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CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 4.3 Green’s Theorem

Our next variant of the fundamental theorem of calculus is Green’s
 1 
George Green (1793–1841) was a British mathematical physicist. He spent much of the early part of his life working in his father’s bakery and grain mill. He was finally admitted as an undergraduate to Cambridge in 1832, aged nearly forty.
theorem, which relates an integral, of a derivative of a (vector-valued) function, over a region in the \(xy\)-plane, with an integral of the function over the curve bounding the region. First we need to define some properties of curves.

Definition 4.3.1.

  1. A curve \(C\) with parametrization \(\vr(t)\text{,}\) \(a\le t\le b\text{,}\) is said to be closed if \(\vr(a)=\vr(b)\text{.}\)
  2. A curve \(C\) is said to be simple if it does not cross itself. More precisely, if \(\vr(t)\text{,}\) \(a\le t\le b\text{,}\) is a parametrization of the curve and if \(a\le t_1,t_2\le b\) obey \(t_1\ne t_2\) and \(\{t_1,t_2\} \ne\{a,b\}\text{,}\) then \(\vr(t_1)\ne \vr(t_2)\text{.}\) That is, if \(\vr(t_1)=\vr(t_2)\text{,}\) then either \(t_1=t_2\) or \(t_1=a\text{,}\) \(t_2=b\text{,}\) or \(t_1=b\text{,}\) \(t_2=a\text{.}\)
  3. A curve \(C\) is piecewise smooth if it has a parametrization \(\vr(t)\) which
    • is continuous and which
    • is differentiable except possibly at finitely many points with
    • the derivative being continuous and nonzero except possibly at finitely many points.
Here are sketches of some examples.
And here is Green’s theorem.

Warning 4.3.3.

Note that in Theorem 4.3.2 we are assuming that \(F_1\) and \(F_2\) have continuous first partial derivatives at every point of \(R\text{.}\) If that is not the case, for example because \(F_1\) or \(F_2\) is not defined on all of \(R\text{,}\) then the conclusion of Green’s theorem can fail. An example is \(F_1=-\frac{y}{x^2+y^2}\text{,}\) \(F_2=\frac{x}{x^2+y^2}\text{,}\) \(R=\Set{(x,y)}{x^2+y^2\le 1}\text{.}\) See Examples 4.3.7 and 4.3.8.
Here are three notational remarks before we start the proof.
  • One way to remember the integrand on the right hand side is to write it as \((\vnabla\times\vF)\cdot\hk\text{.}\)
  • Many people use \(M\) instead of \(F_1\) and \(N\) instead of \(F_2\text{.}\) Then Green’s theorem becomes \(\oint_{C} \big[M(x,y)\,\dee{x} +N(x,y)\,\dee{y}\big] =\dblInt_{R}\Big(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\Big)\ \dee{x}\dee{y}\)
  • The symbol \(\oint_C\) is just an alternate notation for \(\int_C\) that is sometimes used when \(C\) is a closed curve. See Notation 2.4.1.

Proof.

We prove the result by reformulating it as a divergence theorem statement. To that end, we define
\begin{align*} V&=\Set{(x,y,z)}{ (x,y)\in R,\ \ 0\le z\le 1}\\ \vG(x,y,z) &= F_2(x,y)\,\hi -F_1(x,y)\,\hj \end{align*}
Notice that \(V\) is exactly the volume obtained by expanding \(R\) vertically upward by one unit.
The definition of \(\vG\) does not contain a typo — the \(x\)-component of \(\vG\) really is \(F_2\) and the \(y\)-component of \(\vG\) really is \(-F_1\text{.}\) (More or less the reverse of what you would normally write down.)
These definitions have been rigged so that the divergence theorem applied to \(\vG\) and \(V\text{,}\) namely
\begin{align*} \dblInt_{\partial V} \vG\cdot\hn\,\dee{S} &=\tripInt_V\vnabla\cdot\vG\ \dee{V} \end{align*}
gives us exactly Green’s theorem, as we shall now see.
Since \(\vnabla\cdot\vG = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\text{,}\) the right hand side is just
\begin{align*} \tripInt_V\vnabla\cdot\vG\ \dee{V} &= \dblInt_{R}\dee{x}\dee{y}\int_0^1\dee{z} \ \vnabla\cdot\vG\\ &= \dblInt_{R}\dee{x}\dee{y}\int_0^1\dee{z} \ \left(\frac{\partial F_2}{\partial x}(x,y) - \frac{\partial F_1}{\partial y}(x,y)\right)\\ &= \dblInt_{R}\dee{x}\dee{y}\ \left(\frac{\partial F_2}{\partial x}(x,y) - \frac{\partial F_1}{\partial y}(x,y)\right) \end{align*}
because the integrand is independent of \(z\text{.}\) This is exactly the right hand side of Green’s theorem.
Now for the left hand side. The boundary, \(\partial V\text{,}\) of \(V\) is the union of the (flat) bottom, the (flat) top and the (curved) side. The outward unit normal on the (horizontal, flat) top is \(+\hk\) and the outward unit normal on the (horizontal, flat) bottom is \(-\hk\) so that
\begin{align*} \dblInt_{\partial V} \vG\cdot\hn\,\dee{S} &=\dblInt_{\text{top}} \vG\cdot\hk\,\dee{S} +\dblInt_{\text{bottom}} \vG\cdot(-\hk)\,\dee{S} +\dblInt_{\text{side}} \vG\cdot\hn\,\dee{S}\\ &=\dblInt_{\text{side}} \vG\cdot\hn\,\dee{S} \end{align*}
We have used the fact that the \(\hk\) component of \(\vG\) is exactly zero to discard the integrals over the top and bottom of \(\partial V\text{.}\) To evaluate the integral over the side, we’ll parametrize the side. Suppose that \(\vr(t)=x(t)\,\hi +y(t)\,\hj\text{,}\) \(a\le t\le b\text{,}\) is a parametrization of \(C\text{,}\) with the arrow in the figure above giving the direction of increasing \(t\text{.}\) Then we can use
\begin{gather*} \vR(t,z) = \vr(t) +z\,\hk = x(t)\,\hi +y(t)\,\hj +z\,\hk \qquad a\le t\le b,\ 0\le z\le 1 \end{gather*}
as a parametrization of the side. We’ll use (3.3.1) to determine \(\hn\,\dee{S}\) for the side. Since
\begin{align*} \frac{\partial\vR}{\partial t}(t,z) & = x'(t)\,\hi +y'(t)\,\hj\\ \frac{\partial\vR}{\partial z}(t,z) & = \hk \end{align*}
(3.3.1) gives
\begin{align*} \hn\,\dee{S} &= \frac{\partial\vR}{\partial t}(t,z)\times \frac{\partial\vR}{\partial z}(t,z)\ \dee{t}\dee{z}\\ &= \big(x'(t)\,\hi +y'(t)\,\hj\big)\times \hk\ \dee{t}\dee{z}\\ &= \big(-x'(t)\,\hj +y'(t)\,\hi\big)\ \dee{t}\dee{z} \end{align*}
Note that with this choice of \(\pm\) sign (that is, \(\frac{\partial\vR}{\partial t}\times \frac{\partial\vR}{\partial z}\ \dee{t}\dee{z}\) rather than \(-\frac{\partial\vR}{\partial t}\times \frac{\partial\vR}{\partial z}\ \dee{t}\dee{z}\)), the vector \(\hn\) really is the outward pointing normal, as we see from the sketch
We can now compute the surface integral directly.
\begin{align*} &\dblInt_{\partial V} \vG\cdot\hn\,\dee{S} =\dblInt_{\text{side}} \vG\cdot\hn\,\dee{S}\\ &\hskip0.25in=\int_a^b\dee{t}\int_0^1\dee{z}\ \vG\big(\vR(t,z)\big)\cdot \big(-x'(t)\,\hj +y'(t)\,\hi\big)\\ &\hskip0.25in=\int_a^b\dee{t}\int_0^1\dee{z}\ \big(F_2(x(t),y(t))\,\hi -F_1(x(t),y(t))\,\hj\big)\cdot \big(-x'(t)\,\hj +y'(t)\,\hi\big)\\ &\hskip0.25in=\int_a^b\dee{t}\ \big[F_2(x(t),y(t))\,y'(t) +F_1(x(t),y(t))\,x'(t)\big]\\ &\hskip2in \text{since the integrand is independent of } z\\ &\hskip0.25in=\oint_{C} \big[F_1(x,y)\,\dee{x} +F_2(x,y)\,\dee{y}\big] \end{align*}
This is exactly the left hand side of Green’s theorem.

Example 4.3.4.

Evaluate
\begin{equation*} \oint_C\big[(x-xy)\,\dee{x} + (y^3+1)\,\dee{y}\big] \end{equation*}
where \(C\) is the curve given in the figure
Solution.
Let \(R=\Set{(x,y)}{1\le x\le 2,\ 0\le y\le 1}\text{.}\) By Green’s theorem
\begin{align*} \oint_C\big[(x-xy)\,\dee{x} + (y^3+1)\,\dee{y} &=\dblInt_R\Big[\frac{\partial }{\partial x}(y^3+1) - \frac{\partial }{\partial y}(x-xy)\Big]\dee{x}\dee{y}\\ &=\int_1^2\dee{x}\int_0^1\dee{y}\ x =\frac{x^2}{2}\bigg|_1^2 =\frac{3}{2} \end{align*}
Here is a simple corollary of Green’s theorem that tells how to compute the area enclosed by a curve in the \(xy\)-plane.

Proof.

This is just Green’s theorem applied first with \(\vF = x\,\hj\text{,}\) then with \(\vF= -y\,\hi\) and finally with \(\vF = \frac{1}{2}\big[-y\,\hi +x\,\hj\big]\text{.}\) For all three of these \(\vF\)’s,
\begin{equation*} \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 1 \end{equation*}
so that Green’s theorem gives
\begin{align*} \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} =\dblInt_{R}\ \dee{x}\dee{y}\\ & = \text{Area}(R) \end{align*}

Example 4.3.6.

In this example we will use Green’s theorem to compute the area enclosed by the astroid \(x^{2/3} + y^{2/3} = a^{2/3}\text{.}\)
In Example 1.1.9 we found the parametrization
\begin{equation*} \vr(t) = x(t)\,\hi + y(t)\,\hj = a\cos^3t\,\hi + a\sin^3 t\,\hj \qquad 0\le t\le 2\pi \end{equation*}
for the astroid. So, by Corollary 4.3.5,
\begin{align*} \text{Area} &=\frac{1}{2}\oint_C \big[x\dee{y} -y\dee{x}\big] =\frac{1}{2}\int_0^{2\pi} \big[x(t)y'(t) -y(t)x'(t)\big]\dee{t}\\ &=\frac{3a^2}{2}\int_0^{2\pi} \big[\cos^3t\sin^2t\cos t +\sin^3t\cos^2 t\sin t\big]\dee{t}\\ &=\frac{3a^2}{2}\int_0^{2\pi} \cos^2t\sin^2t\big[\cos^2t +\sin^2t\big]\dee{t}\\ &=\frac{3a^2}{2}\int_0^{2\pi} \cos^2t\sin^2t\ \dee{t}\\ &=\frac{3a^2}{8}\int_0^{2\pi} \sin^2(2t)\ \dee{t} =\frac{3a^2}{16}\int_0^{2\pi} [1-\cos(4t)]\ \dee{t}\\ &=\frac{3}{8}a^2\pi \end{align*}

Example 4.3.7. Trick Question.

Evaluate
\begin{equation*} \oint_C\vB\cdot\dee{\vr} \end{equation*}
where
\begin{equation*} \vB = \frac{-y\,\hi+x\,\hj}{x^2+y^2} \end{equation*}
and \(C\) is the curve
\begin{align*} x(t) &= \sin(\cos t)\\ y(t) &= \sin(\sin t)\\ z(t) &= 0 \end{align*}
with \(0\le t\le 2\pi\text{.}\)
Solution.
First let’s think about the curve \(C\text{.}\) If the curve were just \(X(t)=\cos t\text{,}\) \(Y(t)=\sin t\text{,}\) \(Z(t)=0\text{,}\) it would be the unit circle centred on the origin in the \(xy\)-plane, traversed counterclockwise. For \(-\frac{\pi}{2}\le u\le \frac{\pi}{2}\text{,}\) the function \(\sin u\) increases monotonically with \(u\) and is of the same sign as \(u\) so that, since \(|\sin t|,|\cos t|\le 1 \lt \frac{\pi}{2}\text{,}\)
  • \(x(t) = \sin\big(\cos t)\) has the same sign as \(X(t)=\cos t\) and is increasing at precisely the same \(t\)’s as is \(X(t)\)
  • \(y(t) = \sin\big(\sin t)\) has the same sign as \(Y(t)=\sin t\) and is increasing at precisely the same \(t\)’s as is \(Y(t)\)
So the extra sine in our parametrization of \(C\) just distorts the circle, straightening the sides a little as depicted here.
It looks like our problem is a straightforward Green’s theorem problem like Example 4.3.4. Let’s just try using the strategy of Example 4.3.4. Because
\begin{align*} \frac{\partial \vB_2}{\partial x} - \frac{\partial \vB_1}{\partial y} &=\frac{\partial }{\partial x}\frac{x}{x^2+y^2} - \frac{\partial }{\partial y}\frac{-y}{x^2+y^2}\\ &= \frac{1}{x^2+y^2} - \frac{2x^2}{{(x^2+y^2)}^2} +\frac{1}{x^2+y^2} - \frac{2y^2}{{(x^2+y^2)}^2}\\ &= \frac{(x^2+y^2) - 2x^2 + (x^2+y^2) - 2y^2}{{(x^2+y^2)}^2}\\ &=0 \end{align*}
it looks like Green’s theorem gives us, trivially,
\begin{equation*} \oint_C\vB\cdot\dee{\vr} =\oint_C\big[\vB_1\,\dee{x} +\vB_2\,\dee{y}\big] = \dblInt_{R}\left(\frac{\partial B_2}{\partial x} - \frac{\partial B_1}{\partial y}\right)\ \dee{x}\dee{y} =0 \end{equation*}
where \(R\) is the region inside our curve \(C\text{.}\)
That was easy — but it’s also very wrong! Our next steps are to
  • verify that \(\oint_C\vB\cdot\dee{\vr}\ne 0\text{,}\) and
  • explain why we got the wrong answer, and
  • modify our computation so as to give the correct answer. We’ll do this in Example 4.3.8.
Verification that \(\oint_C\vB\cdot\dee{\vr}\ne 0\text{:}\)}
Since
\begin{align*} x'(t) &= -\cos(\cos t)\ \sin t\\ y'(t) &= \cos(\sin t)\ \cos t\\ z'(t) &= 0 \end{align*}
our integral is
\begin{align*} \oint_C\vB\cdot\dee{\vr} &=\oint_C\big[\vB_1\,\dee{x} +\vB_2\,\dee{y}\big]\\ &=\int_0^{2\pi}\big[\vB_1\big(x(t),y(t)\big)\,x'(t) +\vB_2\,\big(x(t),y(t)\big)\,y'(t)\big]\dee{t}\\ &=\int_0^{2\pi} \frac{\sin(\sin t)\,\cos(\cos t)\,\sin t +\sin(\cos t)\,\cos(\sin t)\,\cos t} {\sin^2(\cos t) +\sin^2(\sin t)} \dee{t} \end{align*}
This is a very ugly looking integral
 2 
Indeed!
. But even if we can’t evaluate the integral, we can see that the integrand is strictly positive, and that forces \(\oint_C\vB\cdot\vr \gt 0\text{.}\) Because
\begin{equation*} 0\le|\sin t|,|\cos t|\le 1 \lt \frac{\pi}{2} \end{equation*}
  • \(\cos(\cos t) \gt 0\text{,}\) and \(\sin(\sin t)\) has the same sign as \(\sin t\text{,}\) and \(\sin(\sin t)\) is zero if and only if \(\sin t=0\text{.}\) So the first term in the numerator,
    \begin{equation*} \cos(\cos t)\,\sin(\sin t)\,\sin t\ge 0 \end{equation*}
    and is zero if and only if \(\sin t=0\)
  • \(\cos(\sin t) \gt 0\text{,}\) and \(\sin(\cos t)\) has the same sign as \(\cos t\text{,}\) and \(\sin(\cos t)\) is zero if and only if \(\cos t=0\text{.}\) So the second term in the numerator,
    \begin{equation*} \cos(\sin t)\,\sin(\cos t)\,\cos t\ge 0 \end{equation*}
    and is zero if and only if \(\cos t=0\text{.}\)
  • There is no \(t\) for which both \(\sin t\) and \(\cos t\) are simultaneously zero. So the whole numerator
    \begin{equation*} \sin(\sin t)\,\cos(\cos t)\,\sin t +\sin(\cos t)\,\cos(\sin t)\,\cos t \gt 0 \end{equation*}
    is strictly positive.
Since the integrand is strictly positive, the integral is strictly positive.
Why we got the wrong answer:
In our initial and wrong calculation above, we assumed that \(\frac{\partial B_2}{\partial x}(x,y) - \frac{\partial B_1}{\partial y}(x,y)=0\) at all points \((x,y)\) of the region \(R\) inside \(C\text{.}\) That’s not true. While it is true for most points, it is not true for all points. The vector field \(\vB(x,y)\) is not defined at \((x,y)=(0,0)\text{.}\) So \(\frac{\partial B_2}{\partial x}(x,y) - \frac{\partial B_1}{\partial y}(x,y)\) is also not defined at \((x,y)=(0,0)\text{.}\) That’s enough to invalidate Green’s theorem. Read the statement of Theorem 4.3.2 again carefully.

Example 4.3.8. Example 4.3.7, again.

Evaluate
\begin{equation*} \oint_C\vB\cdot\dee{\vr} \end{equation*}
where
\begin{equation*} \vB = \frac{-y\,\hi+x\,\hj}{x^2+y^2} \end{equation*}
and \(C\) is the curve
\begin{align*} x(t) &= \sin(\cos t)\\ y(t) &= \sin(\sin t)\\ z(t) &= 0 \end{align*}
with \(0\le t\le 2\pi\text{.}\)
Solution.
This is the same integral that we computed incorrectly in Example 4.3.7. We’ll use two ingredients to compute \(\oint_C\vB\cdot\dee{\vr}\) correctly.
  • Let \(a \gt 0\) and denote by \(C_a\) the counterclockwise oriented circle in the \(xy\)-plane that is of radius \(a\) and is centered on the origin. We can explicitly compute \(\oint_{C_a}\vB\cdot\dee{\vr}\text{.}\) To do so just parametrize \(C_a\) by \(x(t)=a\cos t\text{,}\) \(y(t) = a\sin t\text{,}\) \(z(t)=0\text{.}\) Then \(x'(t)=-a\sin t\text{,}\) \(y'(t) = a\cos t\) and
    \begin{align*} \oint_{C_a}\vB\cdot\dee{\vr} &=\int_0^{2\pi}\Big[\frac{-a\sin t\,\hi+a\cos t\,\hj} {a^2\cos^2t+a^2\sin^2t}\Big] \cdot\big[-a\sin t\,\hi+a\cos t\,\hj\big] \dee{t}\\ &=\int_0^{2\pi}\dee{t}=2\pi \end{align*}
  • Pick an \(a\) that is small enough that \(C_a\) lies entirely inside \(C\) and apply Green’s theorem with the region, \(R_a\text{,}\) that is between \(C\) and \(C_a\text{.}\)
    The curve bounding \(R_a\) has two components — \(C\) and \(C_a\text{,}\) but now \(C_a\) is oriented clockwise. (Recall that, in Green’s theorem, when you walk along a boundary curve in the direction of the arrow, \(R_a\) has to be on your left.). Use \(-C_a\) to denote \(C_a\) oriented clockwise. \(\frac{\partial B_2}{\partial x}(x,y) - \frac{\partial B_1}{\partial y}(x,y)\) really is zero at all points \((x,y)\) of the region \(R_a\text{.}\) So Green’s theorem gives
    \begin{align*} 0&= \dblInt_{R_a}\Big(\frac{\partial B_2}{\partial x} - \frac{\partial B_1}{\partial y}\Big)\ \dee{x}\dee{y} = \oint_{C}\vB\cdot\dee{\vr} + \oint_{-C_a}\vB\cdot\dee{\vr}\\ & = \oint_{C}\vB\cdot\dee{\vr} - \oint_{C_a}\vB\cdot\dee{\vr} \end{align*}
    and so
    \begin{gather*} \oint_{C}\vB\cdot\dee{\vr} = \oint_{C_a}\vB\cdot\dee{\vr} =2\pi \end{gather*}

Exercises Exercises

Exercise Group.

Exercises — Stage 1
1.
Let \(R\) be the square
\begin{gather*} R=\Set{(x,y)}{0\le x\le 1,\ 0\le y\le 1} \end{gather*}
and let \(f(x,y)\) have continuous first partial derivatives.
  1. Use the fundamental theorem of calculus to show that
    \begin{gather*} \dblInt_R\frac{\partial f}{\partial y}(x,y)\ \dee{x}\,\dee{y} =\int_0^1 f(x,1)\ \dee{x} - \int_0^1 f(x,0)\ \dee{x} \end{gather*}
  2. Use Green’s theorem to show that
    \begin{gather*} \dblInt_R\frac{\partial f}{\partial y}(x,y)\ \dee{x}\,\dee{y} =\int_0^1 f(x,1)\ \dee{x} - \int_0^1 f(x,0)\ \dee{x} \end{gather*}
2.
Let \(R\) be a finite region in the \(xy\)-plane, whose boundary, \(C\text{,}\) consists of a single, piecewise smooth, simple closed curve that is oriented counterclockwise. “Simple” means that the curve does not intersect itself. Use Green’s theorem to show that
\begin{equation*} \dblInt_R\vnabla\cdot\vF\ \dee{x}\,\dee{y}=\oint_C\vF\cdot\hn\,\dee{s} \end{equation*}
where \(\vF=F_1\,\hi+F_2\,\hj\text{,}\) \(\hn\) is the outward unit normal to \(C\) and \(s\) is the arclength along \(C\text{.}\)
3.
Integrate \(\ds\frac{1}{2\pi}\oint_C \frac{x\,\dee{y}-y\,\dee{x}}{x^2+y^2}\) counterclockwise around
  1. the circle \(x^2+y^2=a^2\)
  2. the boundary of the square with vertices \((-1,-1)\text{,}\) \((-1,1)\text{,}\) \((1,1)\) and \((1,-1)\)
  3. the boundary of the region \(1\le x^2+y^2\le 2,\ y\ge0\)
4.
Show that
\begin{gather*} \pdiff{}{x}\Big( \frac{x}{x^2+y^2}\Big) =\pdiff{}{y}\Big( \frac{-y}{x^2+y^2}\Big) \end{gather*}
for all \((x,y)\ne (0,0)\text{.}\) Discuss the connection between this result and the results of Q[4.3.3].

Exercise Group.

Exercises — Stage 2
5.
Evaluate \(\int_C\vF\cdot d\vr\) where \(\vF = x^2y^2\,\hi + 2xy\,\hj\) and \(C\) is the boundary of the square in the \(xy\)-plane having one vertex at the origin and diagonally opposite vertex at the point \((3, 3)\text{,}\) oriented counterclockwise.
6.
Evaluate \(\ds\oint_C (x\sin y^2 -y^2)\,\dee{x}+(x^2y\cos y^2+3x)\,\dee{y}\) where \(C\) is the counterclockwise boundary of the trapezoid with vertices \((0,-2),\ (1,-1),\ (1,1)\) and \((0,2)\text{.}\)
7. (✳).
Evaluate \(I= \ds\oint_{\cC} \Big(\frac{1}{3}x^2y^3-x^4y\Big)\,\dee{x} +\big(xy^4+x^3y^2\big)\,\dee{y}\) counterclockwise around the boundary of the half-disk \(0\le y\le \sqrt{4-x^2}\text{.}\)
8. (✳).
Let \(\cC\) be the counterclockwise boundary of the rectangle with vertices \((1,0)\text{,}\) \((3,0)\text{,}\) \((3,1)\) and \((1,1)\text{.}\) Evaluate
\begin{equation*} \oint_\cC\big(3y^2+2xe^{y^2}\big)\,\dee{x} + \big(2yx^2 e^{y^2}\big)\,\dee{y} \end{equation*}
9. (✳).
Consider the closed region enclosed by the curves \(y = x^2 + 4x + 4\) and \(y = 4 - x^2\text{.}\) Let \(C\) be its boundary and suppose that \(C\) is oriented counter-clockwise.
  1. Draw the oriented curve \(C\) carefully in the \(xy\)-plane.
  2. Determine the value of
    \begin{equation*} \oint_C xy\, \dee{x} + (e^y + x^2 ) \dee{y} \end{equation*}
10. (✳).
Let
\begin{equation*} \vF(x, y) = \big(y^2 - e^{-y^2} + \sin x\,,\, 2xye^{-y^2} + x\big) \end{equation*}
Let \(C\) be the boundary of the triangle with vertices \((0, 0)\text{,}\) \((1, 0)\) and \((1, 2)\text{,}\) oriented counter-clockwise. Compute
\begin{equation*} \int_C \vF\cdot\dee{\vr} \end{equation*}
11. (✳).
Suppose the curve \(C\) is the boundary of the region enclosed between the curves \(y = x^2 - 4x+3\) and \(y = 3 - x^2 + 2x\text{.}\) Determine the value of the line integral
\begin{equation*} \int_C \big(2xe^y + \sqrt{2 + x^2}\big)\, \dee{x} + x^2 (2 + e^y)\, \dee{y} \end{equation*}
where \(C\) is traversed counter-clockwise.
12. (✳).
Let
\begin{gather*} \vF(x,y) = \big(\tfrac{3}{2}y^2 + e^{-y} +\sin x\big)\,\hi +\big(\tfrac{1}{2}x^2+x-x e^{-y}\big)\,\hj \end{gather*}
Find \(\int_C\vF\cdot\dee{\vr}\text{,}\) where \(C\) is the boundary of the triangle \((0,0)\text{,}\) \((1,-2)\text{,}\) \((1,2)\text{,}\) oriented anticlockwise.
13. (✳).
  1. Use Green’s theorem to evaluate the line integral
    \begin{equation*} \int_C \frac{-y}{x^2+y^2}\dee{x} + \frac{x}{x^2+y^2}\dee{y} \end{equation*}
    where \(C\) is the arc of the parabola \(y = \frac{1}{4}x^2 + 1\) from \((-2, 2)\) to \((2, 2)\text{.}\)
  2. Use Green’s theorem to evaluate the line integral
    \begin{equation*} \int_C \frac{-y}{x^2+y^2}\dee{x} + \frac{x}{x^2+y^2}\dee{y} \end{equation*}
    where \(C\) is the arc of the parabola \(y = x^2 -2\) from \((-2, 2)\) to \((2, 2)\text{.}\)
  3. Is the vector field
    \begin{equation*} \vF=\frac{-y}{x^2+y^2}\hi + \frac{x}{x^2+y^2}\hj \end{equation*}
    conservative? Provide a reason for your answer based on your answers to the previous parts of this question.
14. (✳).
Suppose the curve \(C\) is the boundary of the region enclosed between the curves \(y = x^2 - 4x + 3\) and \(y = 3 - x^2 + 2x\text{.}\) Determine the value of the line integral
\begin{equation*} \int_C \big(2xe^y + \sqrt{2} + x^2\big)\dee{x} + x^2 \big(2 + e^y)\dee{y} \end{equation*}
where \(C\) is traversed counter-clockwise.
15. (✳).
Let \(\vF(x, y) = P \,\hi + Q\,\hj\) be a smooth plane vector field defined for \((x,y) \ne (0, 0)\text{,}\) and suppose \(Q_x = P_y\) for \((x,y) \ne (0, 0)\text{.}\) In the following \(I_j = \int_{C_j} \vF\cdot\dee{\vr}\) for integer \(j\text{,}\) and all \(C_j\) are positively oriented circles. Suppose \(I_1 = \pi\) where \(C_1\) is the circle \(x^2 + y^2 = 1\text{.}\)
  1. Find \(I_2\) for \(C_2 : (x - 2)^2 + y^2 = 1\text{.}\) Explain briefly.
  2. Find \(I_3\) for \(C_3 : (x - 2)^2 + y^2 = 9\text{.}\) Explain briefly.
  3. Find \(I_4\) for \(C_4 : (x - 2)^2 + (y-2)^2 = 9\text{.}\) Explain briefly.
16. (✳).
Consider the vector field \(\vF = P\,\hi + Q\,\hj\text{,}\) where
\begin{equation*} P=\frac{x+y}{x^2+y^2},\qquad Q=\frac{y-x}{x^2+y^2} \end{equation*}
  1. Compute and simplify \(Q_x - P_y\text{.}\)
  2. Compute the integral \(\int_{C_R} \vF \cdot \dee{\vr}\) directly using a parameterization, where \(C_R\) is the circle of radius \(R\text{,}\) centered at the origin, and oriented in the counterclockwise direction.
  3. Is \(\vF\) conservative? Carefully explain how your answer fits with the results you got in the first two parts.
  4. Use Green’s theorem to compute \(\int_C \vF \cdot \dee{\vr}\) where \(C\) is the triangle with vertices \((1, 1)\text{,}\) \((1, 0)\text{,}\) \((0, 1)\) oriented in the counterclockwise direction.
  5. Use Green’s theorem to compute \(\int_C \vF \cdot \dee{\vr}\) where \(C\) is the triangle with vertices \((-1, -1)\text{,}\) \((1, 0)\text{,}\) \((0, 1)\) oriented in the counterclockwise direction.
17. (✳).
  1. Evaluate
    \begin{equation*} \int_C \sqrt{1+x^3}\,\dee{x} +\big(2xy^2 + y^2\big)\,\dee{y} \end{equation*}
    where \(C\) is the unit circle \(x^2+y^2 = 1\text{,}\) oriented counterclockwise.
  2. Evaluate
    \begin{equation*} \int_C \sqrt{1+x^3}\,\dee{x} +\big(2xy^2 + y^2\big)\,\dee{y} \end{equation*}
    where \(C\) is now the part of the unit circle \(x^2+y^2 = 1\text{,}\) with \(x\ge 0\text{,}\) still oriented counterclockwise.

Exercise Group.

Exercises — Stage 3
18. (✳).
Evaluate the line integral
\begin{equation*} \int_C (x^2 + y e^x ) \,\dee{x} + (x \cos y + e^x ) \,\dee{y} \end{equation*}
where \(C\) is the arc of the curve \(x = \cos y\) for \(-\pi/2 \le y \le \pi/2\text{,}\) traversed in the direction of increasing \(y\text{.}\)
19. (✳).
Use Green’s theorem to establish that if \(C\) is a simple closed curve in the plane, then the area \(A\) enclosed by \(C\) is given by
\begin{equation*} A=\frac{1}{2}\oint_C x\,\dee{y}-y\,\dee{x} \end{equation*}
Use this to calculate the area inside the curve \(x^{2/3}+y^{2/3}=1\text{.}\)
20. (✳).
Let \(\vF(x,y)=(x+3y)\,\hi+(x+y)\,\hj\) and \(\vG(x,y)=(x+y)\,\hi+(2x-3y)\,\hj\) be vector fields. Find a number \(A\) such that for each circle \(C\) in the plane
\begin{equation*} \oint_C\vF\cdot \dee{\vr}=A\oint_C\vG\cdot \dee{\vr} \end{equation*}
21. (✳).
Let \(\vF(x,y) = \frac{y^3}{ {(x^2+y^2)}^2}\hi -\frac{xy^2}{ {(x^2+y^2)}^2}\hj\text{,}\) \((x,y)\ne (0,0)\text{.}\)
  1. Compute \(\oint_C\vF\cdot \dee{\vr}\) where \(C\) is the unit circle in the \(xy\)-plane, positively oriented.
  2. Use (a) and Green’s theorem to find \(\oint_{C_0}\vF\cdot \dee{\vr}\) where \(C_0\) is the ellipse \(\frac{x^2}{16}+\frac{y^2}{25}=1\text{,}\) positively oriented.
22. (✳).
Let \(\cC_1\) be the circle \((x-2)^2+y^2=1\) and let \(\cC_2\) be the circle \((x-2)^2+y^2=9\text{.}\) Let \(\vF=-\frac{y}{x^2+y^2}\,\hi+\frac{x}{x^2+y^2}\,\hj\text{.}\) Find the integrals \(\oint_{\cC_1}\vF\cdot \dee{\vr}\) and \(\oint_{\cC_2}\vF\cdot \dee{\vr}\text{.}\)
23. (✳).
Let \(R\) be the region in the first quadrant of the \(xy\)-plane bounded by the coordinate axes and the curve \(y=1-x^2\text{.}\) Let \(\cC\) be the boundary of \(R\text{,}\) oriented counterclockwise.
  1. Evaluate \(\int_\cC x\,\dee{s}\text{.}\)
  2. Evaluate \(\int_\cC \vF\cdot \dee{\vr}\text{,}\) where \(\vF(x,y) =\big(\sin(x^2)-xy\big)\,\hi+\big(x^2+\cos(y^2)\big)\,\hj\text{.}\)
24. (✳).
Let \(C\) be the curve defined by the intersection of the surfaces \(z = x + y\) and \(z = x^2 + y^2\text{.}\)
  1. Show that \(C\) is a simple closed curve.
  2. Evaluate \(\oint_C \vF \cdot \dee{\vr}\) where
    1. \(\vF = x^2\,\hi + y^2\,\hj + 3 e^z\,\hk\text{.}\)
    2. \(\vF = y^2\,\hi + x^2\,\hj + 3 e^z\,\hk\text{.}\)
25.
Find a smooth, simple, closed, counterclockwise oriented curve, \(C\text{,}\) in the \(xy\)-plane for the which the value of the line integral \(\oint_C(y^3-y)\,\dee{x}-2x^3\,dy\) is a maximum among all smooth, simple, closed, counterclockwise oriented curves.
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