The Amazing Hole in Sphere Problem
A cylindrical hole, 6 inches long, has been drilled straight through the center of a solid sphere. What is the volume remaining in the sphere? If the answer is
, type in the value of
as your answer.
Note- No other data is needed to solve this.
This brilliant problem was first made famous by Martin Gardner with his ‘Mathematical Games’ column published in Scientific American.
The answer is 36.
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1 solution
Try to picture a small sphere, maybe with diameter 7 inches. When you drill through it with a small drill, you will be cutting out a narrow cylinder of height 6 inches, plus two shallow caps.
Now picture a huge sphere, maybe the size of the earth. When you drill through this with a drill almost as wide as the earth, you will be cutting out a very wide cylinder of height 6 inches, plus two caps that take out most of the remaining volume from the sphere.
It just happens that even though the sphere is so much larger, the drill has to take out just as much more in order to make the height of the hole the same, so that the volume left doesn't depend on the size of the sphere itself, or of the hole itself - only on their relation, which is forced by requiring the hole to be 6 inches long.
John W. Campbell, Jr., editor of Astounding Science Fiction, was one of several readers who solved the sphere problem quickly by reasoning adroitly as follows: The problem would not be given unless it has a unique solution. If it has a unique solution, the volume must be a constant which would hold even when the hole is reduced to zero radius. Therefore the residue must equal the volume of a sphere with a diameter of six inches, namely 36π!
The following solution is borrowed from Mathworld ...
A spherical ring is a sphere with a cylindrical hole cut so that the centers of the cylinder and sphere coincide, also called a napkin ring. Let the sphere have radius R and the cylinder radius r.
The hole cut out consists of a cylindrical portion plus two spherical caps. The volume of the entire cylinder is V_(cylinder)=piLr^2,
and the volume of the upper segment is V_(cap)=1/3pih^2(3R-h).
The volume removed upon drilling of a cylindrical hole is then V (hole) = V (cylinder)+2V_(cap)
where the expressions R^2 = r^2+(1/2L)^2
R = 1/2L+h
obtained from trigonometry have been used to re-express the result.
The volume of the spherical ring itself is then given by V (ring) = V (sphere)-V_(hole)
By the final equation, the remaining volume of any center-drilled sphere can be calculated given only the length of the hole. In particular, if the sphere gets bigger while L remains constant, then the circumference of the ring gets bigger, increasing the volume, but the ring gets narrower, decreasing it. The two effects exactly cancel each other out, leading Gardner to term this an "incredible problem."