Find volume above cone within sphere

Question

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?

Answer 1

First of all, the Jacobian of the spherical coordinat transformation is $\rho^2\sin\phi$ not $\rho\sin\phi$.

Also, the upper bound for $\rho$ should have been $6\cos\phi$ not $3$. The equation $\rho = 3$ describes a sphere of radius $3$ centered at the origin while the equation $\rho = 6\cos\phi$ describes a sphere of radius $3$ centered at $(x,y,z) = (0,0,3)$.

After fixing those two errors, you should get an integral which evaluates to the correct answer.

Answer 2

Just spotted it:

It should be $\rho^2\sin(\phi)$ and the integral along $\rho$ should be from $0$ to $6\cos\phi$, not $0$ to $3$.