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I'm trying to evaluate the integral $$ \int\int_R e^{-(x^2+y^2)}dxdy $$ where $R$ is the circular region centered at the origin of radius $2$. I convert to polar coordinates with $x=2\cos\theta$, $y=2\sin\theta$, to get $$ \int_0^{2\pi}\int_0^2 e^{-4}rdrd\theta. $$ Carrying this out, I get a final answer of $\frac{4\pi}{e^4}$, but apparently the correct answer is $\pi(1-e^{-4})$. I'm pretty sure I evaluated the integral I wrote down correctly, so maybe I have set it up incorrectly? Does anyone see?

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You want $$ \int_0^{2\pi}\int_0^2 e^{-r^2}r\,drd\theta, $$ because $x^2+y^2=r^2$. (You are interested in the region where $r\le 2$, not in the region where $r=2$.)

The integration with respect to $r$ should be easy. You can probably spot an antiderivative. If not, let $u=r^2$.

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Thanks, I was able to get the right answer now. Was the mistake taking $x=2\cos\theta$, instead of $x=r\cos\theta$? – Shao Lin Aug 28 '12 at 6:04
Yes, $x=r\cos\theta$, $y=r\sin\theta$, so $x^2+y^2=r^2(\cos^2\theta+\sin^2\theta)=r^2$. – André Nicolas Aug 28 '12 at 6:06
Thanks Dr. Dré! – Shao Lin Aug 28 '12 at 6:13

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