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

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Yes it is. $ $ $ $ – Did Nov 1 '12 at 18:41
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
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.

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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

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