The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Cutting a Cylinder out of a Sphere

Date: 02/25/99 at 06:45:37
From: Minesh Gajjar
Subject: Remaining Volume of a cut Sphere

I have a geometry problem:

A cylindrical hole has been drilled directly through the centre of a 
sphere. The length of the cylinder is 6 inches. What is the volume 
remaining in the sphere?

Thank you.

Date: 02/25/99 at 15:11:10
From: Doctor Rob
Subject: Re: Remaining Volume of a cut Sphere

There does not seem to be enough data to solve this problem, yet it 
does have a solution. In order for this to be true, the solution must 
be independent of the radius of the cylindrical hole. That means that 
we can assume that the cylindrical hole has radius zero, and compute 
the correct answer. Then the diameter of the sphere is 6 inches, and 
the volume of the sphere will give you the answer.

There is a more direct approach using the following diagram, with the
cylindrical hole bored horizontally with axis PQ through the center O
of the sphere:

                    ,' |\     6      | `.
                  ,'   | \           |   `.
                 /     |  \R         |r    \
                /     r|   \         |      \
               .       |    \        |       .
               |  R-3  |  3  \   3   |  R-3  |
               |       |      O      |       |
               .       |             |       '
                \     r|             |r     /
                 \     |             |     /
                  `.   |             |   ,'
                    `. |      6      | ,'

O is the center of the sphere, R its radius, and r the radius of the
cylindrical hole. Then by the Pythagorean Theorem, r^2 = R^2 - 9.



for the formulas used below. The volume of the sphere is 4*Pi*R^3/3.
The two missing spherical caps have volume (Pi/6)*(3*r^2+[R-3]^2)*(R-
3), and the cylinder has volume Pi*r^2*6. The remaining volume is then

   V = 4*Pi*R^3/3 - (Pi/3)*(3*r^2+[R-3]^2)*(R-3) - Pi*r^2*6,
     = 4*Pi*R^3/3 - (Pi/3)*(3*[R^2-9]+[R-3]^2)*(R-3) - Pi*(R^2-9)*6,
     = (Pi/3)*(4*R^3 - [R-3]^2*[4*R+6] - 18*[R^2-9]),

which simplifies to the correct answer, independent of R or r.

For other similar questions in the Dr. Math archives see:

  What is the Volume of the Sphere?   

  Hole in a Sphere   

- Doctor Rob, The Math Forum   
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Puzzles

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- The Math Forum at NCTM. All rights reserved.