I know this is a really basic question, but I seem to be kind of rusty.
$C$ is the boundary of the circle $x^2+y^2=4$
$$\int_C y^3dx-x^3 dy = \int_A -3x^2-3y^2 dA= \int_0^{2 \pi} \int_0^2 -3 r^2 r dr d \theta = -12 \pi$$
Did I make a mistake? My book says it's $-24 \pi$
As noted in the comments, the last calculation is incorrect:
$$\int_0^2 -3r^2 rdr = -3 \int_0^2 r^3 dr = -\frac{3 r^4}{4} \Huge{\vert_{\small 0}^{\small 2}}$$
This is equal to $-12$. So your integral is $\int_0^{2\pi} -12 d\theta = -24\pi$.
Using the usual notation for Green's Theorem,
$$ P(x,y) = y^3 \rightarrow \frac{\partial P}{\partial y} = 3y^2 \\ Q(x,y) = -x^3 \rightarrow \frac{\partial Q}{\partial x} = -3x^2 $$
$$ \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \ dA = \iint_D \left( -3x^3 - 3y^2 \right) \ dA \\ = -3 \iint_D \left( x^2 + y^2 \right) \ dA $$
Remembering $dA = r \ dr \ d\theta$ in polar coordinates,
$$ = -3 \int_{0}^{2\pi} \int_{0}^{2} r^3 \ dr \ d\theta \\ = -6\pi \int_{0}^{2} r^3 \ dr \\ = -6\pi \left( \frac{2^4}{4} \right) \\ = -24\pi $$
