Evaluate Multiple Integrals with Polar Coordinates

[Blamo_slamo]

Homework Statement

By changing to polar coordinates, evaluate:

[tex]\int[/tex][tex]\int[/tex] e ^(-[tex]\sqrt{x^2 + y^2}[/tex]) dx dy

Both integrals go from 0 --> infinity

Homework Equations

r = [tex]\sqrt{x^2 + y^2}[/tex]

x = r cos[tex]\theta[/tex]

y = r sin[tex]\theta[/tex]

Using the Jacobian to switch to polar coord we get:

J = r d[tex]\theta[/tex] dr

The Attempt at a Solution

[tex]\int[/tex][tex]\int[/tex] e ^ (-r) r d[tex]\theta[/tex] dr

I have my integral set up, but I have no clue what the integration limits are. My prof. said this is always the hardest part of multiple integrals, any help hinting in the right direction would be great!

Thanks!

 

[tiny-tim]

Hi Blamo_slamo!

(have a square-root: √ and a theta: θ and an infinity: ∞ and try using the X² tag just above the Reply box )

Both integrals go from 0 --> infinity
…
I have my integral set up, but I have no clue what the integration limits are. My prof. said this is always the hardest part of multiple integrals, any help hinting in the right direction would be great!

x and y each go from 0 to ∞.

So what does the region look like?

Then where does r go from?

And then where does θ go from?

 

[Blamo_slamo]

So what does the region look like?

So this region has a peak at x = y = 0, and it's at 1. The region also slopes off to 0, as both x, and y go to ∞

Then where does r go from?

I'm still not entirely sure on this, but technically the original function runs off to infinity, as it gets closer and closer to 0, so would r then be from ∞ --> 1? This is an even function, so I can just say it runs over that interval and take a 2 out in front right?

And then where does θ go from?

This one just completes the circle right? So it would go from 0 --> 2π

 

[tiny-tim]

Hi Blamo_slamo!

So this region has a peak at x = y = 0, and it's at 1. The region also slopes off to 0, as both x, and y go to ∞
…
… technically the original function runs off to infinity, as it gets closer and closer to 0, so would r then be from ∞ --> 1? …

no no no no no … forget the function, for the limits, you're only interested in the region

the region is the whole of the first quadrant … 0 ≤ x ≤ ∞ and 0 ≤ y ≤ ∞

so what is that in terms of r and θ ?

 

[Blamo_slamo]

Would it be reliable, if I just subbed in the bounds of x and y, in r?

e.g. r² = x² + y²
so as 0 ≤ x ≤ ∞ and 0 ≤ y ≤ ∞; r would then go from 0 ≤ r ≤ ∞ ?

and as you explained, we're only looking at the first quadrant, so 0 ≤ θ ≤ π/2 ?

 

[tiny-tim]

Hi Blamo_slamo!

(just got up :zzz: …)

Would it be reliable, if I just subbed in the bounds of x and y, in r?

e.g. r² = x² + y²
so as 0 ≤ x ≤ ∞ and 0 ≤ y ≤ ∞; r would then go from 0 ≤ r ≤ ∞ ?

and as you explained, we're only looking at the first quadrant, so 0 ≤ θ ≤ π/2 ?

Yup!

That's exactly the way to do it!

 

[Blamo_slamo]

Thanks a lot tiny-tim!