# Volume of revolution where am i going wrong?

1. Jul 7, 2013

### helpmeplz!

Hey guys I'm stuck on this problem. Its an easy one but I need some help..

It's asking for the volume of the solid obtained by rotating the region bounded by y=x^2, y=4, x=0 about the line x= -2 using the shell method.

I got the answer correct using the disk method (answer is 136 pi/3)..

With shell method I get integral 2pi (2+x) (4-x^2) dx from x=0 to x=2 which is 88pi/3. Can someone please help me out of this mess?

2. Jul 7, 2013

### mathman

You need to supply your work details.

3. Jul 7, 2013

### helpmeplz!

What do you mean? My question is what part of the integral that I set up is wrong and why? Not the evaluating of the integral which i'm using a calculator for anyway.

4. Jul 7, 2013

### helpmeplz!

Using slices, I get integral pi * (2+sqrty)^2 dy from y=0 to 4. which gives 136 pi/3?

5. Jul 7, 2013

### jbrussell93

There should be a hole in the center due to the gap from x=-2 to x=0. I'll leave the rest up to you

6. Jul 7, 2013

### helpmeplz!

Right, but the shell method takes that into account doesn't it? My integral only foes from 0 to 2..

7. Jul 7, 2013

### jbrussell93

Yes the shell method does, so I think you got the correct answer there. Think about the disk method and what your volume would look like. (Yours is missing the hole)

8. Jul 7, 2013

9. Jul 7, 2013

### jbrussell93

Yeah, it looks like it... Unless I'm wrong, which is entirely possible :P

10. Jul 7, 2013

### jbrussell93

Actually, I don't think he ever specifies x=0 as a bound. So assuming x=-2 is the bound instead, it would be correct.

11. Jul 7, 2013

### Redbelly98

Staff Emeritus
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