Answer
$-8 \pi$
Work Step by Step
Consider the Green's Theorem: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$
Here, $D$ is the region enclosed inside the counter-clockwise oriented loop $C$.
Then, we have $\int_C x^2ydx-xy^2 dy=\iint_D (-y^2-x^2) dA$
Here, the region $D$ is inside the circle $x^2+y^2=4$.
Now, in polar co-ordinates we have
$\int_C x^2ydx-xy^2 dy=\int_0^{2}\int_0^{2\pi} -r^2(r) d\theta dr$
or, $[\theta]_0^{2\pi} [\dfrac{-r^4}{4}]_0^{2}=(2\pi) \cdot (-4)$
Hence, $\int_C x^2ydx-xy^2 dy=-8 \pi$
