Answer
The coordinates of the center of mass: $\left( {0,\frac{{8{{\sin }^3}{\theta _0}}}{{15\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}} \right)$.
Work Step by Step
We have the mass density $\delta \left( {x,y} \right) = {x^2}$.
In polar coordinates: $\delta \left( {r\cos \theta ,r\sin \theta } \right) = {r^2}{\cos ^2}\theta $.
Let ${\cal D}$ denote the region of the shaded sector in Figure 5. We see that $\theta $ varies from $\theta = \frac{\pi }{2} - {\theta _0}$ to $\theta = \frac{\pi }{2} + {\theta _0}$. Thus, the description of the region ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,\frac{\pi }{2} - {\theta _0} \le \theta \le \frac{\pi }{2} + {\theta _0}} \right\}$
Evaluate the mass of ${\cal D}$:
$M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \mathop \smallint \limits_{r = 0}^1 {r^3}{\cos ^2}\theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \left( {{r^4}|_0^1} \right){\cos ^2}\theta {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} {\cos ^2}\theta {\rm{d}}\theta $
Using the Double-angle formulas in Section 1.4:
${\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)$
we get
$M = \frac{1}{8}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \left( {1 + \cos 2\theta } \right){\rm{d}}\theta $
$ = \frac{1}{8}\left( {\theta + \frac{1}{2}\sin 2\theta } \right)|_{\frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}}$
$ = \frac{1}{8}\left( {\frac{\pi }{2} + {\theta _0} + \frac{1}{2}\sin \left( {\pi + 2{\theta _0}} \right) - \frac{\pi }{2} + {\theta _0} - \frac{1}{2}\sin \left( {\pi - 2{\theta _0}} \right)} \right)$
$ = \frac{1}{8}\left( {2{\theta _0} - \frac{1}{2}\sin 2{\theta _0} - \frac{1}{2}\sin 2{\theta _0}} \right)$
$ = \frac{1}{8}\left( {2{\theta _0} - \sin 2{\theta _0}} \right)$
So, $M = \frac{1}{4}\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)$.
Evaluate ${x_{CM}}$:
${x_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A$
Since ${\cal D}$ is symmetric with respect to the $y$-axis, and $x\delta \left( {x,y} \right) = {x^3}$ is an odd function, we have ${x_{CM}} = 0$.
However, we verify this by direct calculation:
${x_{CM}} = \frac{4}{{{\theta _0} - \sin {\theta _0}\cos {\theta _0}}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \mathop \smallint \limits_{r = 0}^1 {r^4}{\cos ^3}\theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{4}{{5\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \left( {{r^5}|_0^1} \right){\cos ^3}\theta {\rm{d}}\theta $
$ = \frac{4}{{5\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} {\cos ^3}\theta {\rm{d}}\theta $
Using Eq. (6) of the Table of Trigonometric Integrals in Section 8.2:
$\smallint {\cos ^{n - 3}}{\rm{d}}x = \frac{{{{\cos }^{n - 1}}x\sin x}}{n} + \frac{{n - 1}}{n}\smallint {\cos ^{n - 2}}x{\rm{d}}x$
we get
$\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} {\cos ^3}\theta = \frac{{{{\cos }^2}x\sin x}}{3}|_{\frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} + \frac{2}{3}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \cos \theta $
$ = \frac{{{{\sin }^2}{\theta _0}\cos {\theta _0}}}{3} - \frac{{{{\sin }^2}{\theta _0}\cos {\theta _0}}}{3} + \sin \theta |_{\frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}}$
$ = \cos {\theta _0} - \cos {\theta _0} = 0$
So, ${x_{CM}} = 0$.
Evaluate ${y_{CM}}$:
${y_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y\delta \left( {x,y} \right){\rm{d}}A$
$ = \frac{4}{{{\theta _0} - \sin {\theta _0}\cos {\theta _0}}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \mathop \smallint \limits_{r = 0}^1 {r^4}{\cos ^2}\theta \sin \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{4}{{5\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} \left( {{r^5}|_0^1} \right){\cos ^2}\theta \sin \theta {\rm{d}}\theta $
(1) ${\ \ \ \ \ }$ ${y_{CM}} = \frac{4}{{5\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} {\cos ^2}\theta \sin \theta {\rm{d}}\theta $
Consider the integral on the right-hand side: $\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} {\cos ^2}\theta \sin \theta {\rm{d}}\theta $.
Write $u = \cos \theta $. So, $du = - \sin \theta d\theta $. Thus,
$\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} {\cos ^2}\theta \sin \theta {\rm{d}}\theta = - \mathop \smallint \limits_{u = \sin {\theta _0}}^{ - \sin {\theta _0}} {u^2}{\rm{d}}u$
$ = - \frac{1}{3}\left( {{u^3}|_{\sin {\theta _0}}^{ - \sin {\theta _0}}} \right) = - \frac{1}{3}\left( { - {{\sin }^3}{\theta _0} - {{\sin }^3}{\theta _0}} \right)$
$ = \frac{2}{3}{\sin ^3}{\theta _0}$
So, $\mathop \smallint \limits_{\theta = \frac{\pi }{2} - {\theta _0}}^{\frac{\pi }{2} + {\theta _0}} {\cos ^2}\theta \sin \theta {\rm{d}}\theta = \frac{2}{3}{\sin ^3}{\theta _0}$.
Substituting the result back in equation (1) gives:
${y_{CM}} = \frac{4}{{5\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}\left( {\frac{2}{3}{{\sin }^3}{\theta _0}} \right) = \frac{{8{{\sin }^3}{\theta _0}}}{{15\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}$
So, the coordinates of the center of mass: $\left( {0,\frac{{8{{\sin }^3}{\theta _0}}}{{15\left( {{\theta _0} - \sin {\theta _0}\cos {\theta _0}} \right)}}} \right)$.
