# Sphere with a hole drilled in center

1. Dec 6, 2012

### ParoXsitiC

1. The problem statement, all variables and given/known data

Find volume of a sphere with radius R with a hole drilled through the center of radius b. (b<R)

2. Relevant equations

cos thera = adj / hyp

3. The attempt at a solution

$4\,\int _{0}^{\arccos \left( {\frac {b}{R}} \right) }\!\int _{0}^{\pi }\!\int _{b}^{R}\!{\rho}^{2}\sin \left( \phi \right) {d\rho}\,{d\phi} \,{d\theta}$

This is what I am getting with spherical coords but I am unsure if theta goes from 0 to arccos(b/R) or if the rho going from b to R takes care of that gap and instead of I should go from 0 to Pi/2 (multipled by 4).

See picture:

Edit:

Upon looking at the picture and how it would spin around with theta, can this even be done spherically? It seems like it would create spherical hole in the sphere and not a cylindrical drill.

With that in mind here are polar:

$\int _{0}^{2\,\pi }\!\int _{b}^{R}\!2\,\int _{0}^{ \sqrt{-{r}^{2}+{R}^ {2}}}\!r{dz}{dr}\,{d\theta}$

Again the problem arises, should I go from the green line or from the red line?

Last edited: Dec 6, 2012
2. Dec 6, 2012

### pasmith

Work out the volume of the hole, and subtract that from the volume of a sphere of radius R.

3. Dec 6, 2012

### ParoXsitiC

Nevermind, when evaulating both possible ways using b = 0 (just getting volume of a sphere no drill) I get 4/3 Pi R^2 when z goes from +- sqrt(R^2 - r^2) - which is the volumne of a sphere. So obviously that one is right, it makes sense visually but I know in some cases you must use the intersection as the limits.

4. Dec 6, 2012

### haruspex

It is a curious fact that the volume remaining in the sphere can be determined purely from the length of the cylindrical hole. Varying the sphere's radius while keeping that fixed increases the volume drilled out by the same amount as it increases the volume of the whole sphere.