Answer
The coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,\frac{9}{8}} \right)$.
Work Step by Step
The region ${\cal W}$ bounded by the cone $\phi = \frac{\pi }{3}$ and the sphere $\rho = 2$ is defined in spherical coordinates by
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|0 \le \rho \le 2,0 \le \phi \le \frac{\pi }{3},0 \le \theta \le 2\pi } \right\}$
Evaluate the volume of ${\cal W}$:
$V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /3} \mathop \smallint \limits_{\rho = 0}^2 {\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /3} \sin \phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^2 {\rho ^2}{\rm{d}}\rho } \right)$
$ = - 2\pi \left( {\cos \phi |_0^{\pi /3}} \right)\left( {\frac{1}{3}{\rho ^3}|_0^2} \right)$
$ = - 2\pi \left( {\frac{1}{2} - 1} \right)\left( {\frac{8}{3}} \right) = \frac{8}{3}\pi $
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}{{8\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /3} \mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}{\sin ^2}\phi \cos \theta {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{3}{{8\pi }}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /3} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}{\rm{d}}\rho } \right)$
$ = \frac{3}{{8\pi }}\left( {\sin \theta |_0^{2\pi }} \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /3} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}{\rm{d}}\rho } \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}{{8\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /3} \mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}{\sin ^2}\phi \sin \theta {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{3}{{8\pi }}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /3} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}{\rm{d}}\rho } \right)$
$ = - \frac{3}{{8\pi }}\left( {\cos \theta |_0^{2\pi }} \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /3} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}{\rm{d}}\rho } \right)$
Since $\left( {\cos \theta |_0^{2\pi }} \right) = 0$, 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}{{8\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /3} \mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}\sin \phi \cos \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{3}{{8\pi }}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /3} \sin \phi \cos \phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}{\rm{d}}\rho } \right)$
$ = \frac{3}{4}\left( {\frac{1}{2}\mathop \smallint \limits_{\phi = 0}^{\pi /3} \sin 2\phi {\rm{d}}\phi } \right)\left( {\frac{1}{4}{\rho ^4}|_0^2} \right)$
$ = \frac{3}{8}\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /3} \sin 2\phi {\rm{d}}\phi } \right)\left( 4 \right)$
$ = - \frac{3}{4}\left( {\cos 2\phi |_0^{\pi /3}} \right) = - \frac{3}{4}\left( { - \frac{1}{2} - 1} \right) = \frac{9}{8}$
So, the coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,\frac{9}{8}} \right)$.
