The question:
Find the volume of the solid enclosed by the sphere $x^2 + y^2 + z^2 - 6z = 0$ , and the hemisphere $x^2 + y^2 + z^2 = 49 , z ≥ 0$
I set up the triple integral
$\int_0^{2\pi}\int_0^\pi\int_{6cos(\phi)}^7 \rho^2sin(\phi)~d\rho~d\phi~d\theta$
And evaluated it to get $\frac{1372\pi}{3}$, but I am unsure about the integration limits I used. This is one of my first spherical problems and I am a little confused about how to find the $\phi$ and $\rho$ limits.