF\cdot \dee{\vr}=1+54-16=39 \end{equation*}

Solution 2.

This time we shall apply Stokes’ Theorem. The curl of $\vF$ is

The curve $C$ is a triangle and so is contained in a plane. Any plane has an equation of the form $Ax+By+Cz=D\text{.}$ Our plane does not pass through the origin (look at the figure above) so the $D$ must be nonzero. Consequently we may divide $Ax+By+Cz=D$ through by $D$ giving an equation of the form $ax+by+cz=1\text{.}$

Because $(2,0,0)$ lies on the plane, $a=\frac{1}{2}\text{.}$
Because $(0,3,0)$ lies on the plane, $b=\frac{1}{3}\text{.}$
Because $(0,0,6)$ lies on the plane, $c=\frac{1}{6}\text{.}$

So the triangle is contained in the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1\text{.}$ It is the boundary of the surface $S$ that consists of the portion of the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ that obeys $x\ge 0\text{,}$ $y\ge 0$ and $z\ge 0\text{.}$ Rewrite the equation of the plane as $z=6-3x-2y\text{.}$ For this surface

\begin{equation*} \hn\ \dee{S}=(3\hi+2\hj+\hk)\,\dee{x}\,\dee{y} \end{equation*}

by 3.3.2, and we can write

\begin{align*} S&=\Set{(x,y,z)}{x\ge0,\ y\ge0,\ z\ge0,\ z=6-3x-2y}\\ &=\Set{(x,y,z)}{x\ge0, y\ge0,\ 6-3x-2y\ge0,\ z=6-3x-2y} \end{align*}

As $(x,y,z)$ runs over $S\text{,}$ $(x,y)$ runs over the triangle

\begin{align*} R&=\Set{(x,y,z)}{x\ge0,\ y\ge0,\ 3x+2y\le 6}\\ &=\Set{(x,y,z)}{x\ge0,\ 0\le y \le \tfrac{3}{2}(2-x)} \end{align*}

Using horizontal strips as in the figure on the left below,

\begin{align*} \oint_C\vF\cdot \dee{\vr}&=\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{S}\\ &=\dblInt_R [2(y+1)\hi+\hk]\cdot[3\hi+2\hj+\hk]\ \dee{x}\,\dee{y}\cr &=\dblInt_R [6y+7]\ \dee{x}\,\dee{y}\\ & =\int_0^3 \dee{y}\int_0^{{1\over 3}(6-2y)} \dee{x}\ [6y+7]\\ &=\int_0^3 \dee{y}\ \frac{1}{3}[6y+7][6-2y]\\ & =\frac{1}{3}\int_0^3 \dee{y}\ [-12y^2+22y+42]\\ &=\frac{1}{3}\Big[-4y^3+11y^2+42y\Big]_0^3\\ & =\big[-4\times 9 +11\times 3 +42\big] =39 \end{align*}

Alternatively, using vertical strips as in the figure on the right above,

\begin{align*} \oint_C\vF\cdot \dee{\vr} &=\dblInt_R [6y+7]\ \dee{x}\,\dee{y}\\ &=\int_0^2 \dee{x}\int_0^{{3\over 2}(2-x)} \dee{y}\ [6y+7]\\ &=\int_0^2 \dee{x}\ \Big[3\frac{3^2}{2^2}(2-x)^2+7\frac{3}{2}(2-x)\Big]\\ & =\Big[-\frac{27}{4}\,\frac{1}{3}(2-x)^3 -\frac{21}{2}\,\frac{1}{2}(2-x)^2\Big]_0^2\\ &=\frac{9}{4}\,8+\frac{21}{4}\,4 =39 \end{align*}

Example 4.4.5.

Evaluate $\oint_C\vF\cdot\dee{\vr}$ where $\vF= (\cos x +y+z)\,\hi+(x+z)\,\hj +(x+y)\,\hk$ and $C$ is the intersection of the surfaces

\begin{equation*} x^2+\frac{y^2}{2}+\frac{z^2}{3}=1\qquad\text{and}\qquad z=x^2+2y^2 \end{equation*}

oriented counterclockwise when viewed from above.

Solution.

First, let’s sketch the curve. $x^2+\frac{y^2}{2}+\frac{z^2}{3}=1$ is an ellipsoid centred on the origin and $z=x^2+2y^2$ is an upward opening paraboloid that passes through the origin. They are sketched in the figure below. The paraboloid is red.

Their intersection, the curve $C\text{,}$ is the blue curve in the figure. It looks like a deformed

By Salvador Dali?

circle.

One could imagine parametrizing $C\text{.}$ For example, substituting $x^2 = z - 2y^2$ into the equation of the ellipsoid gives $-\frac{3}{2}y^2 + \frac{1}{3}(z+\frac{3}{2}\big)^2 = \frac{7}{4}\text{.}$ This can be solved to give $y$ as a function of $z$ and then $x^2=z-2y^2$ also gives $x$ as a function of $z\text{.}$ However this would clearly yield, at best, a really messy integral. So let’s try Stokes’ theorem.

In fact, since

This $\vF$ is conservative! (In fact $\vF=\vnabla\big(\sin x + xy + xz +yz\big)\text{.}$) As $C$ is a closed curve, $\oint_C\vF\cdot\dee{\vr}=0 \text{.}$

Example 4.4.6.

Evaluate $\dblInt_S\vG\cdot\hn\,\dee{S}$ where $\vG= (2x)\,\hi+(2z-2x)\,\hj +(2x-2z)\,\hk$ and

\begin{equation*} S=\Set{(x,y,z)}{z=\big(1-x^2-y^2\big)(1-y^3)\cos x\ e^{y},\ x^2+y^2\le 1} \end{equation*}

with upward pointing normal

Solution 1.

The surface $S$ is sketched below. It is a pretty weird surface. About the

only simple thing about it is that its boundary, $\partial S\text{,}$ is the circle $x^2+y^2=1\text{,}$ $z=0\text{.}$ It is clear that we should not try to evaluate the integral directly

⁴

That way lies pain.

. In this solution we will combine the divergence theorem with the observation that

\begin{gather*} \vnabla\cdot\vG = \frac{\partial }{\partial x}(2x)+ \frac{\partial }{\partial y}(2z-2x)+ \frac{\partial }{\partial z}(2x-2z) =0 \end{gather*}

to avoid ever having work with the surface $S\text{.}$ Here is an outline of what we will do.

We first select a simple surface $S'$ whose boundary $\partial S'$ is also the circle $x^2+y^2=1\text{,}$ $z=0\text{.}$ A nice simple choice of $S'\text{,}$ and the surface that we will use, is the disk
\begin{equation*} S' =\Set{(x,y,z)}{x^2+y^2=1,\ z=0} \end{equation*}
Then we define $V$ to be the solid whose top surface is $S$ and whose bottom surface is $S'\text{.}$ So the boundary of $V$ is the union of $S$ and $S'\text{.}$
For $S'\text{,}$ we will use the upward pointing normal $\hn=\hk\text{,}$ which is minus the outward pointing normal to $\partial V$ on $S'\text{.}$ So the divergence theorem says that
\begin{equation*} \tripInt_V \vnabla\cdot\vG\,\dee{V} =\dblInt_S\vG\cdot\hn\,\dee{S}-\dblInt_{S'}\vG\cdot\hn\,\dee{S} \end{equation*}
The left hand side is zero because, as we have already seen, $\vnabla\cdot\vG=0\text{.}$ So
\begin{equation*} \dblInt_S\vG\cdot\hn\,\dee{S} =\dblInt_{S'}\vG\cdot\hn\,\dee{S} \end{equation*}
Finally, we compute $\dblInt_{S'}\vG\cdot\hn\,\dee{S}\text{.}$

We saw an argument like this (with $\vG=\vnabla\times\vF$) in the first remark following the proof of Theorem 4.4.1.

So all that we have to do now is compute

\begin{align*} \dblInt_S\vG\cdot\hn\,\dee{S} &=\dblInt_{S'}\vG\cdot\hn\,\dee{S} =\dblInt_{S'}\vG\cdot\hk\,\dee{S} =\dblInt_{\Atop{x^2+y^2\le 1}{z=0}}(2x-2z)\,\dee{x}\dee{y}\\ &=\dblInt_{\Atop{x^2+y^2\le 1}{z=0}}(2x)\,\dee{x}\dee{y}\\ &=0 \end{align*}

simply because the integrand is odd under $x\rightarrow-x\text{.}$

Solution 2.

In this second solution we’ll use Stokes’ theorem instead of the divergence theorem. To do so, we have to express $\vG$ in the form $\vnabla\times\vF\text{.}$ So the first thing to do is to check if $\vG$ passes the screening test, Theorem 4.1.12, for the existence of vector potentials. That is, to check if $\vnabla\cdot\vG=0\text{.}$ It is. We saw this in Solution 1 above.

Next, we have to find a vector potential. In fact, we have already found, in Example 4.1.15, that

\begin{equation*} \vF = (z^2-2xz) \hi +(x^2-2xz)\hj \end{equation*}

is a vector potential for $\vG\text{,}$ which we can quickly check.

Parametrizing $C$ by $\vr(t) = \cos t\,\hi+\sin t\,\hj\text{,}$ $0\le t\le 2\pi\text{,}$ Stokes’ theorem gives (recalling that $z=0$ on $C$ so that $\vF\big(\vr(t)\big)=x^2\,\hj\Big|_{x=\cos t}=\cos^2t$)

\begin{align*} \dblInt_S\vG\cdot\hn\,\dee{S} &= \dblInt_S\vnabla\times\vF\cdot\hn\,\dee{S} = \oint_C \vF\cdot\dee{\vr} =\int_0^{2\pi} \vF\big(\vr(t)\big)\cdot\diff{\vr}{t}\ \dee{t}\\ &= \int_0^{2\pi} \big(\cos^2 t\big)(\cos t)\ \dee{t} \end{align*}

Of course this integral can be evaluated by using that one antiderivative of the integrand $\cos^3 t =\big(1-\sin^2t\big)\cos t$ is $\sin t-\frac{1}{3}\sin^3 t$ and that this antiderivative is zero at $t=0$ and at $t=2\pi\text{.}$ But it is easier to observe that the integral of any odd power of $\sin t$ or $\cos t$ over any full period is zero. Look, for example, at the graphs of $\sin^3x$ and $\cos^3x\text{,}$ below.

Either way

\begin{gather*} \dblInt_S\vG\cdot\hn\,\dee{S} =0 \end{gather*}

Example 4.4.7.

In this example we compute, in three different ways, $\oint_C\vF\cdot \dee{\vr}$ where

\begin{equation*} \vF=(z-y)\,\hi-(x+z)\,\hj-(x+y)\,\hk \end{equation*}

and $C$ is the curve $x^2+y^2+z^2=4\text{,}$ $z=y$ oriented counterclockwise when viewed from above.

Solution 1.

Direct Computation:

In this first computation, we parametrize the curve $C$ and compute $\oint_C\vF\cdot \dee{\vr}$ directly. The plane $z=y$ passes through the origin, which is the centre of the sphere $x^2+y^2+z^2=4\text{.}$ So $C$ is a circle which, like the sphere, has radius $2$ and centre $(0,0,0)\text{.}$ We use a parametrization of the form

\begin{equation*} \vr(t)=\vc+\rho\cos t\,\hi'+ \rho\sin t\,\hj' \qquad 0\le t\le 2\pi \end{equation*}

where

$\vc=(0,0,0)$ is the centre of $C\text{,}$
$\rho=2$ is the radius of $C$ and
$\hi'$ and $\hj'$ are two vectors that
1. are unit vectors,
2. are parallel to the plane $z=y$ and
3. are mutually perpendicular.

The trickiest part is finding suitable vectors $\hi'$ and $\hj'\text{:}$

The point $(2,0,0)$ satisfies both $x^2+y^2+z^2=4$ and $z=y$ and so is on $C\text{.}$ We may choose $\hi'$ to be the unit vector in the direction from the centre $(0,0,0)$ of the circle towards $(2,0,0)\text{.}$ Namely $\hi'=(1,0,0)\text{.}$
Since the plane of the circle is $z-y=0\text{,}$ the vector $\vnabla(z-y)=(0,-1,1)$ is perpendicular to the plane of $C\text{.}$ So $\hk'=\frac{1}{\sqrt{2}}(0,-1,1)$ is a unit vector normal to $z=y\text{.}$ Then $\hj'=\hk'\times\hi'=\frac{1}{\sqrt{2}}(0,-1,1)\times(1,0,0) =\frac{1}{\sqrt{2}}(0,1,1)$ is a unit vector that is perpendicular to $\hi'$ and $\hk'\text{.}$ Since $\hj'$ is perpendicular to $\hk'\text{,}$ it is parallel to $z=y\text{.}$

Substituting in $\vc=(0,0,0)\text{,}$ $\rho=2\text{,}$ $\hi'=(1,0,0)$ and $\hj'=\frac{1}{\sqrt{2}}(0,1,1)$ gives

\begin{align*} &\vr(t)=2\cos t\,(1,0,0)+ 2\sin t\,\frac{1}{\sqrt{2}}(0,1,1) =2\Big(\cos t, \frac{\sin t}{\sqrt{2}},\frac{\sin t}{\sqrt{2}}\Big)\\ &0\le t\le 2\pi \end{align*}

To check that this parametrization is correct, note that $x=2\cos t\text{,}$ $y=\sqrt{2}\sin t\text{,}$ $z=\sqrt{2}\sin t$ satisfies both $x^2+y^2+z^2=4$ and $z=y\text{.}$

At $t=0\text{,}$ $\vr(0)=(2,0,0)\text{.}$ As $t$ increases, $z(t)=\sqrt{2}\sin t$ increases and $\vr(t)$ moves upwards towards $\vr\big(\frac{\pi}{2}\big)=(0,\sqrt{2},\sqrt{2})\text{.}$ This is the desired counterclockwise direction (when viewed from above). Now that we have a parametrization, we can set up the integral.

\begin{align*} \vr(t)&=\big(2\cos t, \sqrt{2}\sin t,\sqrt{2}\sin t\big)\cr \vr\,'(t)&=\big(-2\sin t,\sqrt{2}\cos t,\sqrt{2}\cos t\big)\cr \vF\big(\vr(t)\big)&=\big(z(t)-y(t),-x(t)-z(t),-x(t)-y(t)\big)\cr &=\big(\sqrt{2}\sin t-\sqrt{2}\sin t, -2\cos t-\sqrt{2}\sin t,-2\cos t-\sqrt{2}\sin t\big)\cr &=-\big(0, 2\cos t+\sqrt{2}\sin t,2\cos t+\sqrt{2}\sin t\big)\cr \vF\big(\vr(t)\big)\cdot \vr\,'(t) &=-\big[4\sqrt{2}\cos^2 t+4\cos t\sin t\big]\\ &=-\big[2\sqrt{2}\cos(2t)+2\sqrt{2}+2\sin(2t)\big] \end{align*}

by the double angle formulae $\sin(2t)=2\sin t\,\cos t$ and $\cos(2t) = 2\cos^2t-1\text{.}$ So

\begin{align*} \oint_C\vF\cdot \dee{\vr} &=\int_0^{2\pi}\vF\big(\vr(t)\big)\cdot \vr\,'(t)\ \dee{t}\\ &=\int_0^{2\pi}-\big[2\sqrt{2}\cos(2t)+2\sqrt{2}+2\sin(2t)\big]\ \dee{t}\\ &=-\Big[\sqrt{2}\sin(2t)+2\sqrt{2}t-\cos(2t)\Big]_0^{2\pi}\\ &=-4\sqrt{2}\pi \end{align*}

Oof! Let’s do it an easier way.

Solution 2.

Stokes’ Theorem:

To apply Stokes’ theorem we need to express $C$ as the boundary $\partial S$ of a surface $S\text{.}$ As

\begin{equation*} C=\Set{(x,y,z)}{x^2+y^2+z^2=4,\ z=y} \end{equation*}

is a closed curve, this is possible. In fact there are many possible choices of $S$ with $\partial S=C\text{.}$ Three possible $S$’s (sketched below) are

\begin{align*} S&=\Set{(x,y,z)}{x^2+y^2+z^2\le 4,\ z=y}\cr S'&=\Set{(x,y,z)}{x^2+y^2+z^2= 4,\ z\ge y}\cr S''&=\Set{(x,y,z)}{x^2+y^2+z^2= 4,\ z\le y} \end{align*}

The first of these, which is part of a plane, is likely to lead to simpler computations than the last two, which are parts of a sphere. So we choose what looks like the simpler way.

In preparation for application of Stokes’ theorem, we compute $\vnabla\times\vF$ and $\hn\,\dee{S}\text{.}$ For the latter, we apply the formula $\hn\,\dee{S}=\pm(-f_x,-f_y,1)\,\dee{x}dy$ (of Equation 3.3.2) to the surface $z=f(x,y)=y\text{.}$ We use the $+$ sign to give the normal a positive $\hk$ component.

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix}\hi&\hj&\hk \\ \frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ z-y&-x-z&-x-y\end{matrix}\right]\\ &=\hi\big(-1-(-1)\big)-\hj\big(-1-1\big)+\hk\big(-1-(-1)\big)\\ &=2\,\hj\\ \hn\,\dee{S}&=(0,-1,1)\,\dee{x}\dee{y}\\ \vnabla\times\vF\cdot\hn\,\dee{S}&=(0,2,0)\cdot(0,-1,1)\,\dee{x}\dee{y} =-2\,\dee{x}\dee{y} \end{align*}

The integration variables are $x$ and $y$ and, by definition, the domain of integration is

\begin{equation*} R=\Set{(x,y)}{(x,y,z)\text{ is in }S\text{ for some }z} \end{equation*}

To determine precisely what this domain of integration is, we observe that since $z=y$ on $S\text{,}$ $x^2+y^2+z^2\le 4$ is the same as $x^2+2y^2\le 4$ on $S\text{,}$

\begin{equation*} S=\Set{(x,y,z)}{x^2+2y^2\le 4,\ z=y} \implies R=\Set{(x,y)}{x^2+2y^2\le 4} \end{equation*}

So the domain of integration is an ellipse with semimajor axis $a=2\text{,}$ semiminor axis $b=\sqrt{2}$ and area $\pi a b=2\sqrt{2}\pi\text{.}$ The integral is then

\begin{equation*} \oint_C\vF\cdot \dee{\vr} =\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} =\dblInt_R (-2)\,\dee{x}\dee{y} =-2\ \text{Area}\,(R) =-4\sqrt{2}\pi \end{equation*}

Remark (Limits of integration):

If the integrand were more complicated, we would have to evaluate the integral over $R$ by expressing it as an iterated integrals with the correct limits of integration. First suppose that we slice up $R$ using thin vertical slices. On each such slice, $x$ is essentially constant and $y$ runs from $-\sqrt{(4-x^2)/2}$ to $\sqrt{(4-x^2)/2}\text{.}$ The leftmost such slice would have $x=-2$ and the rightmost such slice would have $x=2\text{.}$ So the correct limits with this slicing are

\begin{equation*} \dblInt_R f(x,y)\,\dee{x}\dee{y} =\int_{-2}^2\dee{x}\int_{-\sqrt{(4-x^2)/2}}^{\sqrt{(4-x^2)/2}} \dee{y}\ f(x,y) \end{equation*}

If, instead, we slice up $R$ using thin horizontal slices, then, on each such slice, $y$ is essentially constant and $x$ runs from $-\sqrt{4-2y^2}$ to $\sqrt{4-2y^2}\text{.}$ The bottom such slice would have $y=-\sqrt{2}$ and the top such slice would have $y=\sqrt{2}\text{.}$ So the correct limits with this slicing are

\begin{equation*} \dblInt_R f(x,y)\,\dee{x}\dee{y} =\int_{-\sqrt{2}}^{\sqrt{2}}\dee{y}\int_{-\sqrt{4-2y^2}}^{\sqrt{4-2y^2}} \dee{x}\ f(x,y) \end{equation*}

Note that the integral with limits

\begin{equation*} \int_{-\sqrt{2}}^{\sqrt{2}}\dee{y}\int_{-2}^{2} \dee{x}\ f(x,y) \end{equation*}

corresponds to a slicing with $x$ running from $-2$ to $2$ on {\bf every} slice. This corresponds to a rectangular domain of integration, not what we have here.

Stokes’ Theorem, Again:

Since the integrand is just a constant (after Stoking — not the original integrand) and $S$ is so simple (because we chose it wisely), we can evaluate the integral $\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S}$ without ever determining $\dee{S}$ explicitly and without ever setting up any limits of integration. We already know that $\vnabla\times\vF=2\,\hj\text{.}$ Since $S$ is the level surface $z-y=0\text{,}$ the gradient $\vnabla(z-y)=-\hj+\hk$ is normal to $S\text{.}$ So $\hn = \frac{1}{\sqrt{2}}(-\hj+\hk)$ and

\begin{align*} \oint_C\vF\cdot \dee{\vr} &=\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} =\dblInt_S (2\hj)\cdot \frac{1}{\sqrt{2}}(-\hj+\hk)\,\dee{S}\\ &=\dblInt_S -\sqrt{2}\,\dee{S} =-\sqrt{2}\ {\rm Area}\,(S) \end{align*}

As $S$ is a circle of radius $2\text{,}$ $\oint_C\vF\cdot \dee{\vr}=-4\sqrt{2}\pi\text{,}$ yet again.

Example 4.4.8.

In Warning 4.1.17, we stated that if a vector field fails to pass the screening test $\vnabla\cdot\vB=0$ at even a single point, for example because the vector field is not defined at that point, then $\vB$ can fail to have a vector potential. An example is the point source

\begin{equation*} \vB(x,y,z) = \frac{\hat\vr(x,y,z)}{r(x,y,z)^2} \end{equation*}

of Example 3.4.2. Here, as usual,

\begin{equation*} r(x,y,z) = \sqrt{x^2+y^2+z^2}\qquad \hat\vr(x,y,z) = \frac{x\hi + y\hj + z\hk}{\sqrt{x^2+y^2+z^2}} \end{equation*}

This vector field is defined on all of $\bbbr^3\text{,}$ except for the origin, and its divergence

\begin{align*} \vnabla\cdot\vB &=\frac{\partial }{\partial x} \left(\frac{x}{(x^2+y^2+z^2)^{3/2}}\right) +\frac{\partial }{\partial y} \left(\frac{y}{(x^2+y^2+z^2)^{3/2}}\right)\\ &\hskip1in +\frac{\partial }{\partial z} \left(\frac{z}{(x^2+y^2+z^2)^{3/2}}\right)\\ &=\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right)\\ &\hskip1in +\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3y^2}{(x^2+y^2+z^2)^{5/2}}\right)\\ &\hskip1in +\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3z^2}{(x^2+y^2+z^2)^{5/2}}\right)\\ &=\frac{3}{(x^2+y^2+z^2)^{3/2}}-\frac{3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}} \end{align*}

is zero everywhere except at the origin, where it is not defined.

This vector field cannot have a vector potential on its domain of definition, i.e. on $\bbbr^3\setminus\{(0,0,0)\} =\Set{(x,y,z)}{(x,y,z)\ne(0,0,0)} \text{.}$ To see this, suppose to the contrary that it did have a vector potential $\vA\text{.}$ Then its flux through any closed surface

⁵

If you are uncomfortable with the surface not having a boundary, poke a very small hole in the surface, giving it a very small boundary. Then take the limit as the hole tends to zero.

(i.e. surface without a boundary) $S$ would be

\begin{gather*} \dblInt_S\vB\cdot\hn\,\dee{S} = \dblInt_S\vnabla\times\vA\cdot\hn\,\dee{S} =\oint_{\partial S}\vA\cdot\dee{\vr} =0 \end{gather*}

by Stokes’ theorem, since $\partial S$ is empty. But we found in Example 3.4.2, with $m=1\text{,}$ that the flux of $\vB$ through any sphere centred on the origin is $4\pi\text{.}$

Subsection 4.4.1 The Interpretation of Div and Curl Revisited

In sections 4.1.4 and 4.1.5 we derived interpretations of the divergence and of the curl. Now that we have the divergence theorem and Stokes’ theorem, we can simplify those derivations a lot.

Subsubsection 4.4.1.1 Divergence

Let $\veps \gt 0$ be a tiny positive number, and then let

\begin{equation*} B_\veps(x_0,y_0,z_0) =\Set{(x,y,z)}{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 \lt \veps^2} \end{equation*}

be a tiny ball of radius $\veps$ centred on the point $(x_0,y_0,z_0)\text{.}$ Denote by

\begin{equation*} S_\veps(x_0,y_0,z_0) =\Set{(x,y,z)}{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=\veps^2} \end{equation*}

its surface. Because $B_\veps(x_0,y_0,z_0)$ is really small, $\vnabla\cdot \vv$ is essentially constant in $B_\veps(x_0,y_0,z_0)$ and we essentially have

\begin{equation*} \tripInt_{B_\veps(x_0,y_0,z_0)} \vnabla\cdot \vv\ \dee{V} =\vnabla\cdot \vv(x_0,y_0,z_0)\ \text{Vol}\big(B_\veps(x_0,y_0,z_0)\big) \end{equation*}

Of course we are really making an approximation here, based on the assumption that $\vv(x,y,z)$ is continuous and so takes values very close to $\vv(x_0,y_0,z_0)$ everywhere on the domain of integration. The approximation gets better and better as $\veps\rightarrow 0$ and a more precise statement is

\begin{equation*} \vnabla\cdot\vv(x_0,y_0,z_0)=\lim_{\veps\rightarrow 0} \frac{\tripInt_{B_\veps(x_0,y_0,z_0)} \vnabla\cdot \vv\ \dee{V}} {\text{Vol}\big(B_\veps(x_0,y_0,z_0)\big)} \end{equation*}

By the divergence theorem, we also have

\begin{equation*} \tripInt_{B_\veps(x_0,y_0,z_0)} \vnabla\cdot \vv\ \dee{V} =\dblInt_{S_\veps(x_0,y_0,z_0)} \vv\cdot\hn \ \dee{S} \end{equation*}

Think of the vector field $\vv$ as the velocity of a moving fluid which has density one. We have already seen, in §3.4, that the flux integral for a velocity field has the interpretation

\begin{align*} \dblInt_{S_\veps(x_0,y_0,z_0)}\hskip-10 pt \vv\cdot\hn \ \dee{S} &=\left\{ \begin{array}{l} \text{the volume of fluid leaving $B_\veps(x_0,y_0,z_0)$}\\ \text{through $S_\veps(x_0,y_0,z_0)$ per unit time} \end{array} \right. \end{align*}

We conclude that, as we said in 4.1.19,

\begin{align*} \vnabla\cdot\vv(x_0,y_0,z_0) &=\lim_{\veps\rightarrow 0} \frac{\text{the rate at which fluid is exiting $B_\veps(x_0,y_0,z_0)$}} {\text{Vol}\big(B_\veps(x_0,y_0,z_0)\big)}\\ &=\left\{ \begin{array}{l} \text{rate at which fluid is exiting an }\\ \text{infinitesimal sphere centred at $(x_0,y_0,z_0)$, }\\ \text{per unit time, per unit volume} \end{array} \right.\\ &=\text{strength of the source at $(x_0,y_0,z_0)$} \end{align*}

If our world is filled with an incompressible fluid, a fluid whose density is constant and so never expands or compresses, we will have $\vnabla\cdot\vv=0\text{.}$

Subsubsection 4.4.1.2 Curl

Again let $\veps \gt 0$ be a tiny positive number and let $D_\veps(x_0,y_0,z_0)$ be a tiny flat circular disk of radius $\veps$ centred on the point $(x_0,y_0,z_0)$ and denote by $C_\veps(x_0,y_0,z_0)$ its boundary circle. Let $\hn$ be a unit normal vector to $D_\veps\text{.}$ It tells us the orientation of $D_\veps\text{.}$ Give the circle $C_\veps$ the corresponding orientation using the right hand rule. That is, if the fingers of your right hand are pointing in the corresponding direction of motion along $C_\veps$ and your palm is facing $D_\veps\text{,}$ then your thumb is pointing in the direction $\hn\text{.}$

Because $D_\veps(x_0,y_0,z_0)$ is really small, $\vnabla\!\times\vv$ is essentially constant on $D_\veps(x_0,y_0,z_0)$ and we essentially have

\begin{align*} \dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{S} &=\vnabla\!\times \vv(x_0,y_0,z_0)\cdot\hn\ \text{Area}\big(D_\veps(x_0,y_0,z_0)\big)\\ &=\pi\veps^2\ \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn \end{align*}

Again, this is really an approximate statement which gets better and better as $\veps\rightarrow 0\text{.}$ A more precise statement is

\begin{gather*} \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn =\lim_{\veps\rightarrow 0} \frac{\dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{S}} {\pi\veps^2} \end{gather*}

By Stokes’ theorem, we also have

\begin{gather*} \dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{S} =\oint_{C_\veps(x_0,y_0,z_0)} \vv\cdot d\vr \end{gather*}

Again, think of the vector field $\vv$ as the velocity of a moving fluid. Then $\oint_{C_\veps} \vv\cdot d\vr$ is called the circulation of $\vv$ around $C_\veps\text{.}$

To measure the circulation experimentally, place a small paddle wheel in the fluid, with the axle of the paddle wheel pointing along $\hn$ and each of the paddles perpendicular to $C_\veps$ and centred on $C_\veps\text{.}$

Each paddle moves tangentially to $C_\veps\text{.}$ It would like to move with the same speed as the tangential speed $\vv\cdot\hvt$ (where $\hvt$ is the forward pointing unit tangent vector to $C_\veps$ at the location of the paddle) of the fluid at its location. But all paddles are rigidly fixed to the axle of the paddle wheel and so must all move with the same speed. That common speed will be the average value of $\vv\cdot\hvt$ around $C_\veps\text{.}$ If $\dee{s}$ represents an element of arc length of $C_\veps\text{,}$ the average value of $\vv\cdot\hvt$ around $C_\veps$ is

\begin{equation*} \overline{v_T}=\frac{1}{2\pi\veps}\oint_{C_\veps} \vv\cdot \hvt\ \dee{s} =\frac{1}{2\pi\veps}\oint_{C_\veps} \vv\cdot \dee{\vr} \end{equation*}

since $\dee{\vr}$ has direction $\hvt$ and length $\dee{s}$ so that $\dee{\vr} =\hvt \dee{s}\text{,}$ and since $2\pi\veps$ is the circumference of $C_\veps\text{.}$ If the paddle wheel rotates at $\Om$ radians per unit time, each paddle travels a distance $\Om\veps$ per unit time (remember that $\veps$ is the radius of $C_\veps$). That is, $\overline{v_T}=\Om\veps\text{.}$ Combining all this information,

\begin{align*} \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn &=\lim_{\veps\rightarrow 0} \frac{\dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{S}} {\pi\veps^2}\\ &=\lim_{\veps\rightarrow 0} \frac{\oint_{C_\veps} \vv\cdot \dee{\vr}} {\pi\veps^2}\\ &=\lim_{\veps\rightarrow 0} \frac{2\pi\veps\ \overline{v_T}} {\pi\veps^2}\\ &=\lim_{\veps\rightarrow 0} \frac{2\pi\veps\ (\Omega \veps)} {\pi\veps^2}\\ &=2\Om \end{align*}

so that

\begin{equation*} \Om = \half \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn \end{equation*}

The component of $\vnabla\times\!\vv(x_0,y_0,z_0)$ in any direction $\hn$ is twice the rate at which the paddle wheel turns when it is put into the fluid at $(x_0,y_0,z_0)$ with its axle pointing in the direction $\hn\text{.}$ The direction of $\vnabla\times\!\vv(x_0,y_0,z_0)$ is the axle direction which gives maximum rate of rotation and the magnitude of $\vnabla\times\!\vv(x_0,y_0,z_0)$ is twice that maximum rate of rotation. For this reason, $\vnabla\times \vv$ is called the “vorticity”.

Subsection 4.4.2 Optional — An Application of Stokes’ Theorem — Faraday’s Law

Magnetic induction refers to a physical process whereby an electric voltage is created (“induced”) by a time varying magnetic field. This process is exploited in many applications, including electric generators, induction motors, induction cooking, induction welding and inductive charging. Michael Faraday

⁶

Michael Faraday (1791–1867) was an English physicist and chemist. He ended up being an extremely influential scientist despite having only the most basic of formal educations.

is generally credited with the discovery of magnetic induction. Faraday’s law is the following. Let $S$ be an oriented surface with boundary $C\text{.}$ Let $\vE$ and $\vB$ be the (time dependent) electric and magnetic fields and define

\begin{align*} \oint_C\vE\cdot \dee{\vr}&=\text{voltage around }C\\ \dblInt_S\vB\cdot \hn \,\dee{S}&=\text{magnetic flux through }S \end{align*}

Then the voltage around $C$ is the negative of the rate of change of the magnetic flux through $S\text{.}$ As an equation, Faraday’s Law is

\begin{equation*} \oint_C\vE\cdot \dee{\vr}=-\frac{\partial }{\partial t}\dblInt_S\vB\cdot\hn\,\dee{S} \end{equation*}

We can reformulate this as a partial differential equation. By Stokes’ Theorem,

\begin{equation*} \oint_C\vE\cdot \dee{\vr}=\dblInt_S(\vnabla\times\vE)\cdot \hn\,\dee{S} \end{equation*}

so Faraday’s law becomes

\begin{equation*} \dblInt_S\Big(\vnabla\times\vE+\frac{\partial\vB}{\partial t}\Big)\cdot\hn\,\dee{S}=0 \end{equation*}

This is true for all surfaces $S\text{.}$ So the integrand, assuming that it is continuous, must be zero.

To see this, let $\vG=\Big(\vnabla\times\vE+\frac{\partial\vB}{\partial t}\Big)\text{.}$ Suppose that $\vG(\vx_0)\ne 0\text{.}$ Pick a unit vector $\hn$ in the direction of $\vG(\vx_0)\text{.}$ Let $S$ be a very small flat disk centered on $\vx_0$ with normal $\hn$ (the vector we picked). Then $\vG(\vx_0)\cdot\hn \gt 0$ and, by continuity, $\vG(\vx)\cdot\hn \gt 0$ for all $\vx$ on $S\text{,}$ if we have picked $S$ small enough. Then $\dblInt_S\Big(\vnabla\times\vE+\frac{\partial\vB}{\partial t}\Big)\cdot \hn\,\dee{S} \gt 0\text{,}$ which is a contradiction. So $\vG=\vZero$ everywhere and we conclude that

\begin{equation*} \vnabla\times\vE+\frac{\partial\vB}{\partial t}=0 \end{equation*}

This is one of Maxwell’s electromagnetic field equations

⁷

For the others, see Example 4.1.2

Exercises 4.4.3 Exercises

Exercise Group.

Exercises — Stage 1

1.

Each of the figures below contains a sketch of a surface $S$ and its boundary $\partial S\text{.}$ Stokes’ theorem says that $\oint_{\partial S}\vF\cdot \dee{\vr} =\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{S}$ if $\hn$ is a correctly oriented unit normal vector to $S\text{.}$ Add to each sketch a typical such normal vector.

(a)

(b)

(c)

2.

Let

$R$ be a finite region in the $xy$-plane,
the boundary, $C\text{,}$ of $R$ consist of a single piecewise smooth, simple closed curve
- that is oriented (i.e. an arrow is put on $C$) consistently with $R$ in the sense that if you walk along $C$ in the direction of the arrow, then $R$ is on your left
$F_1(x,y)$ and $F_2(x,y)$ have continuous first partial derivatives at every point of $R\text{.}$

Use Stokes’ theorem to show that

\begin{equation*} \oint_{C} \big[F_1(x,y)\,\dee{x} +F_2(x,y)\,\dee{y}\big] =\dblInt_{R}\Big(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\Big)\ \dee{x}\dee{y} \end{equation*}

i.e. to show Green’s theorem.

3.

Verify the identity $\ \oint_C\phi\vnabla\psi\cdot \dee{\vr}=-\oint_C\psi\vnabla\phi\cdot \dee{\vr}\ $ for any continuously differentiable scalar fields $\phi$ and $\psi$ and curve $C$ that is the boundary of a piecewise smooth surface.

Exercise Group.

Exercises — Stage 2

4.

Let $C$ be the curve of intersection of the cylinder $x^2+y^2=1$ and the surface $z=y^2$ oriented in the counterclockwise direction as seen from $(0,0,100)\text{.}$ Let $\vF=(x^2-y\,,\,y^2+x\,,\,1)\text{.}$ Calculate $\oint_C\vF\cdot \dee{\vr}$

by direct evaluation
by using Stokes’ Theorem.

5.

Evaluate $\oint_C \vF\cdot \dee{\vr}$ where $\vF=ye^x\,\hi+(x+e^x)\,\hj+z^2\,\hk$ and $C$ is the curve

\begin{equation*} \vr(t)=(1+\cos t)\,\hi+(1+\sin t)\,\hj+(1-\sin t-\cos t)\,\hk\qquad 0\le t\le 2\pi \end{equation*}

6. (✳).

Find the value of $\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{S}$ where $\vF = \big(z - y\,,\, x\,,\, -x\big)$ and $S$ is the hemisphere

\begin{equation*} \Set{(x, y, z) \in\bbbr^3 }{ x^2 + y^2 + z^2 = 4,\ z \ge 0} \end{equation*}

oriented so the surface normals point away from the centre of the hemisphere.

7. (✳).

Let $\cS$ be the part of the surface $z=16-{(x^2+y^2)}^2$ which lies above the $xy$-plane. Let $\vF$ be the vector field

\begin{equation*} \vF=x\ln(1+z)\,\hi+x(3+y)\,\hj+y\cos z\,\hk \end{equation*}

Calculate

\begin{equation*} \dblInt_{\cS}\vnabla\times\vF\cdot\hn\,\dee{S} \end{equation*}

where $\hn$ is the upward normal on $\cS\text{.}$

8. (✳).

Let $\cC$ be the intersection of the paraboloid $z=4-x^2-y^2$ with the cylinder $x^2+(y-1)^2=1\text{,}$ oriented counterclockwise when viewed from high on the $z$-axis. Let $\vF=xz\,\hi+x\,\hj+yz\,\hk\text{.}$ Find $\oint_{\cC}\vF\cdot \dee{\vr}\text{.}$

9.

Let $\vF = - ye^z\,\hi + x^3\cos z\,\hj + z\sin(xy)\,\hk\text{,}$ and let $S$ be the part of the surface $z = (1-x^2)(1-y^2)$ that lies above the square $-1\le x\le 1\text{,}$ $-1\le y\le 1$ in the $xy$-plane. Find the flux of $\vnabla\times \vF$ upward through $S\text{.}$

10.

Evaluate the integral $\oint_C \vF\cdot \dee{\vr}\text{,}$ in which $\vF = (e^{x^2} - yz\,,\,\sin y - yz \,,\,xz + 2y)$ and $C$ is the triangular path from $(1, 0, 0)$ to $(0, 1, 0)$ to $(0, 0, 1)$ to $(1, 0, 0)\text{.}$

11. (✳).

Let $\vF(x,y,z)=-z\,\hi+x\,\hj+y\,\hk$ be a vector field. Use Stokes’ theorem to evaluate the line integral $\oint_C\vF\cdot \dee{\vr}$ where $C$ is the intersection of the plane $z=y$ and the ellipsoid $\frac{x^2}{4}+\frac{y^2}{2}+\frac{z^2}{2}=1\text{,}$ oriented counter-clockwise when viewed from high on the $z$-axis.

12. (✳).

Consider the vector field $\vF(x,y, z) = z^2 \,\hi + x^2 \,\hj + y^2\,\hk$ in $\bbbr^3\text{.}$

Compute the line integral $I_1 = \int_{C_1}\vF\cdot\dee{\vr}$ where $C_1$ is the curve consisting of three line segments, $L_1$ from $(2, 0, 0)$ to $(0, 2, 0)\text{,}$ then $L_2$ from $(0, 2, 0)$ to $(0, 0, 2)\text{,}$ finally $L_3$ from $(0, 0, 2)$ to $(2, 0, 0)\text{.}$
A simple closed curve $C_2$ lies on the plane $E\,:\, x + y + z = 2\text{,}$ enclosing a region $R$ on the plane of area $3\text{,}$ and oriented in a counterclockwise direction as observed from the positive $x$-axis. Compute the line integral $I_2 = \int_{C_2}\vF\cdot\dee{\vr}\text{.}$

13. (✳).

Let $C = C_1 + C_2 + C_3$ be the curve given by the union of the three parameterized curves

\begin{alignat*}{2} \vr_1(t) &= \big(2\cos t, 2\sin t, 0\big), &\qquad &0 \le t \le \pi/2\\ \vr_2(t) &= \big(0, 2\cos t, 2\sin t\big), & &0 \le t \le \pi/2\\ \vr_3(t) &= \big(2\sin t, 0, 2\cos t\big), & &0 \le t \le \pi/2 \end{alignat*}

Draw a picture of $C\text{.}$ Clearly mark each of the curves $C_1\text{,}$ $C_2\text{,}$ and $C_3$ and indicate the orientations given by the parameterizations.
Find and parameterize an oriented surface $S$ whose boundary is $C$ (with the given orientations).
Compute the line integral $\int_C \vF\cdot\dee{\vr}$ where
\begin{equation*} \vF = \Big(y + \sin(x^2)\,,\, z - 3x + \ln(1 + y^2)\,,\, y + e^{z^2}\Big) \end{equation*}

14. (✳).

We consider the cone with equation $z = \sqrt{x^2 + y^2}\text{.}$ Note that its tip, or vertex, is located at the origin $(0, 0, 0)\text{.}$ The cone is oriented in such a way that the normal vectors point downwards (and away from the $z$ axis). In the parts below, both $S_1$ and $S_2$ are oriented this way.

Let $\vF = \big(-zy, zx, xy \cos(yz)\big)\text{.}$

Let $S_1$ be the part of the cone that lies between the planes $z = 0$ and $z = 4\text{.}$ Note that $S_1$ does not include any part of the plane $z = 4\text{.}$ Use Stokes’ theorem to determine the value of
\begin{equation*} \dblInt_{S_1} \vnabla\times\vF \cdot \hn\,\dee{S} \end{equation*}
Make a sketch indicating the orientations of $S_1$ and of the contour(s) of integration.
Let $S_2$ be the part of the cone that lies below the plane $z = 4$ and above $z = 1\text{.}$ Note that $S_2$ does not include any part of the planes $z = 1$ and $z = 4\text{.}$ Determine the flux of $\vnabla\times\vF$ across $S_2\text{.}$ Justify your answer, including a sketch indicating the orientations of $S_2$ and of the contour(s) of integration.

15. (✳).

Consider the curve $C$ that is the intersection of the plane $z = x + 4$ and the cylinder $x^2 + y^2 = 4\text{,}$ and suppose $C$ is oriented so that it is traversed clockwise as seen from above.

Let $\vF(x, y, z) = \big(x^3 + 2y\,,\, \sin(y) + z\,,\, x + \sin(z^2)\big)\text{.}$

Use Stokes’ Theorem to evaluate the line integral $\oint_C\vF\cdot\dee{\vr}\text{.}$

16. (✳).

Consider the vector field $\vF(x, y, z) = (z^2 , x^2 , y^2)$ in $\bbbr^3\text{.}$ Compute the line integral $\oint_C \vF\cdot\dee{\vr}\text{,}$ where $C$ is the curve consisting of the three line segments, $L_1$ from $(2, 0, 0)$ to $(0, 2, 0)\text{,}$ then $L_2$ from $(0, 2, 0)$ to $(0, 0, 2)\text{,}$ and finally $L_3$ from $(0, 0, 2)$ to $(2, 0, 0)\text{.}$
A simple closed curve $C$ lies in the plane $x + y + z = 2\text{.}$ The surface this curve $C$ surrounds inside the plane $x + y + z = 2$ has area $3\text{.}$ The curve $C$ is oriented in a counterclockwise direction as observed from the positive x-axis. Compute the line integral $\oint_C\vF\cdot \dee{\vr}$ , where $\vF$ is as in (a).

17. (✳).

Evaluate the line integral

\begin{equation*} \int_C\left(z+\frac{1}{1+z}\right)\dee{x} +xz\,\dee{y} +\left(3xy-\frac{x}{(z+1)^2}\right)\dee{z} \end{equation*}

where $C$ is the curve parameterized by

\begin{equation*} \vr(t) = \big(\cos t\,,\, \sin t\,,\, 1 - \cos^2 t \sin t\big)\qquad 0 \le t \le 2\pi \end{equation*}

18. (✳).

A simple closed curve $C$ lies in the plane $x + y + z = 1\text{.}$ The surface this curve $C$ surrounds inside the plane $x + y + z = 1$ has area $5\text{.}$ The curve $C$ is oriented in a clockwise direction as observed from the positive $z$-axis looking down at the plane $x + y + z = 1\text{.}$

Compute the line integral of $\vF (x, y, z) = (z^2 , x^2 , y^2)$ around $C\text{.}$

19. (✳).

Let $C$ be the oriented curve consisting of the 5 line segments which form the paths from $(0, 0, 0)$ to $(0, 1, 1)\text{,}$ from $(0, 1, 1)$ to $(0, 1, 2)\text{,}$ from $(0, 1, 2)$ to $(0, 2, 0)\text{,}$ from $(0, 2, 0)$ to $(2, 2, 0)\text{,}$ and from $(2, 2, 0)$ to $(0, 0, 0)\text{.}$ Let

\begin{equation*} \vF = (-y+e^x\sin x)\,\hi +y^4\,\hj +\sqrt{z}\tan z\,\hk \end{equation*}

Evaluate the integral $\int_C\vF\cdot\dee{\vr}\text{.}$

20. (✳).

Suppose the curve $C$ is the intersection of the cylinder $x^2 + y^2 = 1$ with the surface $z = xy^2\text{,}$ traversed clockwise if viewed from the positive z-axis, i.e. viewed “from above”. Evaluate the line integral

\begin{equation*} \int_C (z + \sin z) \,\dee{x} + (x^3 - x^2 y) \,\dee{y} + (x \cos z - y) \,\dee{z} \end{equation*}

21. (✳).

Evaluate $\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S}$ where $S$ is that part of the sphere $x^2+y^2+z^2=2$ above the plane $z=1\text{,}$ $\hn$ is the upward unit normal, and

\begin{gather*} \vF(x,y,z) = -y^2\,\hi +x^3\,\hj + \big(e^x + e^y +z\big)\,\hk \end{gather*}

22. (✳).

Let

\begin{equation*} \vF = x \sin y\,\hi - y \sin x\,\hj + (x - y)z^2\,\hk \end{equation*}

Use Stokes’ theorem to evaluate

\begin{equation*} \int_C\vF\cdot \dee{\vr} \end{equation*}

along the path consisting of the straight line segments successively joining the points $P_0 = (0, 0, 0)$ to $P_1 = (\pi/2, 0, 0)$ to $P_2 = (\pi/2, 0, 1)$ to $P_3 = (0, 0, 1)$ to $P_4 = (0, \pi/2, 1)$ to $P_5 = (0, \pi/2, 0)\text{,}$ and back to $(0, 0, 0)\text{.}$

23. (✳).

Let

\begin{equation*} \vF=\left(\frac{2z}{1+y}+\sin(x^2)\,,\, \frac{3z}{1+x}+\sin(y^2)\,,\, 5(x+1)(y+2)\right) \end{equation*}

Let $C$ be the oriented curve consisting of four line segments from $(0,0,0)$ to $(2,0,0)\text{,}$ from $(2,0,0)$ to $(0,0,2)\text{,}$ from $(0,0,2)$ to $(0,3,0)\text{,}$ and from $(0,3,0)$ to $(0,0,0)\text{.}$

Draw a picture of $C\text{.}$ Clearly indicate the orientation on each line segment.
Compute the work integral $\int_C\vF\cdot\dee{\vr}\text{.}$

24. (✳).

Evaluate $\displaystyle\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{S}$ where $\vF=y\,\hi+2z\,\hj+3x\,\hk$ and $S$ is the surface $z=\sqrt{1-x^2-y^2}\text{,}$ $z\ge 0$ and $\hn$ is a unit normal to $S$ obeying $\hn\cdot\hk\ge 0\text{.}$

25. (✳).

Let $\cS$ be the curved surface below, oriented by the outward normal:

\begin{equation*} x^2 + y^2 + 2(z-1)^2 = 6,\qquad z\ge 0. \end{equation*}

(E.g., at the high point of the surface, the unit normal is $\hk\text{.}$)

Define

\begin{equation*} \vG = \nabla\times\vF, \quad\text{where}\quad \vF = (xz - y^3\cos z)\,\hi + x^3 e^z\,\hj + xyze^{x^2+y^2+z^2}\,\hk. \end{equation*}

Find $\dblInt_\cS \vG\cdot \hn\dee{S}\text{.}$

26. (✳).

Let $C$ be a circle of radius $R$ lying in the plane $x+y+z=3\text{.}$ Use Stokes’ Theorem to calculate the value of

\begin{equation*} \oint_C \vF\cdot \dee{\vr} \end{equation*}

where $\vF=z^2\hi+x^2\hj+y^2\hk\text{.}$ (You may use either orientation of the circle.)

27.

Let $S$ be the oriented surface consisting of the top and four sides of the cube whose vertices are $(\pm 1,\pm1,\pm1)\text{,}$ oriented outward. If $\vF(x,y,z)=(xyz,xy^2,x^2yz)\text{,}$ find the flux of $\vnabla\times\vF$ through $S\text{.}$

28.

Let $S$ denote the part of the spiral ramp (that is helicoidal surface) parametrized by

\begin{equation*} x=u\cos v,\ y=u\sin v,\ z=v\qquad 0\le u\le 1,\ 0\le v\le 2\pi \end{equation*}

Let $C$ denote the boundary of $S$ with orientation specified by the upward pointing normal on $S\text{.}$ Find

\begin{equation*} \int_C y\,\dee{x}-x\,\dee{y}+ xy\,\dee{z} \end{equation*}

Exercise Group.

Exercises — Stage 3

29.

Let $C$ be the intersection of $x+2y-z=7$ and $x^2-2x+4y^2=15\text{.}$ The curve $C$ is oriented counterclockwise when viewed from high on the $z$-axis. Let

\begin{equation*} \vF=\big(e^{x^2}+yz\big)\,\hi +\big(\cos(y^2)-x^2\big)\,\hj +\big(\sin(z^2)+xy\big)\,\hk \end{equation*}

Evaluate $\oint_C\vF\cdot \dee{\vr}\text{.}$

30. (✳).

Find the curl of the vector field $\vF = \big(2 + x^2 + z\,,\, 0\,,\, 3 + x^2 z\big)\text{.}$
Let $C$ be the curve in $\bbbr^3$ from the point $(0, 0, 0)$ to the point $(2, 0, 0)\text{,}$ consisting of three consecutive line segments connecting the points $(0, 0, 0)$ to $(0, 0, 3)\text{,}$ $(0, 0, 3)$ to $(0, 1, 0)\text{,}$ and $(0, 1, 0)$ to $(2, 0, 0)\text{.}$ Evaluate the line integral
\begin{equation*} \int_C \vF\cdot\dee{\vr} \end{equation*}
where $\vF$ is the vector field from (a).

31. (✳).

Let $S$ be the bucket shaped surface consisting of the cylindrical surface $y^2 + z^2 = 9$ between $x = 0$ and $x = 5\text{,}$ and the disc inside the $yz$-plane of radius $3$ centered at the origin. (The bucket $S$ has a bottom, but no lid.) Orient $S$ in such a way that the unit normal points outward. Compute the flux of the vector field $\vnabla\times\vG$ through $S\text{,}$ where $\vG = (x, -z, y)\text{.}$
Compute the flux of the vector field $\vF = (2 + z, xz^2 , x \cos y)$ through $S\text{,}$ where $S$ is as in (a).

32. (✳).

Let

\begin{equation*} \vF(x, y, z) = \Big(\frac{y}{x} +x^{1+x^2}\,,\, x^2-y^{1+y^2}\,,\, \cos^5(\ln z)\Big) \end{equation*}

Write down the domain $D$ of $\vF\text{.}$
Circle the correct statement(s):
1. D is connected.
2. D is simply connected.
3. D is disconnected.
Compute $\vnabla\times\vF\text{.}$
Let $C$ be the square with corners $(3 \pm 1, 3 \pm 1)$ in the plane $z = 2\text{,}$ oriented clockwise (viewed from above, i.e. down $z$-axis). Compute
\begin{equation*} \int_C \vF \cdot \dee{\vr} \end{equation*}
Is $\vF$ conservative?

33. (✳).

A physicist studies a vector field $\vF(x, y, z)\text{.}$ From experiments, it is known that $\vF$ is of the form

\begin{equation*} \vF(x, y, z) = xz\,\hi + (axe^y z + byz)\,\hj + (y^2 - xe^y z^2 )\,\hk \end{equation*}

for some real numbers $a$ and $b\text{.}$ It is further known that $\vF = \vnabla\times\vG$ for some differentiable vector field $\vG\text{.}$

Determine $a$ and $b\text{.}$
Evaluate the surface integral
\begin{equation*} \dblInt_S\vF\cdot\hn\,\dee{S} \end{equation*}
where $S$ is the part of the ellipsoid $x^2 + y^2 + \frac{1}{4} z^2 = 1$ for which $z \ge 0\text{,}$ oriented so that its normal vector has a positive $z$-component.

34. (✳).

Let $C$ be the curve in the $xy$-plane from the point $(0, 0)$ to the point $(5, 5)$ consisting of the ten line segments consecutively connecting the points $(0,0)\text{,}$ $(0,1)\text{,}$ $(1,1)\text{,}$ $(1,2)\text{,}$ $(2,2)\text{,}$ $(2,3)\text{,}$ $(3,3)\text{,}$ $(3,4)\text{,}$ $(4,4)\text{,}$ $(4,5)\text{,}$ $(5,5)\text{.}$ Evaluate the line integral

\begin{gather*} \int_C \vF \cdot\dee{\vr} \end{gather*}

where

\begin{gather*} \vF = y\,\hi + (2x - 10)\,\hj \end{gather*}

35. (✳).

Let $\vF = \big(\sin x^2\,,\,xz\,,\,z^2\big)\text{.}$ Evaluate $\oint_C\vF\cdot\dee{\vr}$ around the curve $C$ of intersection of the cylinder $x^2+y^2=4$ with the surface $z=x^2\text{,}$ traversed counter clockwise as viewed from high on the $z$-axis.

36. (✳).

Explain how one deduces the differential form

\begin{equation*} \vnabla\times\vE = -\frac{1}{c}\frac{\partial\vH}{\partial t} \end{equation*}

of Faraday’s law from its integral form

\begin{equation*} \oint_C\vE\cdot\dee{\vr} = -\frac{1}{c}\ \diff{ }{t}\dblInt_S\vH\cdot\hn\,\dee{S} \end{equation*}

37. (✳).

Let $C$ be the curve given by the parametric equations:

\begin{equation*} x=\cos t,\ y=\sqrt{2}\sin t,\ z=\cos t,\ 0\le t\le 2\pi \end{equation*}

and let

\begin{equation*} \vF=z\,\hi+x\,\hj+y^3z^3\,\hk \end{equation*}

Use Stokes’ theorem to evaluate

\begin{equation*} \oint_C \vF\cdot \dee{\vr} \end{equation*}

38. (✳).

Use Stokes’ theorem to evaluate

\begin{equation*} \oint_C z\,\dee{x}+x\,\dee{y}-y\,\dee{z} \end{equation*}

where $C$ is the closed curve which is the intersection of the plane $x+y+z=1$ with the sphere $x^2+y^2+z^2=1\text{.}$ Assume that $C$ is oriented clockwise as viewed from the origin.

39. (✳).

Let $S$ be the part of the half cone

\begin{equation*} z=\sqrt{x^2+y^2},\quad y\ge 0, \end{equation*}

that lies below the plane $z=1\text{.}$

Find a parametrization for $S\text{.}$
Calculate the flux of the velocity field
\begin{equation*} \vv=x\,\hi + y\, \hj -2 z\,\hk \end{equation*}
downward through $S\text{.}$
A vector field $\vF$ has curl $\vnabla\times \vF= x\, \hi + y\, \hj -2 z\,\hk\text{.}$ On the $xz$-plane, the vector field $\vF$ is constant with $\vF(x,0,z)=\hj\text{.}$ Given this information, calculate
\begin{equation*} \int_\cC \vF\cdot \dee{\vr}, \end{equation*}
where $\cC$ is the half circle
\begin{equation*} x^2+y^2=1,\ z=1,\ y\ge 0 \end{equation*}
oriented from $(-1,0,1)$ to $(1,0,1)\text{.}$

40.

Consider $\dblInt_S(\vnabla\times\vF)\cdot\hn\,dS$ where $S$ is the portion of the sphere $x^2+y^2+z^2=1$ that obeys $x+y+z\ge 1\text{,}$ $\hn$ is the upward pointing normal to the sphere and $\vF=(y-z)\hi+(z-x)\hj+(x-y)\hk\text{.}$ Find another surface $S'$ with the property that $\dblInt_S(\vnabla\times\vF)\cdot\hn\,\dee{S} =\dblInt_{S'}(\vnabla\times\vF)\cdot\hn\,\dee{S}$ and evaluate $\dblInt_{S'}(\vnabla\times\vF)\cdot\hn\,\dee{S}\text{.}$

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