Double integral of a rational function Consider the region $D$ given by $1\leq x^2+y^2\leq2\land0\leq y\leq x$. Compute $$\iint_D\frac{xy(x-y)}{x^3+y^3}dxdy$$
Attempt: The region $D$ is part of a ring in the first quadrant below the line $y=x$
Any hints are wellcome.
 A: Changing to polar coordinates, $x=\rho \cos\theta$, $y=\rho \sin\theta$, and the Jacobian of the transformation is $J=\rho$. Then:
$$\int_1^\sqrt2 \rho d\rho\int_0^\frac{\pi}{4}\frac{\sin\theta\cos\theta(\cos\theta-\sin\theta)}{\cos^3\theta+\sin^3\theta}d\theta$$
The first integral is immediate and yields $\frac{1}{2}$, so we'll multiply the answer given by the trigonometric integral by one half. For the trigonometric integral, let's use the substitution $u=\cos^3\theta +\sin^3\theta$, $du=(-3\cos^2\theta\sin\theta+3\sin^2\theta\cos\theta)d \theta=-3(\cos^2\theta\sin\theta-\sin^2\theta\cos\theta)d\theta$. The integral becomes:
$$-\frac{1}{3}\int_1^\frac{\sqrt2}{2}\frac{du}{u}=-\frac{1}{3}\log u\bigg|_{u=1}^{u=\frac{\sqrt2}{2}}=-\frac{1}{3} \log \frac{\sqrt 2}{2}$$
Multiplying by one half yields $I=-\frac{1}{6} \log \frac{\sqrt 2}{2}=\frac{\log 2}{12}$.
A: $$x=r\cos \phi$$
$$y=r\sin \phi$$
$$J=r$$
$$\int_{0}^{\pi /4}\int_1^{\sqrt2}\frac{r^4\cos \phi \sin \phi (\cos \phi - \sin \phi)}{r^3(\cos^3 \phi + \sin^3 \phi)}drd\phi=$$
Can you take it from here? 
