Answer
$-8 \pi$
Work Step by Step
Green's Theorem states that $\int_C Adx+B dy=\iint_D (\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y})dA$
Here, $D$ is the region enclosed inside the counter-clockwise oriented loop $C$.
Here, we have $\int_C x^2ydx-xy^2 dy=\iint_D (-y^2-x^2) dA$
We are given that $D$ is the region inside the circle $x^2+y^2=4$
Now, in polar co-ordinates we have
$\int_C x^2ydx-xy^2 dy=\int_0^{2}\int_0^{2\pi} -r^2(r) d\theta dr=[\theta]_0^{2\pi} [\dfrac{-r^4}{4}]_0^{2}$
Thus, $\int_C x^2ydx-xy^2 dy=(2\pi)(-4)=-8 \pi$
