Associated Topics || Dr. Math Home || Search Dr. Math

### Proof for Volume of a Segment of a Sphere

```
Date: 11/19/2001 at 02:56:55
From: S. Kozakavich
Subject: Proof for Volume of a Segment of a Sphere

I am in need of assistance in proving the volume of a truncated
spherical cap (or a segment of a sphere I think it is also called).  I
think I should be subtracting the volume of a smaller top cap from the
volume of the whole spherical cap, but am getting lost.  I know what
the end result should be, but I can't get there.  Perhaps my problem
is with labelling..... I am unsure.  I would greatly appreciate any
assistance you could offer.

```

```
Date: 11/19/2001 at 04:04:05
From: Doctor Jeremiah
Subject: Re: Proof for Volume of a Segment of a Sphere

The shape you're talking about is indeed called a 'spherical cap'.
You'll find an illustration at:

http://www.mathforum.org/dr.math/faq/formulas/faq.sphere.html

(scroll down to Spherical Cap). But I can't answer this question

Do you have the radius of the sphere?
If so, do you have the thicknesses of the big cap and little cap?
If not, do you have the radius of at least one of the caps?

It's important because to generate this proof we need to know what
form our answer must take. I could just assume I know what you want
and show you how, but if I guess wrong then I will have wasted a lot
of time for both of us.

Let me know what measurements you want the answer to depend upon

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/19/2001 at 09:42:32
From: Shelaine Kozakavich
Subject: Re: Proof for Volume of a Segment of a Sphere

Good morning Doctor Jeremiah,

information earlier. Here is what I am given for my question to prove
that the volume of the "truncated" spherical cap is:

(Pi/6)(h)(3a^2 + 3b^2 + h^2)

I am told that the top of a spherical cap is cut off by a plane
parallel to the base. In the truncated cap, the base is a circle with
radius a and the top of the circle with radius b, and the height is h.

I was directed to subtract the volume of the smaller top cap from the
volume of the whole spherical cap, but I do not end up with the above
formula for the "truncated" spherical cap.

Thank-you so much for your assistance here!
S. Kozakavich
```

```
Date: 11/19/2001 at 20:41:23
From: Doctor Jeremiah
Subject: Re: Proof for Volume of a Segment of a Sphere

Hi Shelaine,

It could be done by finding the volume of two spherical caps and
subtracting a smaller one from a larger one, and that might be the
best way to go, but I am going to do it directly.

Unfortunately we don't know the radius of the sphere, but we can
figure it out:

|-------a--------|
|----b---|
+++++
+++         +++
+++----b---+        +++  -----------+-
+     \       |       /     +          |
+        \      |      /        +        | h
+-------a--\-----+     /          + ------+-
+  +         \    |    /         +  +      |
+      +       R   |   R       +      +     | Z
+         R     \  |  /     R         +     |
+             +   \ | /   +             +    |
+                + \|/ +                +    |
+                   +                   +   -+-

We have two congruent right angled triangles.

b^2 + (Z+h)^2 = R^2
a^2 + Z^2 = R^2

What we need to do is combine these and get rid of z:

Z^2 = R^2 - a^2
Z = (R^2 - a^2)^(1/2)

b^2 + (Z+h)^2 = R^2
b^2 + Z^2 + 2hZ + h^2 = R^2
b^2 + (R^2 - a^2) + 2h(R^2 - a^2)^(1/2) + h^2 = R^2
2h(R^2 - a^2)^(1/2) = a^2 - b^2 - h^2
(R^2 - a^2)^(1/2) = (a^2 - b^2 - h^2)/(2h)
R^2 - a^2 = (a^2 - b^2 - h^2)^2/(4h^2)

And that means that:
Z = (R^2 - a^2)^(1/2)
Z = ( (a^2 - b^2 - h^2)^2/(4h^2) )^(1/2)
Z = (a^2 - b^2 - h^2)/(2h)

Now we have a Z that does not depend on R at all and we can use R with
no reference to Z.

To integrate a sphere we need to use a disk of dz thickness that has
an area of Pi*x^2 as we move it from z = Z to z = Z+h

+++++
+++         +++
+++-----------------+++  -----+-
+   |         +         |   +    | dz
+-----+--------/|\--------+-----+ -+-
+                |                +
+                 |                 +
+                                     +
+                                     +
+                                       +
+            R^2 = x^2 + z^2            +
+                                       +
+                                     +
+                                     +
+                                   +
+                                 +
+                               +
+                           +
+++                 +++
+++         +++
+++++

z=Z+h
/
Volume =  |  Pi*x^2 dz   <===   x^2 = R^2 - z^2
/
z=Z

z=Z+h
/
Volume =  |  Pi*(R^2 - z^2) dz
/
z=Z

z=Z+h           z=Z+h
/               /
Volume =  |  Pi*R^2 dz -  |  Pi*z^2 dz
/               /
z=Z             z=Z

z=Z+h       z=Z+h
/           /
Volume = Pi*R^2  |  dz - Pi  |  z^2 dz
/           /
z=Z         z=Z

z=Z+h        z=Z+h
|            |
Volume = Pi*R^2*z | - Pi*z^3/3 |
|            |
z=Z          z=Z

Let's test the results of our integration. At this point, if the
limits were -R and R (the whole sphere) we would have this:

z=R          z=R
|            |
Volume = Pi*R^2*z | - Pi*z^3/3 |
|            |
z=-R         z=-R

Volume = (Pi*R^2*(R) - Pi*R^2*(-R)) - (Pi*(R)^3/3 - Pi*(-R)^3/3)
Volume = (Pi*R^2*R + Pi*R^2*R) - (Pi*R^3/3 + Pi*R^3/3)
Volume = 2*Pi*R^3 - 2*Pi*R^3/3
Volume = 4*Pi*R^3/3

So now let's solve the actual problem. And with limits of Z and Z+h we
get:

Volume = (Pi*R^2*(Z+h) - Pi*R^2*(Z)) - (Pi*(Z+h)^3/3 - Pi*(Z)^3/3)
Volume = Pi*R^2*h - Pi*(Z+h)^3/3 + Pi*Z^3/3
Volume = Pi*R^2*h - Pi*(Z^3+3hZ^2+3Zh^2+h^3)/3 + Pi*Z^3/3
Volume = Pi*R^2*h - Pi*(3hZ^2+3Zh^2+h^3)/3
Volume = Pi*R^2*h - Pi*h*Z^2 - Pi*h^2*Z - Pi*h^3/3

And since  Z^2 = R^2 - a^2:

Volume = Pi*R^2*h - Pi*h(R^2-a^2) - Pi*h^2*Z - Pi*h^3/3
Volume = Pi*R^2*h - Pi*h*R^2 + Pi*h*a^2 - Pi*h^2*Z - Pi*h^3/3
Volume = Pi*h*a^2 - Pi*h^2*Z - Pi*h^3/3

And since  Z = (a^2 - b^2 - h^2)/(2h):

Volume = Pi*h*a^2 - Pi*h^2*(a^2 - b^2 - h^2)/(2h) - Pi*h^3/3
Volume = Pi*h*a^2 - Pi*h*(a^2 - b^2 - h^2)/2 - Pi*h^3/3
Volume = Pi*h*a^2 - Pi*h*a^2/2 + Pi*h*b^2/2 + Pi*h^3/2 - Pi*h^3/3
Volume = Pi*h*a^2/2 + Pi*h*b^2/2 + 3*Pi*h^3/6 - 2*Pi*h^3/6
Volume = Pi*h*a^2/2 + Pi*h*b^2/2 + Pi*h^3/6
Volume = Pi*h*a^2/2 + Pi*h*b^2/2 + Pi*h*h^2/6
Volume = 3*Pi*h*a^2/6 + 3*Pi*h*b^2/6 + Pi*h*h^2/6
Volume = Pi*h*(3a^2 + 3b^2 + h^2)/6

Wasn't that messy!?

Let me know if you want help doing it with two full spherical caps or
if you need more details about the way I did it here.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/23/2001 at 00:17:02
From: Shelaine Kozakavich
Subject: Re: Proof for Volume of a Segment of a Sphere

I want to thank you for all of your time and expertise. You provide a
valuable resource here with this Web site and your assistance with
problem questions for those who ask for your help. It is certainly
appreciated! Keep up the good work.

Thanks again,
Shelaine :)
```
Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search