# Volume of two pieces of a sphere cut by a plane

1. May 9, 2013

### ParoxysmX

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

Consider the unit sphere $x^{2} + y^{2} + z^{2} = 1$

Find the volume of the two pieces of the sphere when the sphere is cut by a plane at $z=a$.

3. The attempt at a solution

My interpretation is that $a$ is a point on the z-axis that the plane cuts at. So the height of the segment is $r-(r-a)$. After that, I'm not sure how to proceed. Should you somehow integrate the volume of the segment between $r=1$ and $r=a$?

Last edited: May 9, 2013
2. May 9, 2013

### VantagePoint72

Volume or area? You say volume in the title and area in the question.

3. May 9, 2013

### ParoxysmX

Volume. Edited accordingly.

4. May 9, 2013

### VantagePoint72

Well, you will have $a<1$ (since it's a unit sphere) so you will want to go the other way around; but, yes, what you are proposing should work. If you give a try and get stuck, show your work in lots of detail and we'll be able to help more.

5. May 9, 2013

### SteamKing

Staff Emeritus
The height of the segment is r-a. Your expression of the height, r - (r - a), equates to just a.

6. May 9, 2013

### HallsofIvy

Obviously a must be between -1 and 1. It is sufficient to assume that $0\le a\le 1$ and calculate the volume above z= a and below the sphere. (Of course, the volume of the entire sphere is $(4/3)\pi$ so the volume low z= a is $(4/3)\pi$ minus the volume above. And if a< 0, just flip it over.)

In cylindrical coordinates the base is given by $r^2+ a^2= 1$ so the cover that base r goes from 0 to $1- a^2$ and $\theta$ from 0 to $2\pi$. For each r and $\theta$, the height is $z- a= \sqrt{1- r^2}- a$

7. May 14, 2013

### ParoxysmX

Ok I believe I have that one figured out. The next problem now is to calculate the surface area of those segments using spherical coordinates. I'm told the formula $S = \int^{b}_{\phi=a}\int^{d}_{\theta=c} sin\theta d\theta d\phi$ should be used. What are a, b, c and d here?