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.

Given $$\iint\limits_{R}e ^{-x^2-y^2}dxdy$$ for $R=\{(x,y):x^2+y^2 \le 9\}$. Is it $$ \int\limits_{0}^{2\pi}\int\limits_{0}^{3}e ^{-r^2} r dr d \theta$$

share|improve this question
    
Yes it is. $ $ $ $ –  Did Nov 1 '12 at 18:41
2  
As I recall the $dxdy$ changes to $rdrd\theta$ in polar. So shouldn't there be an extra factor of $r$ in the transformed integral? If so that would mean one could substitute say $w=-r^2$ and make headway... –  coffeemath Nov 1 '12 at 18:45
    
is the answer pi- pi/(e^9) –  Jack F Nov 1 '12 at 18:45
    
You need to be very careful with missing the $dx$, $dy$, $dr$ and $d\theta$ in your integrals! –  JavaMan Nov 1 '12 at 18:49
    
Actually the outside integral goes to $2\pi$ so I'm getting $2\pi(1-e^{-9})$, i.e. twice your answer. –  coffeemath Nov 1 '12 at 18:50
show 1 more comment

1 Answer

up vote 4 down vote accepted

I assume you're asking whether $$ \int \text{d}x\int \text{d}y\ e^{-x^2-y^2} = \int_0^{2\pi}\text{d}\phi\int_0^3\text{d}r\ e^{-r^2}\ , $$ where the first integral is restricted to $x^2+y^2\leq9$. The answer is 'no', as $\text{d}x\text{d}y=r\text{d}r\text{d}\phi$, so you need a further factor $r$ on the right-hand-side.

share|improve this answer
    
yes i forgot to add r, but I get some weird answer like pi-(pi/e^9) –  Jack F Nov 1 '12 at 18:47
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.