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

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:

I end up getting 0 ≤ ρ ≤ root(2)sinΦ.

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

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 Quote by 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: 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##.

 Quote by 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##.
Oh wow that makes so much sense now! Thanks!!! :DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD DD