
\section{Green's Theorem in Two Dimensions}\label{sec:lecture42}
In this section, we focus on \textbf{Green's Theorem}, which relates circulation integrals on a closed curve to curl integrals on the region bounded by said curve.  

\begin{theorem}
Let $\mathcal{C}$ be a simple, closed, piecewise smooth curve in $\mathbb{R}^2$, oriented counterclockwise, and let $R$ denote the region bounded by $\mathcal{C}$.  If $\mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j}$ is $C^1$ on some open set containing $R$, then
\begin{equation}\label{42eq:greenstheorem1}
\oint_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r} = \iint_{R} \curl \mathbf{F} \cdot \mathbf{k} \, d A.
\end{equation}
In terms of the component functions $F_1$ and $F_2$, this means
\begin{equation}\label{42eq:greenstheorem2}
\oint_{\mathcal{C}} F_1 dx + F_2 dy = \iint_{R} \bigg( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \bigg) dx dy.
\end{equation}
\end{theorem}

We postpone the proof for now, focusing first on the geometric interpretation and applications of the formula above.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .4\linewidth]{figures/42fig1.pdf}
\caption{An example of a curve $\mathcal{C}$ and region $R$ satisfying the first part of the hypothesis of Green's Theorem.}
\end{minipage}
\end{figure}


\subsection{Green's Theorem: Some Special Cases} Let us take some time to consider the following situations where Green's Theorem may be applied.

\begin{itemize}
\item Let's first show that Green's Theorem is consistent with the ``local version'' from the last section.  Suppose that $\mathcal{C} = \mathcal{C}_{\epsilon}$ is a circle of radius $\epsilon > 0$ centred at some point $P \in \mathbb{R}^2$, and let $R_{\epsilon}$ denote its interior.  Applying formula \eqref{42eq:greenstheorem1} to a vector field $\mathbf{F}$ on $R_{\epsilon}$, we have 
$$
\oint_{\mathcal{C}_{\epsilon}} \mathbf{F} \cdot d \mathbf{r} = \iint_{R_{\epsilon}} \curl \mathbf{F} \cdot \mathbf{k} \, d A \approx \big(\textrm{ Area of } R_{\epsilon} \big) \cdot \big(\curl \mathbf{F} (P) \cdot \mathbf{k} \big),
$$
and the approximation should work well so long as $\epsilon > 0$ is assumed to be small enough (so that $\curl \mathbf{F} $ is close to constant on $R_\epsilon$).




\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .3\linewidth]{figures/42fig2.png}
\caption{The circle $\mathcal{C}_{\epsilon}$ about $P$. }
\end{minipage}
\end{figure}
\medskip

\item If $\mathbf{F}$ is \textit{irrotational} on the region $R$ bounded by $\mathcal{C}$ (so that $\nabla \times \mathbf{F} = \mathbf{0}$ at every point in $R$), then formula \eqref{42eq:greenstheorem1} shows that $\oint_{\mathcal{C}} \mathbf{F}  \cdot d \mathbf{r} = 0$. In particular, if $\mathbf{F}$ is irrotational \textit{everywhere} in $\mathbb{R}^2$, then $\oint_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r} = 0$ for all $\mathcal{C} \subset \mathbb{R}^2$ which satisfy the assumption of Green's Theorem. (This is true, for example, whenever $\mathbf{F}$ is conservative!)

\medskip

\item A domain $D \subset \mathbb{R}^2$ is \textbf{simply connected} if every simple, closed curve in $D$ encloses a region \textit{contained in} $D$. 

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .50\linewidth]{figures/42fig3.pdf}
\end{minipage}
\end{figure}

If $\nabla \times \mathbf{F} = \mathbf{0}$ everywhere on a \textit{simply connected} domain $D$, then $\oint_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r} = 0$ for every simple, closed, piecewise smooth curve $\mathcal{C}$ contained in $D$. With an additional argument, we can also drop the assumption that $\mathcal{C}$ is simple. (How would you go about proving this?) Hence, we have the following:
\begin{center}
\fbox{
\begin{minipage}{.9\textwidth}
$$
\nabla \times \mathbf{F} = \mathbf{0} \textrm{ on } D \textrm{ and } D \textrm{ is simply connected } \Rightarrow \mathbf{F} \textrm{ is conservative on } D.
$$
\end{minipage}
}
\end{center}

\smallskip

\end{itemize}

The last implication can fail if $D$ is not simply connected.  The following example illustrates this failure for a specific function $\mathbf{F}$ and region $D$.

\begin{example}\label{42example1}
Let $\mathbf{F} (x,y) = - \frac{y}{x^2 + y^2} \mathbf{i} + \frac{x}{x^2 + y^2} \mathbf{j}$ on $D = \mathbb{R}^2 \setminus \{\mathbf{0}\}$. From Example \ref{example:puncturedPlaneCurl}, we know that $\nabla \times \mathbf{F} = 0$ everywhere on its domain $D$. And yet, 
if $\mathcal{C}_1$ is a circle around the origin, then
$$
\oint_{\mathcal{C}_1} \mathbf{F} \cdot d \mathbf{r} = 2 \pi\neq 0!
$$
(Exercise: parametrize the circle by $\mathbf{r}(t) = a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j}$ with $0 \leq t \leq 2 \pi$, and complete the calculation.)
This can only happen because the domain $D = \mathbb{R}^2 \setminus \{\mathbf{0} \}$ is not simply connected. 
\end{example}


In the last example, if we had instead taken a curve $\mathcal{C}_2 \subset \mathbb{R}^2 \setminus \{\mathbf{0}\}$ satisfying the hypothesis of Green's Theorem and bounding a region that does not contain the origin, then we would indeed have that $\oint_{\mathcal{C}_2} \mathbf{F} \cdot d \mathbf{r} = 0$. This is because $\mathbf{F}$ is \textit{well defined} and \textit{irrotational} for \textit{every point} of the region bounded by $\mathcal{C}_2$---a key difference from $\mathcal{C}_1$. 


\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig4.png}
\caption{The three curves $\mathcal{C}_1,\mathcal{C}_2$ and $\mathcal{C}_3$ discussed in Example \ref{42example1}}
\end{minipage}
\end{figure}


On the other hand, let $\mathcal{C}_3$ be \textbf{any} simple, closed, piecewise smooth curve that circles the origin once counterclockwise. We can show, with an extension of Green's Theorem to be discussed shortly, that 
$$
\oint_{\mathcal{C}_3} \mathbf{F} \cdot d \mathbf{r} = \oint_{\mathcal{C}_1} \mathbf{F} \cdot d \mathbf{r} = 2\pi.
$$
A useful way to write this is as follows. Define the ``delta function'' $\delta_{\mathbf{0}}$ in the plane by saying that
$$
\iint_{R} \delta_{\mathbf{0}} \, d A =
\begin{cases}
1, \textrm{ if } \mathbf{0} \in R, \\
0, \textrm{ if } \mathbf{0} \not \in R.
\end{cases}
$$
(The ``delta function" is not actually a function, and has to be defined rigorously as either a distribution or a measure. We will nonetheless use this terminology, and defer the rigorous definition to more advanced courses such as Math 420.) With that notation, the vector field $\mathbf{F}$ defined above satisfies
$$
\curl \mathbf{F} \cdot \mathbf{k} = 2 \pi \delta_{\mathbf{0}},
$$
since
$$
\iint_{R} \curl \mathbf{F} \cdot \mathbf{k} \, d A  = \oint_{\mathcal{\partial R}} \mathbf{F} \cdot d \mathbf{r} =
\begin{cases}
2 \pi, \textrm{ if } \mathbf{0} \in R, \\
0, \textrm{ if } \mathbf{0} \not\in R
\end{cases}
$$
and $\partial R = \mathcal{C}$ denotes the boundary curve of $R$.



\subsection{Using Green's Theorem to simplify integrals}
In some cases, Green's Theorem allows us to convert a complicated line integral to an area integral that is much easier to evaluate. For instance, let us evaluate the integral
$$
\int_\mathcal{C} \left( \sqrt{x^2-x+16}+xy\right) dx + (2x+x^2+\sin(y^3)) dy,
$$
where $\mathcal{C}$ is the circle $x^2+(y-2)^2=4$, oriented counterclockwise. 
This would be unpleasant to integrate directly, but Green's Theorem can help. 
Let $D$ be the disc $x^2+(y-2)^2\leq 4$. We write
\begin{align*}
&\int_\mathcal{C} \left( \sqrt{x^2-x+16}+xy\right) dx + (2x+x^2+\sin (y^3)) dy\\
&\ \ \ = \iint_D \left( \frac{\partial}{\partial x}(2x+x^2+\sin (y^3)) 
-\frac{\partial}{\partial y}\left( \sqrt{x^2-x+16}+xy\right)\right) dA
\\
&\ \ \ = \iint_D (2 +2x -x) dA
\\
&\ \ \ =2(\hbox{ area of }D)=2\cdot 4\pi = 8\pi,
\end{align*}
where we used that $\iint_D x\,dA=0$ by symmetry.




\subsection{Using Green's Theorem to compute areas}  
Simplifications in the other direction (where the line integral is simpler than the area integral) are less common, but here is one important case.
Suppose that $\mathbf{F}$ is some vector field satisfying $\nabla \times \mathbf{F} = \mathbf{k}$ (so that $\curl \mathbf{F} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{k} = 1$). 
If you prefer a concrete example, you can use $\mathbf{F} (x,y) = x \mathbf{j}$ or $\mathbf{F}(x,y) = - y \mathbf{i}$.  
Then, with $R$ and $\mathcal{C}$ as in Green's Theorem,
\begin{equation}\label{42eq:greensareaformula}
\big(\textrm{Area of } R\big) = \iint_{R} 1 \, dA = \oint_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r} = \oint_{\mathcal{C}} x d y = - \oint_{\mathcal{C}}  y dx.
\end{equation}
This means that we can use the above circulation integrals to compute the area of $R$!
We will show some specific examples of this idea in the Worked Problems below. 


\subsection{An Extension of Green's Theorem} 
There is a version of Green's Theorem for more complicated regions bounded by several curves, as follows.  

\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig8.pdf}
\end{minipage}
\end{figure}



Let $R \subset \mathbb{R}^2$ be bounded by \textit{several} simple, closed, piecewise smooth curves $\mathcal{C}_1,...,\mathcal{C}_k$, \textit{oriented positively with respect to }$R$. If $\mathcal{C}$ is the \textit{total boundary} of $R$ (that is, all of the curves $\mathcal{C}_i$ considered together) then we still have
$$
\oint_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r} = \iint_{R} \curl \mathbf{F} \cdot \mathbf{k} \, d A.
$$






\begin{proof}[A Sketch of the Proof of the Extension]
Essentially, we want to connect each of the component pieces $\mathcal{C}_i$ of the boundary $\mathcal{C}$ of $R$ by adding curves (such as the line segments in the figure above) to 	``convert'' $\mathcal{C}$ into a single curve $\mathcal{C}'$, before applying our standard Green's Theorem to $\mathcal{C}'$.  One can check that each of these added-in line segments are traversed twice, once in each direction.  Hence, the total contribution to the line integral over the added segments is zero! Also, if there are any curves ``cutting out holes" in $D$, then they are counted in $\mathcal{C}'$ with the same orientation as in $\mathcal{C}$. This proves the extension.
\end{proof}

As a special case, let us return to the discussion after Example \ref{42example1}. 
\smallskip

\begin{proposition}
Suppose that $\mathbf{F}$ is smooth and has curl equal to $\mathbf{0}$ on $D = \mathbb{R}^2 \setminus \{\mathbf{0}\}$.  Then, for any two curves $\mathcal{C}_1$ and $\mathcal{C}_2$ encircling the origin which satisfy the hypothesis of Green's Theorem, one has
$$
\oint_{\mathcal{C}_1} \mathbf{F} \cdot d \mathbf{r} = \oint_{\mathcal{C}_2} \mathbf{F} \cdot d \mathbf{r}.
$$
\end{proposition}

\begin{proof}
This follows from a similar idea as the proof of the extension of Green's Theorem.  It suffices to prove this when $\mathcal{C}_1$ is a small circle contained entirely in the region bounded by $\mathcal{C}_2$ (why?). We join the curves $\mathcal{C}_1$ and $\mathcal{C}_2$ together as in the diagram below.
\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig9.pdf}
\end{minipage}
\end{figure}



We then see that $\mathcal{C}_2$ is traversed in the positive direction along $\mathcal{C}$, whereas $\mathcal{C}_1$ is traversed in the negative (i.e. opposite) direction as in the extended version of Green's Theorem. Hence, we have
$$
\oint_{\mathcal{C}_2} \mathbf{F} \cdot d \mathbf{r} - \oint_{\mathcal{C}_1} \mathbf{F} \cdot d \mathbf{r} = \oint_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r} = \iint_{R} \big( \nabla \times \mathbf{F} \big) \cdot \mathbf{k} \, d A = 0,
$$
where at the end we used that $\mathbf{F}$ has zero curl on all of $\mathbb{R}^2 \setminus \{\mathbf{0}\}$.
\end{proof}



\subsection{Sketching a Proof of Green's Theorem} We conclude this section by giving some insight into the proof of Green's Theorem.  The argument in Step 1 will be given in full. For Step 2, however, we will only sketch the argument without providing all details. 


\medskip
\noindent
\underline{\textit{Step 1}:} Suppose that $R \subset \mathbb{R}^2$ is both $x$-simple and $y$-simple. This means that 
$$
R := \{(x,y) : \ a \leq x \leq b, \  f(x) \leq y \leq g(x) \},
$$
for some real numbers $a < b$ and continuous functions $f, g : [a,b] \rightarrow \mathbb{R}$, and that $R$ also has a similar representation with the roles of $x$ and $y$ reversed.  

\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig10.pdf}
\end{minipage}
\end{figure}

This means that $\mathcal{C}$, the boundary of $R$, consists of four curves $\mathcal{C}_i$ as indicated (note that $\mathcal{C}_2$ and $\mathcal{C}_4$ could be just single points).  To prove Green's Theorem for the region $R$, then, we must show that
\begin{equation}\label{42eq:greensforproof1}
\oint_{\mathcal{C}} F_1 dx + F_2 dy = \iint_{R} \bigg(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \bigg) dx d y.
\end{equation}
We will prove that
\begin{equation}\label{42eq:greensforproof2}
\oint_{\mathcal{C}} F_1 dx = \iint_{R} - \frac{\partial F_1}{\partial y} dx dy;
\end{equation}
the calculation for the $F_2$ parts is similar but with the  $x$ and $y$ variables interchanged.

\smallskip

We first remark that, since $\mathcal{C} = \mathcal{C}_1 + \mathcal{C}_2 + \mathcal{C}_3 + \mathcal{C}_4$,  we can break the integral on the left-hand side of \eqref{42eq:greensforproof2} into the sum of component integrals
$$
\int_{\mathcal{C}_i} F_1 dx,  \textrm{ for } i=1,2,3,4.
$$
and evaluate each piece one at a time.  First, the easy pieces, which are
$$
\int_{\mathcal{C}_2} F_1 dx = \int_{\mathcal{C}_4} F_1 dx = 0,
$$
which follows since $dx = 0$ uniformly on the line segments $\mathcal{C}_2$ and $\mathcal{C}_4$.  Next, for the curves $\mathcal{C}_1$ and $\mathcal{C}_3$, we use the parametrizations $\mathbf{r}_1 (t) = (t, f(t))$ and $\mathbf{r}_3 (x) = (t, g(t))$ (respectively) to write
$$
\int_{\mathcal{C}_1} F_1 dx = \int_a^b F_1 (t, f(t)) dt \quad \textrm{and} \quad \int_{\mathcal{C}_3} F_1 dx = - \int_a^b F_1 (t, g(t)) dt,
$$
with the minus sign on $\mathcal{C}_3$ to match its orientation---that is, we start at $(b,g(b))$ and move to $(a,g(a))$.  Adding up these integrals, we see that
$$
\oint_{\mathcal{C}} F_1 dx =- \int_a^b \Big[F_1 (t,g(t)) - F_1 (t,f(t))\Big] dt
$$
Now---and \underline{this is the key step}---we apply the Fundamental Theorem of Calculus in the $y$-variable, to rewrite the difference as the integral of the derivative:
$$
F_1 (t,g(t)) - F_1 (t,f(t)) =  \int_{f(t)}^{g(t)} \frac{\partial F_1(t,y)}{\partial y} dy,
$$
so that
$$
\oint_{\mathcal{C}} F_1 dx = -\int_a^b \int_{f(t)}^{g(t)} \frac{\partial F_1(t,y)}{\partial y} dy dt
= - \iint_{R} \frac{\partial F_1}{\partial y} \, dA,
$$
which matches \eqref{42eq:greensforproof2}.
\medskip

\noindent
\underline{\textit{Step 2}:} In general, our region $R$ need not be either $x$-simple or $y$-simple.  In most of the specific cases we will see in this course, we can write $R$ as a union of finitely many pieces $R_1,...,R_m \subset \mathbb{R}^2$, pairwise disjoint except for their boundary curves, such that $R_1,...,R_m$ are all both $x$-simple and $y$-simple. One such piece is shown in the illustration below. The goal is to cover $R$ by such regions so that the 2-dimensional curl integrals add up and the line integrals along the added boundary curves cancel out.

\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig11.pdf}
\end{minipage}
\end{figure}

However, there are regions $R$ that satisfy the assumptions of Green's Theorem but cannot be written as a union of finitely many pieces as above. In those cases, there are two ways we can proceed. One is to approximate $R$ by regions $R'$ that are finite unions of $x$-simple and $y$-simple pieces as above, so that the boundary curve of $R$ is also approximated by the boundary curve of $R'$. We then apply the theorem to $R'$, and take the limit on both sides as $R'\to R$. (See Practice Problems for an example.) Another method involves using a change of variables to transform $R$, or parts of it, to regions that are $x$-simple and $y$-simple. This part of the proof is omitted here.

\subsection{A First Glance at the Divergence Theorem} We are, in fact, within reach of our next major theorem.  The Divergence Theorem works in both 2 and 3 dimensions; the 3-dimensional version will be covered shortly, but the 2-dimensional version can be stated now. 

\begin{theorem}[Two-Dimensional Divergence Theorem]
Let $\mathcal{C}$ be a closed, simple, counterclockwise oriented curve which bounds a region $R \subset \mathbb{R}^2$, and let $\mathbf{N}_0$ denote the unit normal vector to $\mathcal{C}$, oriented \underline{outward} (so, not necessarily the principal normal vector, which could point inward).  

\smallskip

\noindent
Further, assume that $\mathcal{C}$, $R$, and a vector field $\mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j}$ satisfy the hypotheses of  Green's Theorem.  Then
\begin{equation}\label{42eqDiv}
\iint_{R} \divg \mathbf{F} \, d A = \oint_{\mathcal{C}} \mathbf{F} \cdot \mathbf{N}_0 \, ds = \oint_{\mathcal{C}} F_1 dy - F_2 dx.
\end{equation}
\end{theorem}



\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig12.pdf}
\end{minipage}
\end{figure}


The Divergence Theorem is the first equality in (\ref{42eqDiv}). Before giving the proof of it, we prove the second equality, which interprets the last line integral at the end of  (\ref{42eqDiv}) as a 2-dimensional version of a flux integral. 

Let us first examine what it means for $\mathbf{N}_0$ to point outward from $\mathcal{C}$.  Suppose that $\mathcal{C}$ is parametrized counterclockwise as
$$
\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j}, \quad a \leq t \leq b,
$$
then $\mathbf{r}' (t) = x' (t) \mathbf{i} + y' (t) \mathbf{j}$.  We claim that the unit normal vector $\mathbf{N}_0$ satisfies
$$
\mathbf{N}_0 (t) = \frac{1}{\vert \mathbf{r}' (t) \vert} \big(y' (t) \mathbf{i} - x' (t) \mathbf{j} \big).
$$
Indeed, we have $\vert \mathbf{N}_0 \vert = 1$ and $\mathbf{N}_0 \cdot \mathbf{r}' = 0$.  It remains to check that $\mathbf{N}_0$ (and not $-\mathbf{N}_0$) points outward: we determine this by considering sample points on the curve. For example, consider a point $P=\bfr(t_0)$ where $\mathbf{r}'$ points in the positive $y$-direction, so that $\mathbf{r}' =y'\jj$ with $y'>0$. Counterclockwise orientation means that $D$ should lie to the left of $P$, so that ${\bf N}_0$ should point away from it in the direction of $\ii$. This is consistent with the above formula.



With the aforementioned parametrization, we have
\begin{eqnarray*}
\oint_{\mathcal{C}} \mathbf{F} \cdot \mathbf{N}_0 \, ds & = & \int_a^b \frac{1}{\vert \mathbf{r}' (t) \vert} \big(F_1 y' (t) + F_2 (-x' (t)) \big)  \underbrace{\vert \mathbf{r}' (t) \vert}_{ds} dt \\
\quad & = & \int_a^b \big[F_1 y' (t) - F_2 x' (t) \big] dt \\[1ex]
\quad & = & \oint_{\mathcal{C}} F_1 dy - F_2 dx,
\end{eqnarray*}
proving the second equality in (\ref{42eqDiv}).

\begin{proof}[Proof of the Divergence Theorem]
We apply Green's Theorem to the vector field $\mathbf{G} = -F_2 \mathbf{i} + F_1 \mathbf{j}$. First, note that
$$
\big(\curl \mathbf{G} \big) \cdot \mathbf{k} = \frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} = \divg \mathbf{F},
$$
and
$$
 \oint_{\mathcal{C}} \mathbf{G} \cdot d \mathbf{r}
=\oint_{\mathcal{C}} G_1 dx + G_2 dy = =\oint_{\mathcal{C}} F_1dy - F_2 dx.
$$ 

 Hence, Green's Theorem applied to ${\bf G}$ tells us that
$$
\iint_{D} \divg \mathbf{F} \, dA = \iint_{D} \big(\curl \mathbf{G} \big) \cdot \mathbf{k} \, d A = \oint_{\mathcal{C}} \mathbf{G} \cdot d \mathbf{r} = \oint_{\mathcal{C}} \mathbf{F} \cdot \mathbf{N}_0 \, ds.
$$
\end{proof}

We have proven the Divergence Theorem in two dimensions. We will soon show that a similar result continues to hold in three dimensions.  This, for now, concludes our discussion of Green's Theorem.

\section*{Worked Problems}

\begin{enumerate}[1.]
\item
{\it Use Green's Theorem to calculate the area under the cycloid
$$
x(t) = a(t - \sin t), \, y(t) = a (1 - \cos t) \quad \quad 0 \leq t \leq 2 \pi,
$$
and above the $x$ axis.
 }
 
\smallskip
\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig6.png}
\end{minipage}
\end{figure}
\smallskip

Let $\mathcal{C}_1$ be the cycloid and $\mathcal{C}_2$ be the portion of the $x$-axis below the cycloid.  The direction of \textit{increasing} $t$-values corresponds to \textit{clockwise} motion along $\mathcal{C}_1$. In order to apply Green's Theorem, we need to traverse $\mathcal{C}_1 + \mathcal{C}_2$ in the counterclockwise direction.  Let $\mathcal{C}$, then, denote $\mathcal{C}_1 + \mathcal{C}_2$, but traversed in the counterclockwise (i.e. backward) direction.
Using the area formula \eqref{42eq:greensareaformula} gives:
\begin{eqnarray*}
\textrm{Area} & = & - \oint_{\mathcal{C}} y dx = \int_{\mathcal{C}_1} y dx + \int_{\mathcal{C}_2} y dx  \\
\quad & = & \int_0^{2 \pi} \underbrace{a (1 - \cos t)}_{y(t)} \cdot \underbrace{a(1 - \cos t) dt}_{dx} \\
\quad & = &  \int_0^{2 \pi} a^2 (1 - 2 \cos t + \cos^2 t) dt \\
\quad & = & \int_0^{2\pi} a^2 \left(1 - 2 \cos t + \frac{\cos 2t + 1}{2}\right) dt \\
\quad & = & a^2 \bigg[t - 2 \sin t + \frac{\sin 2t}{4} + \frac{t}{2}\bigg]_{t = 0}^{2 \pi} \\
\quad & = & a^2 (2 \pi - 0 + \pi) \\
\quad & = & 3 \pi a^2.
\end{eqnarray*}
On the second line, we used that $ \int_{\mathcal{C}_2} y dx=0$ because $y=0$ on $C_2$.


\medskip

\item{\it (Using Green's Theorem to simplify integrals.) Evaluate the integral
$$
\oint_{\mathcal{C}} \big( y + e^{\sqrt{x}}) \, dx + (2x + \cos (y^2)) \, dy,
$$
if $\mathcal{C}$ is the union of the parabolas $y = x^2$ and $x = y^2$ (for $0 \leq x,y \leq 1$) oriented counter-clockwise.
 }
 
\medskip
We can utilize Green's Theorem to convert the above line integral into a (much simpler) area integral on the region shown below

\smallskip
\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/42fig7.png}
\end{minipage}
\end{figure}
\smallskip

First, we calculate that
$$
\big( \nabla \times \mathbf{F} \big) \cdot \mathbf{k} = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 2-1 = 1.
$$
Then, by Green's Theorem,
\begin{align*}
 \oint_{\mathcal{C}} \big( y + e^{\sqrt{x}}) \, dx + (2x + \cos y^2) \, dy &= \int_{R} 1 \, d A \\
& = \int_0^1 \int_{x^2}^{\sqrt{x}} dy dx = \int_0^1 (x^{1/2} - x^{2}) dx \\
& = \left[\frac{x^{3/2}}{3/2} - \frac{x^3}{3}\right]_{x = 0}^{1}  \\
&  = \frac{2}{3} - \frac{1}{3}=\frac{1}{3}.
\end{align*}
\end{enumerate}









\section*{Practice Problems}
\begin{enumerate}[1.]
\item
{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


\item {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

\item
{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

\item 
{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

\item
{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 counter-clockwise.}


\medskip

\item
{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

\item
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. 


\medskip


\item {  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 $n\to\infty$.}



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


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

\bigskip















\end{enumerate}

