Double Integral Evaluation Using Polar Coordinates

[Chiborino]

Homework Statement

Evaluate over the x,y plane:
∫∫$e^{-\sqrt{x^{2}+4y^{2}}}$dxdy
And I know the answer SHOULD be $\pi$

Homework Equations

Polar-->rectangular identities maybe?
x--> rcos, y--> rsinθ, dxdy--> rdrdθ

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

 

[SammyS]

Homework Statement

Evaluate over the x,y plane:
∫∫$e^{-r*\sqrt{x^{2}+4y^{2}}}$dxdy
And I know the answer SHOULD be $\pi$

Homework Equations

Polar-->rectangular identities maybe?
x--> rcos, y--> rsinθ, dxdy--> rdrdθ

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 that simplifies the exponent into something easier to handle

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

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

 

[HallsofIvy]

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

 

[Chiborino]

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

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

 

[jackmell]

r*e^{-r*\sqrt{cosθ^{2}+4sinθ^{2}}}[/itex]drdθ
But there isn't an identity that simplifies the exponent into something easier to handle

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?

 

[Chiborino]

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?

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.