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# 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 formula of the parent entry, 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.

## Mathematics Subject Classification

51M25*no label found*51-00

*no label found*

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