Question about specific volume integral. So I need to evaluate the following volume integral, and I'm not sure which bounds to use and that makes the question kind of vague for me at least. 
I've tried to solve it myself, but each try leaves my with a different solution, so I was wondering if someone could show me how to do the following question:

Determine the value of $\iiint_Wz^2dV$, where $W$ is the solide region around the cone $z^2=3x^2+3y^2$ and inside the sphere $x^2+y^2+z^2 = 6z$.


How do you find limits of integration?
Inside the sphere... this is asking for spherical coordinates.
$x = \rho \cos\theta\sin\phi\\
y = \rho \sin\theta\sin\phi\\
z = \rho\cos\phi$
now plug these into the equations given.
$z^2 = 3x^2 + 3y^2\\
\rho^2 \cos^2\phi = 3\rho^2 \cos^2\theta\sin^2\phi + 3\rho^2 \sin^2\theta\sin^2\phi\\
\rho^2\cos^2\phi = 3\rho^2\sin^2\phi\\
\tan^2\phi = 3\\
\phi = \arctan\frac 1{\sqrt 3}
$
$x^2 + y^2 +z^2 =6z \\
\rho^2 \cos^2\theta\sin^2\phi + \rho^2 \sin^2\theta\sin^2\phi + \rho^2\cos^2\phi = 6\rho\cos\phi\\
\rho^2 = 6\rho\cos\phi\\
\rho(\rho-6\cos\phi) = 0$
