# Homework Help: Volume of sphere cut by two parrallel planes

1. Apr 9, 2012

### Koranzite

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

A sphere of radius R with centre at the origin is cut by two parallel planes at $z=\pm a$, where a<R. Write, in cylindrical coordinates, a triple integral which gives the volume of that part of the sphere between the two planes. Evaluate the volume by first performing the r,θ integrals and the the remaining z integral.

2. Relevant equations

$dV=rdrdθdz$

3. The attempt at a solution

The main probelm here is the setting up of my integral, as the answer I am getting is independant of R, which is then clearly wrong.

My integral runs from:
$r=\sqrt{R^2-a^2}$ to $r=\sqrt{R^2-z^2}$
$θ=0$ to $θ=2\pi$
$z=-a$ to $z=a$

I would expect the answer to depend on R, but it keeps cancelling out when I evaluate the r integral. I would be grateful if someone could explain what is wrong with my limits.

2. Apr 9, 2012

### Dick

I think r should go from 0 to $\sqrt{R^2-z^2}$. Shouldn't it?

3. Apr 9, 2012

### Koranzite

Well that is certainly true to get the formula for the volume of the whole sphere, but in this case the minimum value that r takes is $\sqrt{R^2-a^2}$. Your proposal is one that I have considered, but I don't see how it can be justified.

4. Apr 9, 2012

### Dick

You already have the z limits from -a to a. Doesn't that take care of the a dependency? You may be visualizing the r coordinate wrong. It's the distance from the z axis to the edge of your solid parallel to the x-y plane.

5. Apr 9, 2012

### Koranzite

Ah yes, just recognised the problem. I was only imagining r as being the distance to the surface from the z axis, neglecting all the interior volume where it can reduce to 0... Thanks!

6. Apr 9, 2012

### Dick

Right. Your original limits would be for the volume of a sphere with a cylinder cut out of it. Interesting that the R cancels, isn't it? You might not guess that to be true looking a picture of it.