Finding Volume Using Polar Coordinates: Inside Sphere and Outside Cylinder

[EV33]

Homework Statement

Use polar coordinates to find the volume of the given solid.

Inside the sphere x²+y²+z²=16 and outside the cylinder x²+y²=4.

Homework Equations

x=rcosΘ,y=rsinΘ, x²+y²=r²

The Attempt at a Solution

2∫∫ (√(16-r²)r)drdΘ R{(r,Θ)l 0<Θ<2∏, 2<r<4}

I was wondering if this is the correct set up. I solved the equation for the sphere for z, and then substituted my x²+y², and I used that as the function which I want to integrate under because Its the function that goes the highest. Then I figured that the radius was going from 2 to 4 because that is the radius of each of the shapes. I chose to go from 0 to 2pi because I know that I need to roatate 360 degrees. I multiplied by two because i know the functions are symmetric.

 

[Pere Callahan]

1) Why does r not vary from 0 to 4? (EDIT: I overlooked the cylinder)

2) Why do you have only two integrals if your integrating over a region in three-dimensional space?

Remember that in general the volume of some solid S in three dimensional space is given by
<br /> V = \int_S dx dy dz = \int_S r drd\Theta dz<br />
So the third question:

3)Why is the function you are integrating - √(16-r²)r) - different from r?
4) What does the capital R in your formula mean?

 

[EV33]

I had it not vary from 0 to 4 because it has to be inside the sphere but outside the cylinder which has a radius of 2. It is in two integrals because the problems at this point in the book only want you using double integrals.

 

[Pere Callahan]

Ok, as I see it now you had implicitly performed the z-integral
<br /> \int_0^{\sqrt{16-r^2}}dz=\sqrt{16-r^2}<br />

already.

Then the set-up looks correct to me.

 

[EV33]

Awesome. Thank you so much.

 

[Pere Callahan]

You're welcome. If you want your final result checked just post it.

 

[EV33]

sounds good. After the first integral I am getting (2/3)*(12)^(3/2), and then when I do the second integral I end up getting (4/3)*(12)^(3/2)*pi.

That answer worries me a little bit because I would expect it to be equal to this (4/3)*pi*4³-pi*2²*2*sqrt(2) but its not.

 

[Pere Callahan]

sounds good. After the first integral I am getting (2/3)*(12)^(3/2), and then when I do the second integral I end up getting (4/3)*(12)^(3/2)*pi.

That answer worries me a little bit because I would expect it to be equal to this (4/3)*pi*4³-pi*2²*2*sqrt(2) but its not.

Why would you expect that? The answer
<br /> \frac{4}{3}12^{3/2}\pi = 32\sqrt{3}\pi<br />
is correct.

The volume you consider is NOT the volume of the sphere minus the volume of the cylinder because at the poles the cylinder is not contained in the sphere.

 

[EV33]

oh I see what you are saying. I don't know why I didn't realize that. Thank you.