They give me two surfaces $$x^2+y^2+z^2=a^2y;\quad \quad z^2\sin^2(b)=(x^2+y^2)\cos^2(b)$$
If I am not mistaken, it is a sphere displaced on the $y$ axis; and a cone with an origin vertex and angle with respect to the $XY$ plane equal to $b$
I am asked, by changing from variable to cylindrical and to spherical, to calculate the volume that comprises the intersection of these figures.
I started with sphericals. It comes to this: if we call the $(r,\phi,\theta)$-coordinates where $r$ is the radius, $\phi$ the angle of projection on the $XY$ and $\theta$ the angle to the $Z$-axis, we would have $r=a^2\sin(\phi)\sin(b)$ and $\theta=b$.
The thing is, what represents the limits of the cut? I mean, I'd like to reason with you, but I just don't see the limits very well. I've tried, but I'm getting nonsense.
$x^2 + y^2 + z^2 = a^2y\\ z^2\sin^2\beta = (x^2+y^2)\cos^2\beta$
$x = \rho\cos\theta\sin\phi\\ y = \rho\sin\theta\sin\phi\\ z = \rho\cos\phi$
Substitute into the original equations.
$\rho^2 = a^2\rho\sin\theta\sin\phi\\ \rho = a^2\sin\theta\sin\phi$
$\rho^2\cos^2\phi\sin^2\beta = \rho^2\sin^2\phi\cos^2\beta\\ \tan^2\beta = \tan^2\phi\\ \phi = \beta, \pi-\beta$
(We only care about solutions for $\phi$ in $[0,\pi]$)
$\int_0^{2\pi}\int_{\beta}^{\pi-\beta}\int_0^{a^2\sin\theta\sin\phi} \rho^2\sin\phi \ d\rho\ d\phi \ d\theta$
And I will leave the integration to you.
This is the single volume inside the sphere outside the double cones. If you want the two volumes top and bottom, subtract this from the volume of the sphere.