Answer
We prove a famous result of Archimedes.
Work Step by Step
Let $S$ be the surface area of the portion of the sphere of radius $R$ between two horizontal planes $z=a$ and $z=b$ as is shown in Figure 22.
Since the sphere is given by ${x^2} + {y^2} + {z^2} = {R^2}$, we obtain the domain ${\cal D}$ of the surface $S$ in the $xy$-plane:
${\cal D} = \left\{ {\left( {x,y} \right):{R^2} - {b^2} \le {x^2} + {y^2} \le {R^2} - {a^2}} \right\}$
Using $x = r\cos \theta $ and $y = r\sin \theta $, we can parametrize the sphere ${x^2} + {y^2} + {z^2} = {R^2}$ in polar coordinates:
$G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,\sqrt {{R^2} - {r^2}} } \right)$
where the domain is ${\cal D} = \left\{ {\left( {r,\theta } \right):0 \le \theta \le 2\pi ,\sqrt {{R^2} - {a^2}} \le r \le \sqrt {{R^2} - {b^2}} } \right\}$.
${{\bf{T}}_r} = \dfrac{{\partial G}}{{\partial r}} = \left( {\cos \theta ,\sin \theta , - \dfrac{r}{{\sqrt {{R^2} - {r^2}} }}} \right)$
${{\bf{T}}_\theta } = \dfrac{{\partial G}}{{\partial \theta }} = \left( { - r\sin \theta ,r\cos \theta ,0} \right)$
${\bf{N}}\left( {r,\theta } \right) = {{\bf{T}}_r} \times {{\bf{T}}_\theta } = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&{\sin \theta }&{ - \dfrac{r}{{\sqrt {{R^2} - {r^2}} }}}\\
{ - r\sin \theta }&{r\cos \theta }&0
\end{array}} \right|$
${\bf{N}}\left( {r,\theta } \right) = \dfrac{{{r^2}\cos \theta }}{{\sqrt {{R^2} - {r^2}} }}{\bf{i}} + \dfrac{{{r^2}\sin \theta }}{{\sqrt {{R^2} - {r^2}} }}{\bf{j}} + r{\bf{k}}$
$||{\bf{N}}\left( {r,\theta } \right)|| = \sqrt {\dfrac{{{r^4}{{\cos }^2}\theta + {r^4}{{\sin }^2}\theta + {R^2}{r^2} - {r^4}}}{{{R^2} - {r^2}}}} = \dfrac{{Rr}}{{\sqrt {{R^2} - {r^2}} }}$
Evaluate the area of $S$:
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {r,\theta } \right)||{\rm{d}}r{\rm{d}}\theta $
$Area\left( S \right) = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = \sqrt {{R^2} - {b^2}} }^{\sqrt {{R^2} - {a^2}} } \dfrac{{Rr}}{{\sqrt {{R^2} - {r^2}} }}{\rm{d}}r{\rm{d}}\theta $
Write $t = {R^2} - {r^2}$. So, $dt = - 2rdr$. The integral becomes
$Area\left( S \right) = - \dfrac{R}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{t = {b^2}}^{{a^2}} {t^{ - 1/2}}{\rm{d}}t{\rm{d}}\theta $
$ = - R\left( {2\pi } \right)\left( {{t^{1/2}}|_{{b^2}}^{{a^2}}} \right) = - 2\pi R\left( {a - b} \right)$
So, $Area\left( S \right) = 2\pi R\left( {b - a} \right)$.
From Figure 22 we see that the surface area of the corresponding portion of the circumscribed cylinder is also $2\pi R\left( {b - a} \right)$.
Therefore, we conclude that "the surface area of the portion of the sphere of radius $R$ between two horizontal planes $z=a$ and $z=b$ is equal to the surface area of the corresponding portion of the circumscribed cylinder".
Thus proves a famous result of Archimedes.
