Line integral of $$\frac{-y^3}{(x^2+y^2)^2} dx + \frac{xy^2}{(x^2+y^2)^2} dy$$
Using greene's theorem, this integral is equal to zero. Am i right? $P(x,y)$ and $Q(x,y)$ are not continuous at $(0,0)$. Since the circle does not contain $(0,0)$, green's theorem applies?
Using Green's Theorem we get:
$$\int_C\frac{-y^3}{(x^2+y^2)^2}dx+\frac{xy^2}{(x^2+y^2)^2}dy=\int\int_D\left(\frac{8xy^3}{(x^2+y^2)^3}-\frac{2x^2y}{(x^2+y^2)^2}\right)dA$$
Then switch to polor coordinates, since $D$ is a unit circle, (just shifted) we get:
$$=\int_0^{2\pi}\int_0^1\left(8\cos\theta \sin^2\theta\cdot r-2\cdot r\cos^2\theta\sin\theta\right)drd\theta=0$$
These are fairly simply integrals to integrate. For the first simply let $u=\sin\theta$ and the second let $u=\cos\theta$, after we integrate with respect to $r$, of course. So yes, you're right.
