Green's Theorem for the vector field $v=\begin{bmatrix} 1&xy \end{bmatrix}^{T}$ Using Green's Theorem for the vector field $v=\begin{bmatrix}  1&xy  \end{bmatrix}^{T}$ over the upper half of the circle with radius 1 and center $(3,3)$ which is split in two by the line y=x. 
Wouldn't be so hard if the circle wasn't offset. I'm not clear on how the bounds on the region for the double integral should look.
Thanks in advance!
 A: Green's Theorem gives $\int_{\partial R} Pdx+Qdy = \int_R \left( Q_x-P_y \right) dx \, dy$. I'd rather not calculate the double integral, so, I'll calculate the line integral around the half circle. In particular, notice
$$ x = 3+\cos t, \ \ \ y = 3+\sin t $$
for $\pi/4 \leq t \leq 5\pi/4$ clearly parametrizes the circular part of the boundary. For the line segment from $p=(3+\cos (5\pi/4), 3+\sin(5\pi/4))$ to $q=(3+\cos (\pi/4), 3+\sin(\pi/4))$ just use $(x,y) = p+t(q-p)$ for $0 \leq t \leq 1$. The given vector field has $P=1$ and $Q=xy$. You know what, scratch the line integral. I'm going with the double integral instead.
$P_y=0$ and $Q_x=y$. Just shift coordinates $\bar{x} = x-3$ and $\bar{y} = y-3$ hence $d\bar{x} = dx$ and $d\bar{y} = dy$ and $y = \bar{y}+3$. In the barred coordinates we define $\bar{r}^2 = \bar{x}^2+\bar{y}^2$ and the polar $\bar{\theta}$ in kind. The region $R$ is described by $\pi/4 \leq t \leq 5\pi/4$ for $0 \leq \bar{r} \leq 1$ hence:
\begin{align}
\int_{\partial R} Pdx+Qdy &= \int_R \left( Q_x-P_y \right) dx \, dy \\
&= \int_R \left( \bar{y}+3 \right) d\bar{x} \, d\bar{y} \\
&= \int_0^1\int_{\pi/4}^{5\pi/4} \left( \bar{r}\sin \bar{\theta}+3 \right) \bar{r}d\bar{\theta} \, d\bar{r} \\
&= \int_{\pi/4}^{5\pi/4} \left( \frac{1}{3}\sin \bar{\theta}+\frac{3}{2} \right) d\bar{\theta}  \\
&= \frac{3\pi}{2}-\frac{1}{3} \cos \bar{\theta} \bigg{|}_{\pi/4}^{5\pi/4} \\
&= \frac{3\pi}{2} - \frac{\sqrt{2}}{3}.
\end{align}
