Answer
$-8 \pi$
Work Step by Step
Apply Green's Theorem as: $\int_C adx+b dy=\iint_D (\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y})dA$
where, $D$ shows the region enclosed inside the counter-clockwise oriented loop $C$.
As $D$ is the region inside the circle $x^2+y^2=4$
we have $\int_C x^2ydx-xy^2 dy=\iint_D (-y^2-x^2) dA$
$\int_C x^2ydx-xy^2 dy=\int_0^{2}\int_0^{2\pi} -r^2(r) d\theta dr$
This implies that
$[\theta]_0^{2\pi} [\dfrac{-r^4}{4}]_0^{2}=(2\pi)(-4)$
Hence, $\int_C x^2ydx-xy^2 dy=-8 \pi$
