Question

double integral of e^(-x^2 - y^2) dA; 1<=r<=infinity; 0<=theta<=pi

Answer 1

\[\int\limits\limits_{0}^{\pi}\int\limits\limits_{1}^{\infty} e^{-x^2 -y^2} dxd \theta\]

Answer 2

this is weird.... do you mean

\(\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2-y^2}dxdy=\int_0^{2\pi}\int_0^\infty e^{-r^2}r*drd\theta\)

Answer 3

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\infty}re^{-r^2}drd\theta\]

Answer 4

use u substation on r^2

Answer 5

It IS weird!
I wasn't sure if the substitution would be negative. It makes sense, though.

Answer 6

I am saying its weird because you have r and things that look polar and things that done. I think maybe you are trying to solve \(\int_{-\infty}^\infty e^{-x^2}dx\)?

Answer 7

done=don't*

Answer 8

the way you have it wrote, we are to treat y as a constant.

Answer 9

It's supposed to be polar. Trying to solve what zarkon posted

Answer 10

\(\int_0^{2\pi}\int_0^\infty e^{-r^2}r*drd\theta\)

let \(u=r^2\), then \(du=2r dr\), so \(dr=\frac{du}{2r}\) So we have \(\int_0^{2\pi}\frac{1}{2}\int_0^\infty e^{-u}r*\frac{du}{r}d\theta

\) = \(\int_0^{2\pi}\frac{1}{2}\int_0^\infty e^{-u}*dud\theta\)
Can you solve now?

Answer 11

\[\int\limits\limits_{0}^{2\pi}\left(\frac{ 1 }{ 2 }\left( -e^-u \right) :o \rightarrow \infty = -e^{-\infty} +1 = 1)\right) d \theta\]
\[\int\limits_{0}^{2\pi}\frac{ 1 }{ 2 }d \theta = \frac{ \theta }{ 2 } : 0 \rightarrow 2\pi = \frac{ 2\pi }{ 2 } = \pi\]

Answer 12

Look right?

Answer 13

yes

Answer 14

thus \(\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}\)

Answer 15

The answer it wanted was \[\frac{ \pi }{ 2e }...\]

Answer 16

that would be a weird answer

the integral i wrote above has a value of \(\pi\)