area of spherical zoneAreaOfSphericalZone2008-08-15 17:03:372008-08-18 11:45:54Derivationparts of a ballarea of sphere% this is the default PlanetMath preamble. as your knowledge
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\PMlinkescapeword{formula}
Let us consider the circle
$$(x-r)^2+y^2 = r^2$$
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
\begin{align}
A = 2\pi\!\int_{a}^{b}y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx
\end{align}
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 \PMlinkescapetext{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
$$A = 2\pi\!\int_a^b\sqrt{rx-x^2}\,\sqrt{1+\left(\frac{r-x}{\sqrt{rx-x^2}}\right)^2}\,dx =
2\pi\!\int_a^br\,dx = 2\pi r(b\!-\!a).$$
Hence the area of a spherical zone (and also of a spherical calotte) is
\begin{align}
A = 2\pi rh.
\end{align}
From here one obtains as a special case \,$h = 2r$\, the area of the whole sphere:
\begin{align}
A = 4\pi r^2.
\end{align}