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# area of spherical zone

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

with radius $r$ and centre $(r,\,0)$. 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

$\displaystyle A\;=\;2\pi\!\int_{{a}}^{{b}}\!y\,\sqrt{1+\left(\frac{dy}{dx}% \right)^{2}}\,dx$ | (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 height 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

$A\;=\;2\pi\!\int_{a}^{b}\sqrt{rx\!-\!x^{2}}\,\sqrt{1+\left(\frac{r\!-\!x}{% \sqrt{rx\!-\!x^{2}}}\right)^{2}}\,dx\;=\;2\pi\!\int_{a}^{b}r\,dx\;=\;2\pi r(b% \!-\!a).$ |

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

$\displaystyle A\;=\;2\pi rh.$ | (2) |

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

$\displaystyle A\;=\;4\pi r^{2}.$ | (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 ($h=\frac{r}{2}$).

## Mathematics Subject Classification

26B15*no label found*53A05

*no label found*51M04

*no label found*

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