Answer
The coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,\frac{{3R}}{8}} \right)$.
Work Step by Step
The volume of a sphere ${x^2} + {y^2} + {z^2} \le {R^2}$ is $\frac{4}{3}\pi {R^3}$. Therefore, the volume of the hemisphere ${x^2} + {y^2} + {z^2} \le {R^2}$ is $\frac{2}{3}\pi {R^3}$.
Since $z \ge 0$, the description of the region ${\cal W}$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|0 \le \rho \le R,0 \le \phi \le \frac{\pi }{2},0 \le \theta \le 2\pi } \right\}$
In spherical coordinates:
$x = \rho \sin \phi \cos \theta $, ${\ \ }$ $y = \rho \sin \phi \sin \theta $, ${\ \ }$ $z = \rho \cos \phi $
Evaluate the average of $x$-coordinate:
$\bar x = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x{\rm{d}}V = \frac{3}{{2\pi {R^3}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^R {\rho ^3}{\sin ^2}\phi \cos \theta {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{3}{{2\pi {R^3}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /2} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^R {\rho ^3}{\rm{d}}\rho } \right)$
Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
$\bar x = \frac{3}{{2\pi {R^3}}}\left( {\sin \theta |_0^{2\pi }} \right)\left( {\frac{1}{2}\left( {1 - \cos 2\phi } \right)|_0^{\pi /2}} \right)\left( {\frac{1}{4}{\rho ^4}|_0^R} \right)$
Since $\left( {\sin \theta |_0^{2\pi }} \right) = 0$, so $\bar x = 0$.
Evaluate the average of $y$-coordinate:
$\bar y = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = \frac{3}{{2\pi {R^3}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^R {\rho ^3}{\sin ^2}\phi \sin \theta {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{3}{{2\pi {R^3}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /2} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^R {\rho ^3}{\rm{d}}\rho } \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = - \left( {\cos \theta |_0^{2\pi }} \right) = 0$, so $\bar y = 0$.
Evaluate the average of $z$-coordinate:
$\bar z = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{3}{{2\pi {R^3}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^R {\rho ^3}\sin \phi \cos \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{3}{{2\pi {R^3}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /2} \sin \phi \cos \phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^R {\rho ^3}{\rm{d}}\rho } \right)$
$ = \frac{3}{{2\pi {R^3}}}\left( {2\pi } \right)\left( {\frac{1}{2}\mathop \smallint \limits_{\phi = 0}^{\pi /2} \sin 2\phi {\rm{d}}\phi } \right)\left( {\frac{1}{4}{\rho ^4}|_0^R} \right)$
$ = \frac{3}{{{R^3}}}\left( { - \frac{1}{4}\cos 2\phi |_0^{\pi /2}} \right)\left( {\frac{1}{4}{R^4}} \right)$
$ = \frac{{3R}}{4}\left( { - \frac{1}{4}\left( { - 1 - 1} \right)} \right) = \frac{{3R}}{8}$
So, the coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,\frac{{3R}}{8}} \right)$.
