Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

up vote 0 down vote accepted

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 $u=r^2$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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