Using Green's theorem I want to calculate $\oint_{\sigma}\left (2xydx+3xy^2dy\right )$, where $\sigma$ is the boundary curve of the quadrangle with vertices $(-2,1)$, $(-2,-3)$, $(1,0)$, $(1,7)$ with positive orientation in relation to the quadrangle.
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I have done the following:
We consider the space $D=\{(x,y)\mid -2\leq x\leq 1, x-1\leq y\leq x+6\}$.
From Green's Theorem we get the following:
\begin{align*}\oint_{\sigma}\left (2xydx+3xy^2dy\right )&=\iint_D\left (\frac{\partial{(3xy^2)}}{\partial{x}}-\frac{\partial{(2xy)}}{\partial{y}}\right )dxdy \\ & = \iint_D\left (3y^2-2x\right )dxdy \\ & = \int_{-2}^1 \int_{x-1}^{x+6}\left (3y^2-2x\right )dydx \\ & = \int_{-2}^1\left (21x^2+91x+217\right )dx \\ & = \frac{1155}{2}=577,5\end{align*}
At the given solution the result should be $332,25$. What have I done wrong?
Best Answer
The integral must be
$$\int_{-2}^1\int_{x-1}^{2x+5}(3y^2-2x)\,dy\,dx=\int_{-2}^1\left(y^3-2xy\right)_{x-1}^{2x+5}dx=$$
$$=\int_{-2}^1\left[(2x+5)^3-(x-1)^3-x(2x+5)^2+x(x-1)^2\right]dx=$$
$$=\int_{-2}^1\left[(x+5)(2x+5)^2+(x-1)^2\right]dx=\ldots$$