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.