Triple Integrals: Spherical Coordinates - Finding the Bounds for ρ

[theBEAST]

Homework Statement

Find the volume of the solid that lies above the cone z = root(x² + y²) and below the sphere x² + y² + x² = z.

Homework Equations

x² + y² + x² = ρ²

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
I end up getting 0 ≤ ρ ≤ root(2)sinΦ.

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

 

[LCKurtz]

Homework Statement

Find the volume of the solid that lies above the cone z = root(x² + y²) and below the sphere x² + y² + x² = z.

Homework Equations

x² + y² + x² = ρ²

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
I end up getting 0 ≤ ρ ≤ root(2)sinΦ.

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

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 $\phi$.

 

[theBEAST]

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 $\phi$.

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