# How to set up double integral for volume by rotation

1. Mar 1, 2012

### MeMoses

1. The problem statement, all variables and given/known data
Find the volume of the solid of revolution of the area bounded by the curves about the given axis.
y= x^2 - 2, y = 0, about y = -1, consider only the area above y=-1

2. Relevant equations

3. The attempt at a solution
So I drew out the problem and figured it would be simplest to use a double integral. I'm not sure how to write this nicely but, I got:
4pi * (integral from -1 to 0 of y(integral from 0 to sqrt(y+2) of dx) dy)
Hopefully you can read that. Note I cut it in half so I multiply by 2 to get the volume I want. I'm fairly positive this is right, I'm just curious as to why i am getting a negative answer. If I just have to switch the integral from (-1 to 0) to (0 to -1) why is that? or did I set the problem up wrong since the axis is y=-1?
Also is there a simple way to cover the problem with a single single intgral (I know i can break it up)?

2. Mar 1, 2012

### LCKurtz

You have$$4\pi \int_{-1}^0\int_0^{\sqrt{y+2}} y\, dxdy$$Your mistake is using y in the integrand; that is not the radius of rotation of the dxdy element. What is the correct distance from the line y = -1 to the dxdy element located at (x,y)?

3. Mar 1, 2012

### MeMoses

so the integrand would just be (y+1) and that should solve my problems?

4. Mar 1, 2012

### LCKurtz

I think so. Try it.

5. Mar 1, 2012

### LCKurtz

And, yes to that last question. Once you do the inner dx integral in your double integral, you will be looking at the single integral you could have used in the first place.