Answer
(a) we prove that $\bar x = 0$.
This implies that the centroid lies on the $y$-axis.
(b) we show that ${M_y} = 0$. Hence, ${x_{CM}} = 0$.
Work Step by Step
(a) Referring to Figure 22, we can consider ${\cal D}$ as a vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)| - a \le x \le a,{g_1}\left( x \right) \le y \le {g_2}\left( x \right)} \right\}$
By definition, the $x$-coordinate of the centroid is given by
$\bar x = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A$
where $A$ is the area of ${\cal D}$.
Using the definition of ${\cal D}$, we evaluate
$\bar x = \frac{1}{A}\mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{A}\mathop \smallint \limits_{x = - a}^a x\left( {y|_{{g_1}\left( x \right)}^{{g_2}\left( x \right)}} \right){\rm{d}}x$
$ = \frac{1}{A}\mathop \smallint \limits_{x = - a}^a x\left( {{g_2}\left( x \right) - {g_1}\left( x \right)} \right){\rm{d}}x$
Since both ${g_1}\left( x \right)$ and ${g_2}\left( x \right)$ are even functions, ${g_2}\left( x \right) - {g_1}\left( x \right)$ is also an even function.
However, the integrand $x\left( {{g_2}\left( x \right) - {g_1}\left( x \right)} \right)$ is an odd function. Therefore, the integral above is zero, that is, $\bar x = 0$. This implies that the centroid lies on the $y$-axis.
(b) The moment of ${\cal D}$ with respect to the $y$-axis is given by
${M_y} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - a}^0 \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x + \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Consider the first integral on the right-hand side. We can transform $x \to - x$ and write:
(1) ${\ \ \ \ }$ $\mathop \smallint \limits_{x = - a}^0 \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{ - x = - a}^0 \mathop \smallint \limits_{y = {g_1}\left( { - x} \right)}^{{g_2}\left( { - x} \right)} \left( { - x} \right)\delta \left( { - x,y} \right){\rm{d}}y{\rm{d}}\left( { - x} \right)$
$ = \mathop \smallint \limits_{x = a}^0 \mathop \smallint \limits_{y = {g_1}\left( { - x} \right)}^{{g_2}\left( { - x} \right)} x\delta \left( { - x,y} \right){\rm{d}}y{\rm{d}}x$
$ = - \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = {g_1}\left( { - x} \right)}^{{g_2}\left( { - x} \right)} x\delta \left( { - x,y} \right){\rm{d}}y{\rm{d}}x$
Since both ${g_1}\left( x \right)$ and ${g_2}\left( x \right)$ are even functions, we have
${g_1}\left( { - x} \right) = {g_1}\left( x \right)$, ${\ \ \ }$ ${g_2}\left( { - x} \right) = {g_2}\left( x \right)$
If the mass density satisfies $\delta \left( { - x,y} \right) = \delta \left( {x,y} \right)$, then equation (1) becomes
$\mathop \smallint \limits_{x = - a}^0 \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x = - \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Substituting this result back in ${M_y}$ gives
${M_y} = - \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x + \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = {g_1}\left( x \right)}^{{g_2}\left( x \right)} x\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x = 0$
By definition, ${x_{CM}} = \frac{{{M_y}}}{M}$. Therefore, ${x_{CM}} = 0$.
