So I think I have solved this problem, however my result is different than the correct one and I am not sure what I am doing wrong.
We have a sphere $x^2+y^2+z^2 = 1$, through this sphere a cylinder-hole with radius $b$ is drilled in the middle. We want to determine $b$ if we would like that the remaining volume of the sphere is half of the original volume of the sphere (which is $4\pi/3$, so we want the new volume to be $2\pi/3$).
My idea is to calculate it by calculating the volume of a sphere where the $xy$ plane projection has the radius $b$ to $1$ instead of $0$ to $1$.
So I describe this "new" sphere with polar coordinates:
$x = rsin(\theta)cos(\phi)$
$y = rsin(\theta)sin(\phi)$
$z = rsin(\theta)$
where $\theta:[0,\pi ], \phi:[0,2\pi]$ and $r:[b,1]$.
We solve this using a triple integral like this:
$\int_0^{2\pi}\int_0^{\pi}\int_b^{1} r^2sin(\theta)drd\theta d\phi$ = $2\pi\int_0^{\pi}sin(\theta)d\theta\int_b^{1} r^2dr$ = $2\pi[-cos(\theta)]_0^\pi[r^3/3]_b^1$
= $4\pi(\frac{1}{3}-\frac{b^3}{3})$
This integral is the volume of the remaining volume of the sphere after drilling. Thus:
$4\pi(\frac{1}{3}-\frac{b^3}{3})$ = $\frac{2\pi}{3}$
Which after solving results in $b=\sqrt[3] \frac{1}{2}$
According to my material, the correct solution is $b=\sqrt{ 1-(\frac{1}{4})^{1/3}}$
In my calculations, if I instead said that $b$ should go from $0$ to $1$ then I get the volume of a sphere, $4\pi/3.$ So I have no idea what I'm doing wrong, or if maybe I have found an error in the material.
Spherical coordinates are not useful for this problem. Your description of the "hole" is not correct. Use instead cylindrical coordinates. The "hole" is then described as $$ 0\le r\le b,\quad 0\le\theta\le2\,\pi,\quad-\sqrt{1-r^2}\le z\le\sqrt{1-r^2}. $$