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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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  • $\begingroup$ Try to reduce it to two integrals of one variable. $\endgroup$
    – pointer
    Jul 15, 2014 at 12:12
  • $\begingroup$ 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... $\endgroup$ Jul 15, 2014 at 12:13
  • $\begingroup$ Don't you think this question is duplicated??math.stackexchange.com/questions/867853/… $\endgroup$
    – Shine
    Jul 15, 2014 at 12:29
  • $\begingroup$ How are they connected? @Shine $\endgroup$ Jul 15, 2014 at 12:47
  • $\begingroup$ @Student, Use the Green formula. But the region of the integral is different. $\endgroup$
    – Shine
    Jul 15, 2014 at 13:13

2 Answers 2

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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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