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 r.
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 x. By the Pythagorean theorem
, half of the side of the square is √r2-x2 and the area of the square is
4(√r2-x2)2. Accordingly, we have the function
A(x):= |
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 |