Using Green's theorem, evaluate the line integral $$I=\int_Cx^2(x^2+y^2)dx+y(x^3+y^3)dy,$$ where $C$ is the parallelogram with vertices $(0,0), (1,0), (2,2), (1,2)$, traversed in that order.
My solution: Suppose $\textbf{g}(x,y)=x^2(x^2+y^2)dx+y(x^3+y^3)dy$. Note that $curl \ g =x^2y$. Green's theorem tells us the line integral is now equal to $\int\int_Rx^2ydxdy$.
To find the region $R$ note that $0\le y\le 2$ and $\frac{y}{2}\le x\le \frac{y+2}{2}$.
Hence the integral is $\int_0^2\int_{\frac{y}{2}}^{\frac{y+2}{2}}x^2ydxdy$.
Evaluating gives and answer of $\frac{19}{48}$. Is this solution correct?