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Using Polar Coordinates integrate over the region bounded by the two circles:

$$x^2+y^2=4$$

$$x^2+y^2=1$$

Evaluate the integral of $\int\int3x+8y^2 dx$

So what I did was said that as $x^2+y^2=4$ and $x^2+y^2=1$

That $1 \le r \le 2$. And as there is a symmetry in the four quadrants

$0 \le \theta \le \frac{\pi}{2}$

which gave me $\int_0^\frac{\pi}{2}\int_1^2 3r^2\cos(\theta) +8r^3\sin^2(\theta) ~dr d\theta$

The answer it gives in the book is $30\pi$.

I'm getting $28 +30\pi$

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There is less symmetry than you think because of the $x$. Well, there is symmetry there too, but it is cancellation symmetry: the contribution of $x$ to the integral is $0$. You can either note that, or integrate from $0$ to $2\pi$, or integrate from $0$ to $\pi$ and double.

Note that we need to replace $dx\,dy$ by $r\,dr\,d\theta$.

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  • $\begingroup$ So how do i need to rewrite this? $\endgroup$
    – Jesse Ross
    May 14 '13 at 7:34
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    $\begingroup$ I have added a few suggestions. I assume you want to integrate $x+8y^2$ (don't know where your $3$ came from). So if you really want to use the $x$m use $(r\cos \theta +8r^2\sin^2\theta)r\,dr\,d\theta$. $\endgroup$ May 14 '13 at 7:37
  • $\begingroup$ Ah. Just clicked, thank you so much. $\endgroup$
    – Jesse Ross
    May 14 '13 at 7:37
  • $\begingroup$ @TheGreatDuck: I assume that somewhere in the edit history it was $x+8y^2$. Luckily it was a side comment, and in any case does not affect the answer (it would be a pity to have to edit and bump up the question). $\endgroup$ May 24 '16 at 22:06

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