I'm trying to find the volume of the region bounded by:
$x=0$ , $\left(y-1\right)^2+z^2=1$ and the outside of $x=\sqrt{y^2+z^2}$
In both the cylindrical and spherical coordinates:
Cylindrical Coordinates: $$\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int _0^{2cos\left(\theta \right)}\:\int _0^r\:r\:dx\:dr\:d\theta = \frac{32}{9}$$
Here $x$ goes from $0$ to the cone, and then we are left to integrate over the shifted circle.
Spherical Coordinates: $$\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\int _0^{2cos\left(\theta \right)}\:r^2sin\left(\theta \right)\:dr\:d\theta \:d\phi$$ which doesn't evaluate to the same answer, here $r$ goes from $0$ to the surface of the cylinder, $\theta$ starts from the cone and goes to $x=0$, $\phi$'s is trivial.
My mistake was thinking of $2\cdot cos\left(\theta \right)$ as the surface of the cylinder, the right way to find this is to solve for $r$ in the equation $\left(y-1\right)^2+z^2=2y$ by converting to spherical coordinates, in this case $r$ becomes equal to $2cos(\phi)csc(\theta)$ and the integral evaluates to the expected value.