Use Green’s Theorem to evaluate the line integral along the given positively oriented curve. (which approach to take) Use Green’s Theorem to evaluate the line integral along 
the given positively oriented curve.
$$\int_c y^3 \, dx - x^3 \, dy, C \text{ is the circle } x^2+y^2=4$$
Ok, so I'm not sure how to approach this problem. I can easily find $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$, but I'm not sure which approach to take after that. 
Am I supposed to parameterize by making $y = 2\sin t$ and $x = 2\cos t$ ? and then do I replace dxdy with $-4\sin t\cos t\, dt$? and when I plug those in does it become a single integral from $0$ to $2\pi$? And how would I integrate that? It seems like this would leave me with an over-complicated integral. Would it be faster to switch it to polar coordinates? if so how? I'm not really sure if this problem is implying I use a certain method and I'm not sure if I'm doing that method correctly. I looked in the book for similar examples but I think they vary slightly so I'm not sure which approach I'm supposed to take. 

Green's theorem tell us that $$\int_C y^3 \, dx- x^3\, dy = \iint_{x^2+y^2 \leq 4} \left(\frac{\partial (-x^3)}{\partial x}\right)- \left(\frac{\partial y^3}{\partial y}\right) \,dx \, dy$$
While it is not a must to use polar coordinate, I strongly Iencourage you to do so as the form becomes elegant once you use polar coordinate.
Note that $\,dx\,dy =r \, dr \, d\theta $ when you change to polar coordiante.
