Multivariable Calculus - Integration Assignment

[ConnorM]

Homework Statement

Here is my assignment, http://imgur.com/1edJ3g5
I figured it would be easier if we know we are both looking at the same thing! I'm looking for help with question 2. I seem to be having trouble with the integration.

Homework Equations

r=acosθ

x^2 + y^2 + z^2 = a^2

Cylindrical Coordinates ~ x^2 + y^2 = r^2

The Attempt at a Solution

So far I found my bounds of integration to be,

-sqrt( a^2 - r^2 ) ≤ z ≤ sqrt( a^2 - r^2 )

0 ≤ r ≤ acosθ

-pi/2 ≤ θ ≤ pi/2

After doing this I came up with the integral,

∫∫∫ r dzdrdθ

Since I converted to cylindrical coordinates I need to add the "r" to my integral correct?

From here I integrated and subbed in my limits and found

∫∫2*r*sqrt(a^2 - r^2) drdθ

After doing this if I integrate again I get a really messy answer which I don't think I could integrate again for dθ. Have I messed up somewhere? If I changed the order of my integral would that help at all?

 

[ConnorM]

After looking into it more,

Using symmetry I can change my bounds to,

0 ≤ z ≤ sqrt( a^2 - r^2 )

0 ≤ r ≤ acosθ

0 ≤ θ ≤ pi/2

I just need to multiply my integral by 4.

Next I solved my integral and found,

4∫∫ r*sqrt(a^2 - r^2) drdθ

4∫[ (-1/3)*(a^2 -r^2)^(3/2) ] 0 --> acosθ dθ

(-4/3 )∫ a^3(sin^3 θ) - a^3 dθ

(-4a^3 )/3 ∫ sin^3 θ - 1 dθ

after integrating using u - substitution,

(4a^3 )/3 [ (θ + cosθ + (cos^3 θ)/3 ) ] 0 --> 2pi

So my answer is,

(2a^3 pi)/3 - (8a^3)/9

Does somebody mind checking this, I'm not sure if this is correct or there is a mistake!

 

[Ray Vickson]

After looking into it more,

Using symmetry I can change my bounds to,

0 ≤ z ≤ sqrt( a^2 - r^2 )

0 ≤ r ≤ acosθ

0 ≤ θ ≤ pi/2

I just need to multiply my integral by 4.

Next I solved my integral and found,

4∫∫ r*sqrt(a^2 - r^2) drdθ

4∫[ (-1/3)*(a^2 -r^2)^(3/2) ] 0 --> acosθ dθ

(-4/3 )∫ a^3(sin^3 θ) - a^3 dθ

(-4a^3 )/3 ∫ sin^3 θ - 1 dθ

after integrating using u - substitution,

(4a^3 )/3 [ (θ + cosθ + (cos^3 θ)/3 ) ] 0 --> 2pi

So my answer is,

(2a^3 pi)/3 - (8a^3)/9

Does somebody mind checking this, I'm not sure if this is correct or there is a mistake!

You may notice that you are receiving no responses. Perhaps the reason is that (contrary to PF standards and policies) you post images instead of typing out your problem. I, for one, cannot read your attachments on some media.

 

[ConnorM]

Ok I will type it out! Sorry about that! Don't know why I didn't just type this question out, I had posted a few days ago a longer question and just used the same picture. So here is the question.

2) Find the volume of the solid that the cylinder r=acosθ cuts out of the sphere of radius "a" centered at the origin.

 

[LCKurtz]

@ConnorM: Your answer is correct.

 

[Ray Vickson]

After looking into it more,

Using symmetry I can change my bounds to,

0 ≤ z ≤ sqrt( a^2 - r^2 )

0 ≤ r ≤ acosθ

0 ≤ θ ≤ pi/2

I just need to multiply my integral by 4.

Next I solved my integral and found,

4∫∫ r*sqrt(a^2 - r^2) drdθ

4∫[ (-1/3)*(a^2 -r^2)^(3/2) ] 0 --> acosθ dθ

(-4/3 )∫ a^3(sin^3 θ) - a^3 dθ

(-4a^3 )/3 ∫ sin^3 θ - 1 dθ

after integrating using u - substitution,

(4a^3 )/3 [ (θ + cosθ + (cos^3 θ)/3 ) ] 0 --> 2pi

So my answer is,

(2a^3 pi)/3 - (8a^3)/9

Does somebody mind checking this, I'm not sure if this is correct or there is a mistake!

I get the same thing, too.