My objective:
Using spherical coordinates, set up and compute an integral to find the volume of the ice-cream-cone shaped solid lying above the cone $z = \sqrt{x^2 + y^2}$ and below the sphere $\rho = 6 \cos(\phi)$.
I think that $\rho = 6 \cos(\phi)$ is a sphere with points at $(0,0,0)$ and $(0,0,6)$, so the radius is 3
So I setup the integral to find the volume as follows: $\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{3} 1\rho\sin{\phi}d\rho d\phi d\theta$ which works out to $2\pi(1 - \frac{\sqrt{2}}{2})\frac{3^3}{3} = (\frac{18\pi}{3}) (2 - \sqrt{2})$ if I'm not mistaken, but this is not correct.
I'm thinking my bounds for $\rho$ are incorrect (because I'm fairly confident about the others), but how are they incorrect?