# area of spherical zone

Let us consider the circle

$${(x-r)}^{2}+{y}^{2}={r}^{2}$$ |

with radius $r$ and centre $(r,\mathrm{\hspace{0.17em}0})$. A spherical zone may be thought to be formed when an arc of the circle rotates around the $x$-axis. For finding the are of the zone, we can use the formula

$A=\mathrm{\hspace{0.33em}2}\pi {\displaystyle {\int}_{a}^{b}}y\sqrt{1+{\left({\displaystyle \frac{dy}{dx}}\right)}^{2}}\mathit{d}x$ | (1) |

of the entry area of surface of revolution. Let the ends of the arc correspond the values $a$ and $b$ of the abscissa^{} such that $b-a=h$ is the of the spherical zone. In the formula, we must use the solved form

$$y=(\pm )\sqrt{rx-{x}^{2}}$$ |

of the equation of the circle. The formula then yields

$$A=\mathrm{\hspace{0.33em}2}\pi {\int}_{a}^{b}\sqrt{rx-{x}^{2}}\sqrt{1+{\left(\frac{r-x}{\sqrt{rx-{x}^{2}}}\right)}^{2}}\mathit{d}x=\mathrm{\hspace{0.33em}2}\pi {\int}_{a}^{b}r\mathit{d}x=\mathrm{\hspace{0.33em}2}\pi r(b-a).$$ |

Hence the area of a spherical zone (and also of a spherical calotte) is

$A=\mathrm{\hspace{0.33em}2}\pi rh.$ | (2) |

From here one obtains as a special case $h=2r$ the area of the whole sphere:

$A=\mathrm{\hspace{0.33em}4}\pi {r}^{2}.$ | (3) |

Remark. The formula (2) implies that the centre of mass of a half-sphere is at the halfway point of the axis of symmetry^{} ($h=\frac{r}{2}$).

Title | area of spherical zone |
---|---|

Canonical name | AreaOfSphericalZone |

Date of creation | 2013-03-22 18:19:05 |

Last modified on | 2013-03-22 18:19:05 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 9 |

Author | pahio (2872) |

Entry type | Derivation |

Classification | msc 26B15 |

Classification | msc 53A05 |

Classification | msc 51M04 |

Synonym | area of spherical calotte |

Related topic | AreaOfTheNSphere |

Related topic | CentreOfMassOfHalfDisc |