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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.

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Try to reduce it to two integrals of one variable. – pointer Jul 15 '14 at 12:12
have you tried polar coordinates? It seems there's enough circular simmetry to be able to try that, even though I'm not sure the trigonometric integral that comes out is going to be that nice... – Gennaro Marco Devincenzis Jul 15 '14 at 12:13
Don't you think this question is duplicated??… – Shine Jul 15 '14 at 12:29
How are they connected? @Shine – Student Jul 15 '14 at 12:47
@Student, Use the Green formula. But the region of the integral is different. – Shine Jul 15 '14 at 13:13
up vote 4 down vote accepted

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 $I=-\frac{1}{6} \log \frac{\sqrt 2}{2}=\frac{\log 2}{12}$.

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$$x=r\cos \phi$$ $$y=r\sin \phi$$ $$J=r$$

$$\int_{0}^{\pi /4}\int_1^{\sqrt2}\frac{r^4\cos \phi \sin \phi (\cos \phi - \sin \phi)}{r^3(\cos^3 \phi + \sin^3 \phi)}drd\phi=$$

Can you take it from here?

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