Hole in the Sphere

 


THIS IS an incredible problem—incredible because it seems to lack sufficient data for a solution. A cylindrical hole six inches long has been drilled straight through the center of a solid sphere. What is the volume remaining 

The answer:


Without resorting to calculus, the problem can be solved as follows. Let R be the radius of the sphere. As the first illustration indicates, the radius of the cylindrical hole will then be the square root of R2 – 9, and the altitude of the spherical caps at each end of the cylinder will be R – 3. To determine the residue after the cylinder and caps have been removed, we add the volume of the cylinder, 6π (R2 – 9), to twice the volume of the spherical cap, and subtract the total from the volume of the sphere, 4πR3/3. The volume of the cap is obtained by the following formula, in which A stands for its altitude and r for its radius: πA(3r 2 + A2)/6. When this computation is made, all terms obligingly cancel out except 36π— the volume of the residue in cubic inches. In other words, the residue is constant regardless of the hole’s diameter or the size of the sphere! The earliest reference I have found for this beautiful problem is on page 86 of Samuel I. Jones’s Mathematical Nuts, self-published, Nashville, 1932. A two-dimensional analog of the problem appears on page 93 of the same volume. Given the longest possible straight line that can be drawn on a circular track of any dimensions [see second figure], the area of the track will equal the area of a circle having the straight line as a diameter. 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. 






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