Answer
Using the definition of the $y$-coordinate of the centroid, we prove Pappus's Theorem.
Work Step by Step
We can consider ${\cal D}$ as a vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)|a \le x \le b,{g_1}\left( x \right) \le y \le {g_2}\left( x \right)} \right\}$
By definition, the $y$-coordinate of the centroid is given by
$\bar y = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A$
where $A$ is the area of ${\cal D}$.
Thus,
$\bar y = \frac{1}{A}\mathop \smallint \limits_{x = a}^b \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} y{\rm{d}}y{\rm{d}}x$
$A\bar y = \frac{1}{2}\mathop \smallint \limits_{x = a}^b \left( {{y^2}|_{{g_1}\left( x \right)}^{{g_2}\left( x \right)}} \right){\rm{d}}x$
(1) ${\ \ \ \ \ }$ $2A\bar y = \mathop \smallint \limits_{x = a}^b \left( {{g_2}{{\left( x \right)}^2} - {g_1}{{\left( x \right)}^2}} \right){\rm{d}}x$
Recall from Section 6.3, the volume of the solid obtained by revolving ${\cal D}$ about the $x$-axis is given by Eq. (2) (Section 6.3):
$V = \pi \mathop \smallint \limits_{x = a}^b \left( {{g_2}{{\left( x \right)}^2} - {g_1}{{\left( x \right)}^2}} \right){\rm{d}}x$
By equation (1), the integral on the right-hand side of $V$ is equal to $2A\bar y$. Hence, $V = 2\pi A\bar y$. This proves the Pappus's Theorem.
