# Disk/Shell Method and Volume

1. Oct 12, 2011

### Rapier

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

Set up the integral (do not evaluate) to find the volume by revolving the region enclosed by y=x^2-2x+2 and y=-x^2+6 about a) x = 3 and b) y = -5.

2. Relevant equations

Shell Method: V = 2∏ ∫ (radius)*(height)dx
Washer Method: V = ∏ ∫ (R^2 - r^2)dx

3. The attempt at a solution

I believe I need to use the Shell Method for part A.
V = 2∏ ∫ (-1→2) (3-x)*(-x^2+6-x^2+2x-2)dx
V = 2∏ ∫ (-1→2) (3-x)*(-2x^2+2x+4)dx

I believe I need to use the Disk/Washer Method for part B.
V = ∏ ∫ (-1→2) [(-x^2 +6+5)^2 - (x^2-2x+2+5)^2]dx
V = ∏ ∫ (-1→2) [(-x^2 +11)^2 - (x^2-2x+7)^2]dx

But I'm not sure. Rotating about the axes are a lot simpler and I'm not sure about how to handle the values for x < 0. In my head, I think the equations will handle that for me, but I'm concerned.

Thanks for the help.

2. Oct 12, 2011

### iceblits

I don't see anything wrong with the set up, assuming I didn't miss anything

3. Oct 12, 2011

### SammyS

Staff Emeritus
Looks good to me too.

4. Oct 13, 2011

### Rapier

Well, look at me go!

(woot woot)

Thanks for putting some eyeballs on this me.