# Triple Integrals: Spherical Coordinates - Finding the Bounds for ρ

1. Jun 10, 2012

### theBEAST

1. The problem statement, all variables and given/known data
Find the volume of the solid that lies above the cone z = root(x2 + y2) and below the sphere x2 + y2 + x2 = z.

2. Relevant equations
x2 + y2 + x2 = ρ2

3. The attempt at a solution
The main issue I have with this question is finding what the boundary of integration is for ρ. I tried to solve for it by:
https://dl.dropbox.com/u/64325990/Photobook/Photo%202012-06-10%207%2041%2030%20PM.jpg [Broken]
I end up getting 0 ≤ ρ ≤ root(2)sinΦ.

However the answer says the 0 ≤ ρ ≤ cosΦ, what am I doing wrong?

Last edited by a moderator: May 6, 2017
2. Jun 10, 2012

### LCKurtz

The outer surface is the sphere $x^2+y^2+z^2=z$. Writing that in spherical coordinates gives $\rho^2=\rho \cos\phi$. Dividing by $\rho$ gives $\rho = \cos\phi$. So $\rho$ goes from $0$ to $\cos\phi$. You get the cone with an appropriate limits for $\phi$.

Last edited by a moderator: May 6, 2017
3. Jun 10, 2012

### theBEAST

Oh wow that makes so much sense now! Thanks!!! DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD