Archimedes’ cylinders in cube
The following problem has been solved by Archimedes:
Two distinct circular cylinders are inscribed in a cube; the axes thus intersect each other perpendicularly. Determine the volume common to both cylinders, when the radius of the base of the cylinders is .
If the solid common to both cylinders is cut with a plane parallel to the axes of both cylinders, the figure of intersection is a square. Denote the distance of the plane from the center of the cube be . By the Pythagorean theorem, half of the side of the square is and the area of the square is . Accordingly, we have the function
for the area of the intersection square. If we let here to grow from to , then half of the given solid is got. By the volume of the parent entry (http://planetmath.org/VolumeAsIntegral), the half volume of the solid is
So the volume in the question is . It is of the volume of the cube.
Title | Archimedes’ cylinders in cube |
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Canonical name | ArchimedesCylindersInCube |
Date of creation | 2013-03-22 17:20:51 |
Last modified on | 2013-03-22 17:20:51 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 7 |
Author | pahio (2872) |
Entry type | Example |
Classification | msc 51M25 |
Classification | msc 51-00 |
Synonym | perpendicular cylinders |
Synonym | cylinders inscribed in cube |
Related topic | SubstitutionNotation |