
\section{Green's Theorem in Two Dimensions}

\begin{enumerate}[1.]
\item
{\it  
The curve $x=\sin t$, $y=\pi^2-t^2$, $-\pi\leq t\leq\pi$, is simple and closed.
Use Green's Theorem to find the area bounded by this curve.}

\medskip

The orientation of the curve is clockwise (this can be checked in various ways, e.g. 
$y$ is increasing for $-\pi\leq t\leq 0$ when $x\leq 0$, and decreasing for $0\leq t\leq \pi$
when $x\geq 0$).
Using Green's theorem, and taking the orientation into account,
we compute the area:

$$\hbox{Area}=-\int_C xdy=-\int_{-\pi}^\pi \sin t\cdot(-2t)dt
=\int_{-\pi}^\pi 2t\sin t\,dt$$
$$=-2t\cos t\Big|_{-\pi}^\pi +\int_{-\pi}^\pi 2\cos t\,dt
=-2(-2\pi)+ 2\sin t\Big|_{-\pi}^\pi =4\pi.$$


\bigskip

\item {\it
Use Green's Theorem to evaluate the line integral
$\int_C (xy^2+y^2) \ dx + (4xy+x^2y + \cos (y^2)) \ dy,$ where $C$ is the boundary of the region
enclosed by the parabola $y = x^2$ and the line $y = x$, oriented counterclockwise.}

\medskip

Let ${\bf F}$ be the vector field ${\bf F}= (xy^2+y^2)\ii + (4xy+x^2y + \cos (y^2))\jj$. Then
$$
(\nabla\times {\bf F})\cdot \kk = 
\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=(4y+2xy)-(2xy+2y)=2y,$$
so that
$$
\int_C (xy^2+y^2) \ dx + (4xy+x^2y + \cos (y^2)) \ dy=\int_0^1 \int_{x^2}^x 2y dy dx
= \int_0^1 y^2\Big|_{x^2}^x dx
$$
$$
= \int_0^1 (x^2-x^4) dx=\frac{x^3}{3}-\frac{x^5}{5}\Big|_0^1=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}.
$$


\bigskip


\item
{\it Use Green's Theorem to evaluate the integral ${\int_C ydx+xdy}$, where $C$ is the arc of the circle $(x-1)^2+y^2=1$
from $(0,0)$ to $(1,1)$ in the first quadrant of the $xy$-plane.}

\medskip

Since $\displaystyle {\bf curl\, F}\cdot\kk
=\frac{\partial x}{\partial x}- \frac{\partial y}{\partial y}=0$, we have
$\int_C = \int_{C_1}+ \int_{C_2}$, where $C_1$ is the line segment from $(0,0)$ to $(1,0)$
and $C_2$ is the line segment from $(1,0)$ to $(1,1)$. On $C_1$, $y=dy=0$, so $\int_{C_1}=0$.
On $C_2$, $dx=0$ and $x=1$, so that
$$
\int_{C_2}ydx +xdy=\int_0^1 1 dy= y\Big|_{y=0}^1 = 1.
$$
Therefore $\int_{C}ydx +xdy = \int_{C_2}ydx +xdy = 1.$


\medskip

{\bf Optional: } For comparison, here is an alternative solution where we compute the integral directly.
Parametrize the arc by $x=1-\cos\theta$, $y=\sin\theta$, $0\leq\theta\leq \pi/2$. Then
\begin{align*}
{\int_C ydx + xdy} & = \int_0^{\pi/2} (\sin\theta)(\sin\theta) + (1-\cos\theta)(\cos\theta)d\theta
\\
&= \int_0^{\pi/2} (\sin^2\theta +\cos\theta - \cos^2\theta) d\theta 
= \int_0^{\pi/2} (\cos\theta - \cos(2\theta)) d\theta 
\\
&= \sin\theta - \frac{\sin(2\theta)}{2} \Big|_0^{\pi/2} = 1 - 0 =1.
\end{align*}

\bigskip


\item 
{\it Use Green's Theorem to evaluate the integral
$\int_C y^3 dx - x^3 dy$, where $C$ is the circle $x^2+y^2=4$ oriented counterclockwise.}

\medskip


Let $D$ be the disc $x^2+y^2\leq 4$, then
\begin{align*}
\int_C y^3 dx - x^3 dy
&= \iint_D \left( \frac{\partial (-x^3)}{\partial x} - \frac{\partial (y^3)}{\partial y}\right) dxdy
\\
&= \iint_D (-3x^2 -3y^2) dxdy\\
&= \int_0^{2\pi} \int_0^2 -3r^2\cdot r drd\theta
\\
&= 2\pi \cdot \frac{-3r^4}{4}\Big|_0^2 = 2\pi \cdot (-12) = -24\pi.
\end{align*}

\bigskip

\item
{\it Let ${\bf F}=(e^{x^2}+4x^2y)\ii + (e^{-4y^2}-y^2x+4x)\jj$.
Use Green's Theorem to evaluate the integral
$\int_{C}{\bf F}\cdot d{\bf r}$,
where $C$ is the boundary of the rectangle $0\leq x\leq 1$, $0\leq y\leq 3$, oriented counterclockwise.}


\medskip

We have
$$ {\bf curl \ F}\cdot \kk = \frac{\partial}{\partial x} (e^{-4y^2}-y^2x+4x)
-\frac{\partial}{\partial y}(e^{x^2}+4x^2y)
=-y^2+4 -4x^2.
$$
Thus by Green's Theorem, the integral in question is equal to
$$
\int_0^1 \int_0^3 (4-4x^2-y^2) dydx = \int_0^1 (4y-4x^2y-\frac{y^3}{3})\Big|_{y=0}^3\, dx
$$
$$=\int_0^1 (12 -12x^2 - 9)dx = \int_0^1 (3-12 x^2) dx = (3x-4x^3)\Big|_0^1 = 3-4=-1.
$$



\bigskip


\item
{\it Let ${\bf F}=(y^3+4x)\ii+(3x-4x^3)\jj$.
Find the simple, closed, oriented counterclockwise curve $C$ for which the integral 
$\int_{C}{\bf F}\cdot d{\bf r}$ is maximized (in the sense that
$\int_{C}{\bf F}\cdot d{\bf r}\geq \int_{C'}{\bf F}\cdot d{\bf r}$ for any other
simple, closed, oriented counterclockwise curve $C'$). If no such curve exists, explain why.}






\medskip
By Green's theorem, we have 
$$\int_C(y^3+4x)dx+(3x-4x^3)dy=\iint_D\Big(
\frac{\partial(3x-4x^3)}{\partial x}-
\frac{\partial(y^3+4x)}{\partial y}\Big)dxdy$$
$$=\iint_D(3-12x^2-3y^2)dxdy,$$
where $D$ is the region bounded by $C$.  The function
$3-12x^2-3y^2=3(1-4x^2-y^2)$ is positive inside the ellipse
$4x^2+y^2=1$ and negative outside it.  Hence to maximize the last integral 
we should take $C$ to be the ellipse 
$4x^2+y^2=1$.




\bigskip

\item
{\it Let
\begin{equation}\label{42-eqHW2026}
g(x)=\begin{cases}
0 & \hbox{ if }x<0\hbox{ or } x>1,
\\
x^3\sin(\frac{\pi}{x}) & \hbox{ if } 0<x\leq 1.
\end{cases}
\end{equation}
Prove that the graph of $y=g(x)$, $x\in\rr$, is a smooth curve, with smooth parametrization 
$\bfr(x) =x\ii+g(x)\jj$. (This means that $\bfr$ is $C^1$ and $\bfr'$ is never zero. The main part is to prove that $\bfr'(x)$ exists, is continuous, and is nonzero at $x=0$.) 
Prove further that the graph of $y=g(x)$ for $-1\leq x\leq 1$ has finite length. }




\bigskip

We parametrize the graph by $\bfr(x) =x\ii+g(x)\jj$ as suggested. We first note that this graph is a continuous curve since
$$
|g(x)|\leq | x^3|\to 0=g(0) \hbox{ as }x\to 0.
$$
Next, $\bfr'(x)=\ii + g'(x)\jj$. This is always nonzero as long as $g'(x)$ exists, but we do need to prove existence and continuity. To prove that $g'(0)$ exists, we need to examine the limit
$$
\lim_{x\to 0} \frac{g(x)-0}{x}.
$$
But since $|g(x)|\leq |x^3|$, we have $|g(x)/x|\leq x^2\to 0$ as $x\to 0$. Thus $g'(0)=0$. 

We now prove that $g'(x)$ is continuous. For $x<0$, we have $g'(x)=0$, so that $g'$ is continuous for $x<0$ and continuous at $0$ from the left. 
For $x>0$,
$$
g'(x)=3x^2\sin(\frac{\pi}{x})+ x^3 \sin(\frac{\pi}{x}) \cdot (-\frac{\pi}{x^2})
= (3x^2-\pi x) \sin(\frac{\pi}{x}).
$$
This is clearly continuous at all $x>0$, and 
\begin{equation}\label{a-e1}
|g'(x)|\leq 3x^2+\pi x\leq 7x \hbox{ for }0<x<1.
\end{equation}
Since $7x\to 0$ as $x\to 0$, $g'(x)$ is also continuous from the right at $x=0$. Finally, the last inequality also gets us an estimate on the length of the graph for $-1\leq x\leq 1$:
\begin{align*}
\hbox{length} & = \int_{-1}^1 \sqrt{1+|g'(x)|^2}dx =1+ \int_{0}^1 \sqrt{1+|g'(x)|^2}dx
\\
&\leq 1+\int_0^1 \sqrt{1+49} \, dx \leq  1+\int_0^1 8 \, dx =9.
\end{align*}
We could get a better estimate, but here we just need to know that the length is finite.


\medskip


\item {  \it Let $g(x)$ be the function in (\ref{42-eqHW2026}). Let $C$ be the closed curve, oriented counterclockwise, consisting of:
\begin{itemize}
\item the graph of $y=g(x)$ for $-1\leq x\leq 1$, oriented in the direction of decreasing $x$,
\item the line segment from $(-1,0)$ to $(-1,-2)$,
\item the line segment from $(-1,-2)$ to $(1,-2)$,
\item the line segment from $(1,-2)$ to $(1,0)$.
\end{itemize}
Let also ${\bf F}$ be a $C^1$ vector field on $\rr^2$, and let $D$ be the planar region enclosed by $C$. (This region cannot be split into finitely many regions that are both $x$-simple and $y$-simple.) Prove that Green's Theorem remains valid in this setting, as follows.


\medskip\noindent
 (a) For $n=1,2,\dots$, let $C_n$ be the closed curve, oriented counterclockwise, consisting of:
\begin{itemize}
\item the graph of $y=g(x)$ for $n^{-1}\leq x\leq 1$, oriented in the direction of decreasing $x$,
\item  the line segment from $(n^{-1},0)$ to $(-1,0)$,
\item the line segment from $(-1,0)$ to $(-1,-2)$,
\item the line segment from $(-1,-2)$ to $(1,-2)$,
\item the line segment from $(1,-2)$ to $(1,0)$.
\end{itemize}
Prove that $\oint_{C_n} {\bf F}\cdot d\bfr \to \oint_C {\bf F}\cdot d\bfr$ as $n\to\infty$.}


\bigskip

After cancelling out the parts where $C$ and $C_n$ overlap, and reversing the orientation, the difference $\oint_{C_n} {\bf F}\cdot d\bfr - \oint_C {\bf F}\cdot d\bfr$ is equal to 
\begin{equation}\label{b-e1}
\int_{C_{1,n}} {\bf F}\cdot d\bfr - \oint_{C_{2,n}} {\bf F}\cdot d\bfr,
\end{equation}
where:
\begin{itemize}
\item $C_{1,n}$ is the graph of $y=g(x)$ for $0\leq x\leq n^{-1}$, 
\item  $C_{2,n}$ is the line segment from $(0,0)$ to $(n^{-1},0)$,
\end{itemize}
both oriented in the direction of increasing $x$. (If you prefer to keep the original orientation with decreasing $x$, you need to switch the signs above.) We need to prove that (\ref{b-e1}) goes to $0$ as $n\to\infty$. We will actually prove that this is true for each integral in (\ref{b-e1}) separately. 

Since ${\bf F}$ is $C^1$ everywhere, it is bounded on any bounded and closed rectangle $R$. Taking $R$ to be any rectangle containing $C$ and $C_n$, say $R=\{(x,y):|x|\leq 10,|y|\leq 10\}$, we see that there is a constant $M>0$ such that $|F_1|\leq M$ and $|F_2|\leq M$ on $C$ and $C_n$. Therefore
\begin{align*}
\left| \int_{C_{1,n}} {\bf F}\cdot d\bfr \right| & =\left| \int_0^{1/n} (F_1+F_2g'(x))dx\right|
\leq \int_0^{1/n} (|F_1|+|F_2||g'(x)|)dx
\\
&\leq \int_0^{1/n} 8M\, dx = 8M/n,
\end{align*}
which goes to 0 as $n\to\infty$. We used (\ref{a-e1}) to estimate $|g'|\leq 7$. The second integral is easier:
\begin{align*}
\left| \int_{C_{2,n}} {\bf F}\cdot d\bfr \right| & =\left| \int_0^{1/n} F_1\,dx\right|
\leq \int_0^{1/n} |F_1|\, dx
\\
&\leq \int_0^{1/n} M\, dx = M/n,
\end{align*}
which also goes to 0 as $n\to\infty$.




\medskip\noindent
{\it (b)  For $n=1,2,\dots$, let $D_n$ be the planar region enclosed by $C_n$. Prove that }
$$
\iint_{D_n}  (\nabla\times {\bf F})\cdot\kk \, dA \to \iint_{D}  (\nabla\times {\bf F})\cdot\kk \, dA \hbox{ as }n\to\infty.
$$


\bigskip

We have
\begin{align*}
& \iint_{D_n}  (\nabla\times {\bf F})\cdot\kk \, dA -  \iint_{D}  (\nabla\times {\bf F})\cdot\kk \, dA
\\
&\ \ \ = \iint_{D_n\setminus D}  (\nabla\times {\bf F})\cdot\kk \, dA - \iint_{D\setminus D_n}  (\nabla\times {\bf F})\cdot\kk \, dA.
\end{align*}
We will prove that the last two integrals go to 0 as $n\to \infty$. Since both $D_n\setminus D$ and $D\setminus D_n$ are subsets of the set 
$$
D'_n:= \left\{(x,y):\ 0\leq x\leq \frac{1}{n},\ |y|\leq 2\right\},
$$
we have
$$
\left| \iint_{D_n\setminus D}  (\nabla\times {\bf F})\cdot\kk \, dA\right|
\leq \iint_{D_n\setminus D} | (\nabla\times {\bf F})\cdot\kk | \, dA
\leq \iint_{D'_n}  |(\nabla\times {\bf F})\cdot\kk |\, dA
$$
and similarly for $D\setminus D_n$. So, it suffices to prove that the last integral above goes to 0 as $n\to\infty$.
Since ${\bf F}$ is $C^1$ everywhere, its partial derivatives are bounded on any bounded and closed rectangle $R$. Taking $R$ to be any rectangle containing $C$ and $C_n$, say $R=\{(x,y):|x|\leq 10,|y|\leq 10\}$, we see that there is a constant $M'>0$ such that 
$$
\left| \frac{\partial F_1}{\partial y}\right|\leq M',\ \ \left| \frac{\partial F_2}{\partial x}\right|\leq M'\ \ \hbox{ on }R.
$$
Since $D'_n\subset R$ for all $n$, we have
$$
\iint_{D'_n}  |(\nabla\times {\bf F})\cdot\kk |\, dA \leq \iint_{D'_n}  2M' \, dA
\leq 2M'\cdot 4n^{-1} =8M'n^{-1},
$$
which goes to 0 as claimed.


\medskip\noindent
{\it (c) Assuming that Green's Theorem holds for each $C_n$ and $D_n$, conclude that it also holds for $C$ and $D$.}

\bigskip


Applying Green's Theorem to each $C_n$ and $D_n$, we get
$$
\oint_{C_n} {\bf F}\cdot d\bfr= \iint_{D_n}  (\nabla\times {\bf F})\cdot\kk \, dA .
$$
Taking the limit as $n\to\infty$ on both sides, and using (a) and (b), we get Green's Theorem for $C$ and $D$, as claimed.

















\end{enumerate}

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