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I'm having some troubles trying to compute the following integral: $$ I = \iint_{D} \frac{x^2}{(x^2 + y^2)^{3/2}} dxdy $$ where $ D = \{ (x,y) \in \mathbb{R} : 2 \leq x^2 + y^2 \leq 2y \} $.

What I've tried so far:

  1. I plotted the domain $ D $: enter image description here

  2. Using the following transformation (polar coordinates): $$ \left\{\begin{matrix}x = r\cos\theta\\y = r\sin\theta\end{matrix}\right. $$ I get the integral: $$ \int_{\pi/4}^{3\pi/4} \int_{\sqrt{2}}^{2\sin\theta} \cos^2(\theta) \, dr d\theta $$

The result I should arrive is: $$ I = \frac{2\sqrt{2} + 3\pi + 6}{6} $$ but according to Wolfram what I get is: $$ \frac{10 - 3\pi}{6\sqrt{2}} $$

Questions:

  1. What am I doing wrong? How should I write the integral?

  2. Could you recommend me a good source with several examples of problems like this one?

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    $\begingroup$ The setup is good. Looks like you made a mistake in evaluating the iterated integral. Without seeing the intermediate steps it’s impossible to say where, but it’s likely when you integrated the $-\sqrt2\cos^2\theta\,d\theta$ term in the outer integral. $\endgroup$ – amd Nov 24 '15 at 22:38
  • $\begingroup$ @amd Actually, I didn't compute the integral for myself, I just trusted Wolfram. Do you think that Wolfram has committed a mistake? $\endgroup$ – francolino Nov 24 '15 at 22:40
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    $\begingroup$ No. I used a different CAS package that also came up with Wolfram’s answer. At this point I’d suspect an error in either the problem statement or the given solution. Everything you did looks right to me. $\endgroup$ – amd Nov 24 '15 at 22:56
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Your setup looks good. First, observe that $I = \frac{2\sqrt{2}+3\pi+6}{6}$ cannot be the area of the given domain because $I \approx 3.04$; however, you can over approximate the area of the shaded region by a rectangle of area $2\cdot (2-\sqrt{2}) \approx 1.17$. I went the calculations of the integral by hand, and assuming that I made no mistakes, arrived at the answer $\frac{10-3\pi}{6\sqrt{2}}$ in accordance with Wolfram.

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