### Green's Theorem

(requires JavaScript)

1. Let $C$ be the rectangle with vertices $\left(0,0\right)$, $\left(2,0\right)$, $\left(2,3\right)$, and $\left(0,3\right)$. Evaluate ${\oint }_{C}x{y}^{2}\phantom{\rule{0.2em}{0ex}}dx+{x}^{3}\phantom{\rule{0.2em}{0ex}}dy$ twice: first directly and then by using Green's Theorem.
$6$
2. Use Green's Theorem to evaluate the integral ${\int }_{C}{e}^{y}\phantom{\rule{0.2em}{0ex}}dx+2x{e}^{y}\phantom{\rule{0.2em}{0ex}}dy$ along the positively oriented square with sides $x=0$, $x=1$, $y=0$, and $y=1$.
$e-1$
3. Use Green's Theorem to evaluate the integral ${\int }_{C}{x}^{2}{y}^{2}\phantom{\rule{0.2em}{0ex}}dx+4x{y}^{3}\phantom{\rule{0.2em}{0ex}}dy$ along the positively oriented triangle with vertices $\left(0,0\right)$, $\left(1,3\right)$, and $\left(0,3\right)$.
$\frac{318}{5}$
4. Use Green's Theorem to evaluate the integral ${\int }_{C}x{e}^{-2x}\phantom{\rule{0.2em}{0ex}}dx+\left({x}^{4}+2{x}^{2}{y}^{2}\right)\phantom{\rule{0.2em}{0ex}}dy$ along the positively oriented boundary of the region between the circles ${x}^{2}+{y}^{2}=1$ and ${x}^{2}+{y}^{2}=4$.
$0$
5. Use Green's Theorem to evaluate ${\int }_{C}\mathbf{F}•\phantom{\rule{0.2em}{0ex}}d\mathbf{r}$ if $\mathbf{F}\left(x,y\right)=⟨{y}^{2}\mathrm{cos}\left(x\right),{x}^{2}+2y\mathrm{sin}\left(x\right)⟩$ and $C$ is the triangle from $\left(0,0\right)$ to $\left(2,6\right)$ to $\left(2,0\right)$ to $\left(0,0\right)$. Make sure to check the orientation of the curve.
6. Let $D$ be a region bounded by a simple closed path $C$ in the $xy\text{-plane}$. Use Green's Theorem to prove that the coordinates of the centroid $\left(\stackrel{‾}{x},\stackrel{‾}{y}\right)$ of $D$ are

$\stackrel{‾}{x}=\frac{1}{2A}{\oint }_{C}{x}^{2}\phantom{\rule{0.2em}{0ex}}dy$ and $\stackrel{‾}{y}=-\frac{1}{2A}{\oint }_{C}{y}^{2}\phantom{\rule{0.2em}{0ex}}dx$,

where $A$ is the area of $D$.

7. Use a path integral to find the centroid of the triangle with vertices $\left(0,0\right)$, $\left(1,0\right)$, and $\left(0,1\right)$.
$\left(\frac{1}{3},\frac{1}{3}\right)$