Question on Green's Theorem Consider the vector field $\textbf{f}(x,y)=(ye^{xy}+y^2\sqrt{x})\textbf{i}+(xe^{xy}+\frac{4}{3}yx^{\frac{3}{2}})\textbf{j}$.
Use Green's Theorem to evaluate $\int_C\textbf{f} \dot d\textbf{r}$, where $C$ is the ellipse given by $(x-1)^2+\frac{y^2}{9}=1$, oriented counterclockwise. 
$\int_C\textbf{f} \dot d\textbf{r} = \int \int_R (\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y}) \ dx \ dy$
$=\int \int_R (xe^{xy}+ye^{xy}+2yx^\frac{1}{2})-(ye^{xy}+xe^{xy}+2\sqrt{x}y) \ dx \ dy$
$=\int \int_R 2y\sqrt{x}-2\sqrt{x}y \ dx \ dy$
What would the bounds be on the integral? I think I should parametrize to polar coordinates, but I'm not sure how to do that when the ellipse's center isn't the origin. Or would I just use $0\leq x \leq 2$ and $-\sqrt{9(1-(x-1)^2)} \leq y \leq  +\sqrt{9(1-(x-1)^2)}$ ? 
Thanks. 

After fixing the partial derivative typo, we have 
$$\int \int_R 2y\sqrt{x}-2\sqrt{x}y \ dx \ dy$$
$$\int \int_R 0 \ dx \ dy$$
It does not matter what the bounds are, the answer is zero. 
