Evaluating triple integral with spherical coordinates

[Alex Bard]

Homework Statement

Evaluate the iterated integral ∫ (from 0 to 1) ∫ [from -sqrt(1-x^2) to sqrt(1-x^2) ] ∫ (from 0 to 2-x^2-y^2)

the function given as √(x^2 + y^2) dz dy dx

The Attempt at a Solution

I changed the coordinates and I got the new limits as

∫(from 0 to pi) ∫(from (3pi)/2 to pi/2) ∫(from 0 to √2) √(x^2 + y^2) ρ^2 sin phi dρ dphi dθ

What I'm having problems is with changing the function I need to integrate into spherical coordinates. Should I replace the values of x and y for it's spherical counterparts or is there an easier way via u-sub, etc?

When i try to sub in x = ρ sin phi sin ρ and y = ρ sin phi cos ρ I get a mess. Can anyone nudge me in the right direction please?

 

[mfb]

When i try to sub in x = ρ sin phi sin ρ and y = ρ sin phi cos ρ I get a mess.

What do you get? This should give a nice result.

 

[Alex Bard]

Thanks for answering mfb,

First I noticed I made a mistake, in spherical coordinates

x= ρ sin\phicosθ
y= ρ sin\phisinθ

so now when I substitute that into the formula, I will get

√[(ρ sin \phi cosθ)^2 + (ρ sin \phi sinθ)^2]

What I get stuck with is I can't get rid of the square root. My process, so far, has been to expand everything, using the identity of cos/sin^2(x) to expand to either 1/2(1 +/- cos(2x) etc.

At the end, before the first integration or even getting ρ^2 or sin phi involved, i get the answer below.

√[ρ^2(\frac{1-cos(2 \phi)}{2})]

I can probably take a picture of my work to post if that would help. Otherwise its a lot of writing. Please let me know what information you need to see where i made the mistake.

 

[LCKurtz]

Thanks for answering mfb,

First I noticed I made a mistake, in spherical coordinates

x= ρ sin\phicosθ
y= ρ sin\phisinθ

so now when I substitute that into the formula, I will get

√[(ρ sin \phi cosθ)^2 + (ρ sin \phi sinθ)^2]

What I get stuck with is I can't get rid of the square root. My process, so far, has been to expand everything, using the identity of cos/sin^2(x) to expand to either 1/2(1 +/- cos(2x) etc.

At the end, before the first integration or even getting ρ^2 or sin phi involved, i get the answer below.

√[ρ^2(\frac{1-cos(2 \phi)}{2})]

I can probably take a picture of my work to post if that would help. Otherwise its a lot of writing. Please let me know what information you need to see where i made the mistake.

You're making it way too hard:$$
\sqrt{(ρ sin \phi cosθ)^2 + (ρ sin \phi sinθ)^2}=\sqrt{\rho^2\sin^2\phi(
\cos^2\theta +\sin^2\theta)}$$Now do you see how to simplify it?