Archimedes’ cylinders in cube


The following problem has been solved by Archimedes:

Two distinct circular cylinders are inscribedMathworldPlanetmath 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 parallelMathworldPlanetmathPlanetmath to the axes of both cylinders, the figure of intersection is a square.  Denote the distanceMathworldPlanetmath of the plane from the center of the cube be x.  By the Pythagorean theoremMathworldPlanetmathPlanetmath, 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):= 4(r2-x2)

for the area of the intersection square.  If we let x here to grow from 0 to r, 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

12V=0r4(r2-x2)𝑑x= 4/x=0r(r2x-x33)=83r3.

So the volume in the question is 163r3.  It is 23 of the volume of the cube.

Title Archimedes’ cylinders in cube
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