Given $$\iint\limits_{R}e ^{-x^2-y^2}dxdy$$ for $R=\{(x,y):x^2+y^2 \le 9\}$. Is it $$ \int\limits_{0}^{2\pi}\int\limits_{0}^{3}e ^{-r^2} r dr d \theta$$
double integral $e^{-x^2-y^2}$
1
$\begingroup$
calculus
-
0Yes it is. $ $ $ $ – 2012-11-01
-
2As 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... – 2012-11-01
-
0is the answer pi- pi/(e^9) – 2012-11-01
-
0You need to be very careful with missing the $dx$, $dy$, $dr$ and $d\theta$ in your integrals! – 2012-11-01
-
0Actually the outside integral goes to $2\pi$ so I'm getting $2\pi(1-e^{-9})$, i.e. twice your answer. – 2012-11-01
-
0@coffeemath But does the inner integral equal $(1-e^{-9})$ or $\frac{1}{2}(1-e^{-9})$? – 2012-11-01
1 Answers
4
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.
-
0yes i forgot to add r, but I get some weird answer like pi-(pi/e^9) – 2012-11-01