QUESTION: Verify Green's Theorem for $$\oint (x^2-2xy)dx+(x^2y+3)dy$$ around the curve $y^2=8x$ and $x=2$.
My attempt:
L.H.S. comes out to be $\frac{128}{5}$ which is correct acc. to the book.
For R.H.S. we have $$\int \int \left(\frac{\partial}{\partial x}(x^2y+3) -\frac{\partial}{\partial y}(x^2-2xy)\right)dx \,\ dy$$
$$=\int \int (2xy+2x)dx \,\ dy$$
$$=\int \int 2x(y+1)dx \,\ dy$$
$$=\int (y+1) \left(\int_0^{\frac{y^2}{8}} 2xdx\right) \,\ dy$$
$$=\int \frac{y^4}{64}(y+1) \,\ dy$$
$$=\frac{1}{64} \int_{-4}^4 (y^5+y^4) \,\ dy$$
$$=\frac{2}{64}\frac{4^5}{5}$$
$$=\frac{32}{5}$$
It seems you made a mistake in identifying the integration limits. :) In fact, you should have written
$$\eqalign{ & \int\!\!\!\int\limits_\Omega {\left( {{\partial \over {\partial x}}\left( {{x^2}y + 3} \right) - {\partial \over {\partial y}}\left( {{x^2} - 2xy} \right)} \right)da} \cr & = \int\!\!\!\int\limits_\Omega {\left( {2xy + 2x} \right)da} = \int\!\!\!\int\limits_\Omega {2x\left( {y + 1} \right)da} \cr & = \int_0^2 {\int_{ - \sqrt x }^{\sqrt x } {2x\left( {y + 1} \right)dydx} } = \int_0^2 {2x\left( {\int_{ - \sqrt {8x} }^{\sqrt {8x} } {\left( {y + 1} \right)dy} } \right)} dx \cr & = ... \cr} $$
or this which is the solution you tried
$$\eqalign{ & \int\!\!\!\int\limits_\Omega {\left( {{\partial \over {\partial x}}\left( {{x^2}y + 3} \right) - {\partial \over {\partial y}}\left( {{x^2} - 2xy} \right)} \right)da} \cr & = \int\!\!\!\int\limits_\Omega {\left( {2xy + 2x} \right)da} = \int\!\!\!\int\limits_\Omega {2x\left( {y + 1} \right)da} \cr & = \int_{ - 4}^4 {\int_{{{{y^2}} \over 8}}^2 {2x\left( {y + 1} \right)dxdy} } = \int_{ - 4}^4 {\left( {y + 1} \right)\left( {\int_{{{{y^2}} \over 8}}^2 {2xdx} } \right)dy} \cr & = ... \cr} $$