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, 6Ï€ (R2 – 9), to twice the volume of the spherical cap, and subtract the total from the volume of the sphere, 4Ï€R3/3. The volume of the cap is obtained by the following formula, in which A stands for its altitude and r for its radius: Ï€A(3r 2 + A2)/6. When this computation is made, all terms obligingly cancel out except 36Ï€— 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 36Ï€.

Tags:
test iq