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



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 $28 +30\pi$


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 $r\,dr\,d\theta$.

  • $\begingroup$ So how do i need to rewrite this? $\endgroup$
    – Jesse Ross
    May 14 '13 at 7:34
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.