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I've met a double integral which seems to be calcultated in polar system.

$\iint_D \frac {y} {x^2+y^2} dxdy$ , where $D$ is the region bounded by the following conditions:

$-2y \le x^2+y^2 \le -4y$;

$\frac {x} {\sqrt3} \le y \le 0$

I am not sure how to use the circle $x^2+y^2$ to calculate the double integral.

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  • $\begingroup$ You should write $dD$ at the end of this integral. $\endgroup$
    – Peter Foreman
    Apr 20, 2019 at 11:10
  • $\begingroup$ @Peter Foreman Sorry, it's typo, corrected. $\endgroup$
    – Mikhail Gaichenkov
    Apr 20, 2019 at 11:11
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    $\begingroup$ Set $r^2 = x^2+y^2$, and $\frac{y}{x} = \tan\theta$. Use the Jacobian $dx dy = r dr d\theta$. Change the bounds to be in terms of $r, \theta$. Then crack on! $\endgroup$
    – fGDu94
    Apr 20, 2019 at 11:17

1 Answer 1

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In the polar coordinate system; $$x=r\cos{(\theta)}$$ $$y=r\sin{(\theta)}$$ $$x^2+y^2=r^2$$ $$dxdy=rdrd\theta$$ So the integral bounds are equivalent to $$-2\sin{(\theta)}\le r\le-4\sin{(\theta)}$$ $$\frac1{\sqrt{3}}\cos{(\theta)}\le \sin{(\theta)}\le 0\implies -\pi\le\theta\le-\frac56\pi$$ Hence the integral becomes $$\begin{align} \int_{-\pi}^{-\frac56\pi}\int_{-2\sin{(\theta)}}^{-4\sin{(\theta)}}\sin{(\theta)}drd\theta &=\int_{-\pi}^{-\frac56\pi}-2\sin^2{(\theta)}d\theta\\ &=\int_{-\pi}^{-\frac56\pi}\left(\cos{(2\theta)}-1\right)d\theta\\ &=\left[\frac12\sin{(2\theta)}-\theta\right]_{-\pi}^{-\frac56\pi}\\ &=\frac{\sqrt{3}}4+\frac56\pi-\left(0+\pi\right)\\ &=\frac{\sqrt{3}}{4}-\frac16\pi\\ \end{align}$$

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