Answer
The coordinates of the centroid: $\left( {\bar x,\bar y} \right) \simeq \left( {0.56,0} \right)$
Work Step by Step
We have the shaded region as is shown in Figure 12 bounded by ${r^2} = \cos 2\theta $ for $x \ge 0$.
Notice that $r=0$ when $\theta = - \frac{\pi }{4}$ and $\theta = \frac{\pi }{4}$. As $\theta$ varies from $ - \frac{\pi }{4}$ to $\frac{\pi }{4}$, the ray intersects the curve at $r = \sqrt {\cos 2\theta } $. Thus, the description of the shaded region ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le \sqrt {\cos 2\theta } , - \frac{\pi }{4} \le \theta \le \frac{\pi }{4}} \right\}$
The area of ${\cal D}$ in polar coordinates is given by
$A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \mathop \smallint \limits_{r = 0}^{\sqrt {\cos 2\theta } } r{\rm{d}}r{\rm{d}}\theta $
Using a computer algebra system we obtain $A = \frac{1}{2}$.
The average of $x$-coordinate in polar coordinates
$\bar x = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = 2\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \mathop \smallint \limits_{r = 0}^{\sqrt {\cos 2\theta } } {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta $
Using a computer algebra system we obtain $\bar x = \frac{\pi }{{4\sqrt 2 }}$.
The average of $y$-coordinate in polar coordinates
$\bar y = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = 2\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \mathop \smallint \limits_{r = 0}^{\sqrt {\cos 2\theta } } {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta $
Using a computer algebra system we obtain $\bar y = 0$.
So, the coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {\frac{\pi }{{4\sqrt 2 }},0} \right) \simeq \left( {0.56,0} \right)$.
