# Volume using Triple Integrals

1. Nov 17, 2008

### seichan

1. The problem statement, all variables and given/known data
A volcano fills the volume between the graphs z = 0 and z =1/(x^2+y^2)^24, and outside the cylinder x^2+y^2=1 Find the volume of this volcano.

2. Relevant equations
This is a triple integral to be evaluated in cylindrical coordinates.

3. The attempt at a solution
Alright, so, this is what I've done. I don't understand what I'm doing wrong. I set up a triple integral, the limits of integration as follows:
z goes from 0 to 1
theta goes from 0 to 2pi
and r goes from 1 to z^-1/48 (outside the unit circle, but between the other graphs)

I first integrate with respect to r. I get r^2/2, evaluated between z^-1/48 and 1. This gives the integrand (z^-1/2304)/2-1/2. Then, I integrated with respect to z. This gives the answer ((2304z^(2303/2304)/2303)/2)-z/2. Then, this is evaluated from 0 to 1. This gives (2304/2303)/2-1/2, or, (1/2303)/2. This is integrated with respect to theta, giving (theta/2303)/2. Evaluated at 2pi, this results in (2pi/2303)/2. However, my homework software has told me this is wrong multiple times. Where am I going wrong? Thank you for any assistance.

2. Nov 17, 2008

### HallsofIvy

Staff Emeritus
z does NOT go from 0 to 1. It goes from 0 up to 1/r48. r goes from 1 to infinity:
$$\int_{\theta= 0}^{2\pi}\int_{r= 0}^\infty\int_{z= 0}^{1/r^{48}} r dz dr d\theta$$