I'm trying to find the volume between the surfaces defined by the following equations: $$ x^2 + y^2 + z^2 = b^2 $$ $$ y^2 \tan^2 a = x^2 + z^2 $$ I need to find the volume using two different methods: the first one is by spherical coordinates and the second one by polar coordinates. The problem is I get two different results:
Using spherical coordinates I found: $$ x = \rho \sin \phi \cos \theta \\ z = \rho \sin \phi \sin \theta \\ y = \rho \cos \phi $$ $$ V= \int_0^{2\pi}\int_0^{a}\int_0^{b}\rho^2\sin\phi \:d\rho d\phi d\theta = \frac{2\pi}{3}b^3(-\cos(a)+1) $$ and using polar: $$ x = r \cos \theta \\ z = r \sin \theta \\ y = y $$ $$ V = \int_0^{2\pi}\int_0^{b\sin a}\int_0^{\sqrt{b^2-r^2}}r \:dy dr d\theta = -\frac{2\pi}{3}b^3(\cos^3(a)-1) $$ So I know I have made a mistake at some point but I can't find it. If anyone can help I'd appreciate it. Thanks!
Your working for spherical coordinates is correct. In cylindrical coordinates, the lower bound of $y$ is incorrect. Rest everything is correct.
The lower bound of $y$ as zero for all values of $r$ will mean you are integrating over a cylinder. Please note that $y$ is bound below by the surface of the cone and hence,
$y^2 \tan^2 a = r^2 \implies y = r \cot a$.
So the integral should be,
$ \displaystyle \int_0^{2\pi}\int_0^{b\sin a}\int_{ \color {blue} {r \cot a}}^{\sqrt{b^2-r^2}} r \ dy \ dr \ d\theta$