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Homework question, so just hints please

Sketch the region $$ R=\{ (x,y) \in \mathbb{R} : y \geq x, 1 \leq x^2 + y^2 \leq 2 \} $$ and, by changing to polar coordinates, compute $$ \iint \limits_R \frac{xy}{x^2 + y^2} \mathrm{d}x \, \mathrm{d}y $$

So I have the sketch, as from WolframAlpha:

http://www4c.wolframalpha.com/Calculate/MSP/MSP23281g329ghb4ef13b1f000019cf3e27f69cigib?MSPStoreType=image/gif&s=58&w=200.&h=193.&cdf=Coordinates&cdf=Tooltips

and so, in polar coordinates, the region is described by $$ R=\{ (r,\theta) \in \mathbb{R} : 1 \leq r \leq 2, \frac{\pi}{4} \leq \theta \leq \frac{5 \pi}{4} \} $$

and thus we have the integral $$ \begin{align*} \iint \limits_R \frac{xy}{x^2 + y^2} \mathrm{d}x \, \mathrm{d}y &= \int_1^2 \int_\frac{\pi}{4}^\frac{5 \pi}{4} \! \frac{r\cos\theta \, r\sin\theta}{(r\cos\theta)^2 + (r\sin\theta)^2} \mathrm{d}\theta \, \mathrm{d}r\\ &= \int_1^2 \int_\frac{\pi}{4}^\frac{5 \pi}{4} \! \sin\theta \, \cos\theta \, \mathrm{d}\theta\, \mathrm{d}r\\ \end{align*} $$ but then clearly this is wrong as it evaluates to $0$. However, I don't see where I have made a mistake.

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    $\begingroup$ You forgot to multiply by the determinant of the Jacobian $\left|\frac{\partial(x,y)}{\partial(\theta ,r)}\right| = r$. $\endgroup$ – Winther Mar 4 '15 at 12:42
  • $\begingroup$ btw the total result should be zero by symmetry. If you look at the four parts contained in the regions $(r,\theta)$ with 1) $\theta\in [\pi/4,\pi/2]$, 2) $\theta \in [\pi/2,3\pi/4]$, 3) $\theta \in [3\pi/4,\pi]$ and 4) $\theta \in [\pi,5\pi/4]$ then the integral over 1) + 2) is zero (symmetric about the $x$ axis and $xy/(x^2+y^2)$ is odd in $x$) and 3)+4) is zero (symmetric about the $y$ axis and $xy/(x^2+y^2)$ is odd in $y$). $\endgroup$ – Winther Mar 4 '15 at 13:05
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What you have done seems to be correct except that the area element $dxdy$ becomes $r dr d\theta$ in polar coordinates, and the maximum value of r is $\sqrt{2}$. However, this will still give 0 but what's wrong with that?

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