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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.
Minesh


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  |
              P+-------+------+------+-------+Q
               |       |      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.

See 

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

and

 http://mathforum.org/dr.math/faq/formulas/faq.cylinder.html   

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?
  http://mathforum.org/dr.math/problems/gameshow.html   

  Hole in a Sphere
  http://mathforum.org/dr.math/problems/klein12.30.96.html   

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Puzzles

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