Answer
${\rm{Area}}\left( S \right) = 2\pi \bar y \cdot L$
Work Step by Step
By Eq. (14) the surface area of the graph $z = g\left( y \right)$ for $c \le y \le d$, rotated about the $z$-axis is given by:
$Area\left( S \right) = 2\pi \mathop \smallint \limits_c^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$
If $C$ is the graph $z = g\left( y \right)$ in the $yz$-plane, we can parametrize it by ${\bf{c}}\left( y \right) = \left( {y,g\left( y \right)} \right)$. So,
${\bf{c}}'\left( y \right) = \left( {1,g'\left( y \right)} \right)$
$||{\bf{c}}'\left( y \right)|| = \sqrt {1 + g'{{\left( y \right)}^2}} $
Since the line element is $ds = ||{\bf{c}}'\left( y \right)||dy = \sqrt {1 + g'{{\left( y \right)}^2}} dy$, the integral becomes
${\rm{Area}}\left( S \right) = 2\pi \mathop \smallint \limits_c^d y{\rm{d}}s$
But $\bar y = \dfrac{1}{L}\mathop \smallint \limits_C^{} y{\rm{d}}s$. So,
${\rm{Area}}\left( S \right) = 2\pi \bar y \cdot L$
We know that $2\pi \bar y$ is the distance traversed by the center of mass of the curve when it is rotated about the $z$-axis. Thus, the equation tells us that the area of a surface of revolution $S$ is equal to the length $L$ of the generating curve times the distance traversed by the center of mass. Hence, it proves the Pappus's Theorem.
