Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
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.  

Thank you for your time.


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 
without more information.

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
and then I will be happy to help you.

- 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,

Thank you for your quick reply! Sorry about not supplying all the 
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

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/