sleepaholic
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A hole of radius r is bored through the center of a sphere of radius R > r. Find the volume of the remaining portion of the sphere.
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volumes: A hole of radius r is bored through the center of a
- Thread starter sleepaholic
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Re: volumes: A hole of radius r is bored through the center
John W. Cambell Jr., editor of Astounding Science Fiction, probably offered the best explanation of the hole in the sphere problem for the untrained mathematical eye. The problem would never have been created unless it had 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 36Pi. In other words, the residue is constant regardless of the hole's diameter or the size of the sphere.
Geometry or calculus will lead you to the correct answer. Geometrically, let the radius of the sphere be R. The height of the remaining volume is given as 6. The radius of the cylindrical hole created is then sqrt(R^2 - 9). The height of the spherical end caps at the ends of the cylindrical hole is (R - 3). The remaining volume is then the total volume of the sphere minus the volume of the cylindrical hole minus the volume of the two end caps. The volume of the cylindrical hole is Vh = 6Pi(R^2 - 9). The volume of the end cap is Vc = Pih(3R^2 + h^2)/6. Substituting, expanding, and simplifying, the remaining volume becomes 36Pi, a quantity totally independent of the radius of the sphere. Working it out in general terms with a hole length L, the remaining volume becomes Vr = PiL^3/6.sleepaholic said:A hole of radius r is bored through the center of a sphere of radius R > r. Find the volume of the remaining portion of the sphere.
John W. Cambell Jr., editor of Astounding Science Fiction, probably offered the best explanation of the hole in the sphere problem for the untrained mathematical eye. The problem would never have been created unless it had 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 36Pi. In other words, the residue is constant regardless of the hole's diameter or the size of the sphere.