Answer
The coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {\frac{{4R}}{{3\pi }},\frac{{4R}}{{3\pi }}} \right)$.
Work Step by Step
We have the region ${\cal D}$, a quarter circle defined by ${x^2} + {y^2} \le {R^2}$, $x \ge 0$, $y \ge 0$.
In polar coordinates, the description of ${\cal D}$ is
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le R,0 \le \theta \le \frac{\pi }{2}} \right\}$
Since the area of a disk of radius $R$ is $\pi {R^2}$, the area of a quarter circle is $A = \frac{1}{4}\pi {R^2}$.
Evaluate 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 = \frac{4}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^R {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{4}{{3\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {{r^3}|_0^R} \right)\cos \theta {\rm{d}}\theta $
$ = \frac{{4R}}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \cos \theta {\rm{d}}\theta $
$ = \frac{{4R}}{{3\pi }}\left( {\sin \theta |_0^{\pi /2}} \right) = \frac{{4R}}{{3\pi }}$
Evaluate 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 = \frac{4}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^R {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{4}{{3\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {{r^3}|_0^R} \right)\sin \theta {\rm{d}}\theta $
$ = \frac{{4R}}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \sin \theta {\rm{d}}\theta $
$ = - \frac{{4R}}{{3\pi }}\left( {\cos \theta |_0^{\pi /2}} \right) = \frac{{4R}}{{3\pi }}$
So, the coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {\frac{{4R}}{{3\pi }},\frac{{4R}}{{3\pi }}} \right)$.
