# Evaluating a double integral

1. Nov 6, 2011

### Chiborino

1. The problem statement, all variables and given/known data
Evaluate over the x,y plane:
∫∫$e^{-\sqrt{x^{2}+4y^{2}}}$dxdy
And I know the answer SHOULD be $\pi$

2. Relevant equations
Polar-->rectangular identities maybe?
x--> rcos, y--> rsinθ, dxdy--> rdrdθ

3. The attempt at a solution
I tried using polar coordinates, but it doesn't simplify to something nice. When I did I got:
∫∫$r*e^{-r*\sqrt{cosθ^{2}+4sinθ^{2}}}$drdθ
But there isn't an identity that simplifies the exponent into something easier to handle

Last edited: Nov 7, 2011
2. Nov 7, 2011

### SammyS

Staff Emeritus
$\displaystyle\sqrt{x^{2}+4y^{2}}=r\sqrt{\cos^{2}( \theta)+4sin^{2}(\theta)}$

So that last integral should be $\displaystyle\int\int r\, e^{r^2\sqrt{\cos^{2}( \theta)+4sin^{2}(\theta)}}dr\,d\theta$
.

3. Nov 7, 2011

### HallsofIvy

Staff Emeritus
Since the original integral is in xy coordinates, is that "r" a variable or a constant? Is r just short for $\sqrt{x^2+ y^2}$?

4. Nov 7, 2011

### Chiborino

The r I put inthe original shouldn't be there, my apologies.

5. Nov 7, 2011

### jackmell

What's with this something nice and easy to handle thing? That don't happen in the real world. Try and learn to muscle through things. Suppose I just write it as:

$$\int_0^{2\pi}\left(\int_0^{\infty}re^{-ar}dr\right)d\theta,\quad a>0$$

Can you do that part in parenthesis? Alright, then switch to:

$$a=\sqrt{\cos^2(t)+4\sin^2(t)}$$

and muscle through that one. We can get rid of one of them right?

$$\cos^2(t)+4\sin^2(t)=1+3\sin^2(t)$$

Then what about reducing it further with other trig identities?

Last edited: Nov 7, 2011
6. Nov 7, 2011

### Chiborino

I ultimately wound up getting $\int_0^{2\pi}\frac{1}{5-3cos(2\theta}d\theta$

Upon evaluating the integral I got $\frac{1}{2}arctan(2tan(\theta))$, evaluated between 0 and 2$\pi$, which is 0. But after splitting the boundaries at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, I got simply $\pi$ for an answer.

Thank you for the help.