Hole in a SphereDate: 12/25/96 at 13:16:17 From: Vern Klein Subject: A mathematical oddity... Is anyone there acquainted with the mathematical oddity implied by this exchange that appeared in the "Ask Marilyn" column in _Parade_ magazine on 12-15-96? We hope someone may be able to state the "proposition" in unambiguous terms. Considering the mail volume, who knows when an answer might be had from Marilyn? Quoting from the column of 12-15-96: Recently you published a question about boring through a sphere and leaving a 6-inch cylindrical hole through the center. What is the volume of the remainder of the sphere? You replied, Amazing! The problem appears to be missing enough data - like the size of the sphere - for a solution, but it turns out that the volume of the remainder of the sphere is always the same, no matter what! Whether you bore a slim hole or a fat one IN YOUR 6-INCH SPHERE, you'll always wind up with 113.09724 cubic inches of sphere remaining. This is incorrect. You inadvertently referred to the sphere as a 6-inch sphere. But as you said yourself, the size of the sphere is unknown. It is the cylindrical hole that is 6 inches long. This factor accounts for the curious constancy of the net volume of the remainder of the sphere. Marilyn responds: The sentence should have read, "Whether you bore a slim hole or a fat one IN THIS CASE [meaning a 6-inch hole, not a 6-inch sphere], you'll always wind up with 113.09724 cubic inches of sphere remaining. I (Vern) can make no sense of this whatever and feel I must not understand the problem as originally presented. Any comment or reference to a source where we might find a statement of the problem will be much appreciated. Thank you for your consideration and may the new year be good to all of you there! Vern and Klein Date: 12/26/96 at 13:46:30 From: Doctor Rob Subject: Re: A mathematical oddity... Vern, This is one of my favorite oddities. I have known about this one for about 20 years, so it isn't very new. The idea is that the six inches is measuring the height of the cylindrical surface of the hole in the piece that is left, not the height of the part removed. A typical diagram might be something like the following (pardon the attempt at ASCII Art!): |<----2r--->| A A ,- - - - - -. ----- /| /| |\ ^ / | / | | \ | / | , | | . | / | | | .O | | 6" / | | | | | | 2R/ |6 ` | | ' | / | \ | | / | / | \| |/ v / | (angle C = 90 degrees) `- - - - - -' ----- ---------- B C B 2r C |<------2R------->| Here I am using R for the radius of the sphere and r for the radius of the cylindrical hole. There is a relationship between r and R given by the Pythagorean theorem in triangle ABC: (2r)^2 + 6^2 = (2R)^2, or r^2 + 9 = R^2 The volume of the remaining piece can be computed by integration since it is a solid of rotation obtained by revolving a segment of a circle (say the one bounded by the chord AC and the nearby arc of the circle in the above diagram) about the axis of the cylinder. The result is V = 6*Pi*(R^2 - r^2 - 3), and when you use R^2 - r^2 = 3^2 = 9, you get V = 36*Pi = 113.09724 cubic inches. This is completely independent of both R and r, somewhat surprisingly. Another way of computing the volume of the remaining piece is to start with the volume of the sphere, then subtract the volumes of the two caps cut off the top and bottom of the sphere, then subtract the volume of the right circular cylinder. The cylinder has volume 6*Pi*r^2, and each cap has volume Pi*((2/3)R^3 - 3*R^2 + 9), and the sphere has volume (4/3)*Pi*r^3. You get the same result: V = 6*Pi*(R^2 - r^2 - 3) = 36*Pi If you are trying to find the volume of a 6-inch diameter sphere with no hole, this corresponds to a sphere with a hole with diameter zero. In fact, when this problem was posed to a famous mathematician, he gave the correct answer of 36*Pi almost immediately. His reasoning was that if the problem made sense, the answer must be independent of the hole diameter, which he then set to zero. The remaining volume was then the volume of a sphere with diameter 6 inches, which is (1/6)*Pi*6^3 = 36*Pi. If you need more help, write back again. Happy New Year to you too! -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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