# 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

$$A(x):=\mathrm{\hspace{0.33em}4}({r}^{2}-{x}^{2})$$ |

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

$$\frac{1}{2}V={\int}_{0}^{r}4({r}^{2}-{x}^{2})\mathit{d}x=\mathrm{\hspace{0.33em}4}\underset{x=0}{\overset{r}{/}}\left({r}^{2}x-\frac{{x}^{3}}{3}\right)=\frac{8}{3}{r}^{3}.$$ |

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