# Homework Help: Solid Of Revolution Problem - Washer

1. Mar 22, 2012

### Bellwether

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

Find the volume of a solid created when the area between the function y=x^2+1 and the x-axis (for 0<x<2) is rotated about the line y=-2.

2. Relevant equations

Vs = ∏*r^2*h

3. The attempt at a solution

I can't seem to set this up correctly and am thrown by the inner radius being the x-axis. I originally set up the integral as : Int {0,2} ∏((x^2+1) + 3)^2 dx

I think this creates a full solid around y=-2 -- would I just want to add 1 to the radius instead of 3?

Many thanks!

2. Mar 22, 2012

### tiny-tim

Welcome to PF!

Hi Bellwether! Welcome to PF!

(try using the X2 button just above the Reply box )
That's for the outer radius …

you now need to subtract the volume for the inner radius.

3. Mar 22, 2012

### QuarkCharmer

For problems like this, I used to prefer adding 2 to the function to shift the whole thing up 2 units of y. Then you can simply revolve the function around the line y = 0, and the volume result will be the same.

4. Mar 22, 2012

### Bellwether

Thanks for the welcome!

So it's, ∫∏((x2+1)+3)2 - 22) ?

5. Mar 23, 2012

### tiny-tim

Hi Bellwether!

(just got up :zzz:)
∫ π((x2+1) + 2)2 - 22) dx

6. Mar 23, 2012

### Bellwether

I'm confused -- why is the outer radius just +2 , instead of 3? Isn't (x2 + 1) three units away from the axis of revolution?

7. Mar 23, 2012

### tiny-tim

no, you've counted the "1" twice …

(x2 + 1) - (-2 ) = x2 + 3

8. Mar 23, 2012

### Bellwether

Of course, thank you.