# Drilling a cylinder-hole with radius $b$ through a sphere $x^2+y^2+z^2 = 1$

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.

• I'm not sure if there is only one answer to this as it is not clear from the question if the drilled hole must go through the centre of the sphere. If we assume it is then I'd look at this as a 2D problem. Imagine a circle and remove a centred slot from it. What area would you need to reduce the circle by to half the volume and what $d$ do you need to achieve that? – Warren Hill Aug 18 '17 at 11:19
• It's a neat fact that the volume of the napkin ring that remains after you drill the hole depends on the height of the ring but not on the radius of the sphere you started with. – Jaap Scherphuis Aug 18 '17 at 11:21
• Turns out that I have made an error, and I was instead describing cutting out a sphere inside of my sphere instead of a cylinder through the sphere. – osk Aug 18 '17 at 15:27

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 $$0\le r\le b,\quad 0\le\theta\le2\,\pi,\quad-\sqrt{1-r^2}\le z\le\sqrt{1-r^2}.$$
• I do not describe the hole, I describe the area outside of the hole. The projections of the final shape are the sectors of the sphere between $b$ and $1$. Or am I describing cutting out a sphere inside of the sphere with a radius $b$? That might be where my error is. – osk Aug 18 '17 at 15:16
• Yes, you are describing cutting a sphere inside a sphere. To describe the volumen outside the hole just change $0\le r\le b$ to $b\le r\le 1$. – Julián Aguirre Aug 18 '17 at 15:30