# Homework Help: Polar Coordinate problem

1. Jul 1, 2010

### EV33

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

x=rcosΘ,y=rsinΘ, x²+y²=r²

3. 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.

2. Jul 1, 2010

### 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
$$V = \int_S dx dy dz = \int_S r drd\Theta dz$$
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?

Last edited: Jul 1, 2010
3. Jul 1, 2010

### 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.

4. Jul 1, 2010

### Pere Callahan

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

Then the set-up looks correct to me.

5. Jul 1, 2010

### EV33

Awesome. Thank you so much.

6. Jul 1, 2010

### Pere Callahan

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

7. Jul 1, 2010

### 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.

8. Jul 1, 2010

### Pere Callahan

Why would you expect that? The answer
$$\frac{4}{3}12^{3/2}\pi = 32\sqrt{3}\pi$$
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.

9. Jul 1, 2010

### EV33

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