# 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 $\sqrt{r^{2}\!-\!x^{2}}$ and the area of the square is $4(\sqrt{r^{2}\!-\!x^{2}})^{2}$. Accordingly, we have the function

 $A(x)\;:=\;4(r^{2}\!-\!x^{2})$

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

 $\frac{1}{2}V\;=\;\int_{0}^{r}\!4(r^{2}\!-\!x^{2})\,dx\;=\;4\!% \operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{\,\quad r}\!\left(r^{2}x-\frac{x% ^{3}}{3}\right)\;=\;\frac{8}{3}r^{3}.$

So the volume in the question is $\frac{16}{3}r^{3}$.  It is $\displaystyle\frac{2}{3}$ of the volume of the cube.

Title Archimedes’ cylinders in cube ArchimedesCylindersInCube 2013-03-22 17:20:51 2013-03-22 17:20:51 pahio (2872) pahio (2872) 7 pahio (2872) Example msc 51M25 msc 51-00 perpendicular cylinders cylinders inscribed in cube SubstitutionNotation