# Triple integral in cylindrical coordinates

1. Feb 10, 2013

### Locoism

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

Find the volume of the solid that lies between

z=x2+y2 and
x2+y2+z2=2

2. Relevant equations

z=r2
z=√(2-r2)

3. The attempt at a solution

So changing this into cylindrical coordinates, I get

z goes from r2 to √(2-r2)
r goes from 0 to √2
theta goes from 0 to 2π

so we get $\int_0^{2π} \int_0^{\sqrt{2}} \int_{r^2}^{\sqrt{2-r^2}} r dzdrdt$

I dunno... something feels off about this. I think this way it would take into account the negative area on the outside of the curve... How else could I set up the integral?

2. Feb 10, 2013

### Dick

I would reconsider your statement that r goes from 0 to sqrt(2).

3. Feb 10, 2013

### Locoism

How so? I was thinking of splitting it into 2 parts and having it go from 0 to the function in terms of z, but then I would have to integrate r first, and I run into the same problem...

ie: z from r^2 to 1, then from 1 to √(2-r^2), and r from 0 to √z and then from 0 to √(2-z^2)...

Is there a way to write it in terms of theta???

4. Feb 10, 2013

### Dick

The integral you have is fine. Except that the range of r is off. r^2 is x^2+y^2, not x^2+y^2+z^2. You need to figure out the value of r where r^2=z and r^2+z^2=2 intersect.

5. Feb 10, 2013

### Locoism

Awesome, r from 0 to 1, and z from r^2 to √(2-r^2).
Thank you so much, I see it perfectly now.