Answer
The centroid of ${\cal W}$ is $\left( {0,0,\frac{{3R\left( {1 - \cos \left( {2{\phi _0}} \right)} \right)}}{{16\left( {1 - \cos {\phi _0}} \right)}}} \right)$.
Work Step by Step
First, we describe of the region ${\cal W}$ in spherical coordinates. Since ${\cal W}$ is bounded by $\phi = {\phi _0}$ and the sphere $\rho = R$, we have
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|0 \le \rho \le R,0 \le \phi \le {\phi _0},0 \le \theta \le 2\pi } \right\}$
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}^{{\phi _0}} \mathop \smallint \limits_{\rho = 0}^R {\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_{\rho = 0}^R {\rho ^2}{\rm{d}}\rho } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{{\phi _0}} \sin \phi {\rm{d}}\phi } \right)$
$ = 2\pi \left( {\frac{1}{3}{\rho ^3}|_0^R} \right)\left( { - \cos \phi |_0^{{\phi _0}}} \right)$
$ = \frac{2}{3}\pi {R^3}\left( {1 - \cos {\phi _0}} \right)$
Recall the definition of the centroid in Section 16.5:
$\bar x = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x{\rm{d}}V$, ${\ \ \ }$ $\bar y = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V$, ${\ \ \ }$ $\bar z = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V$
In spherical coordinates:
$x = \rho \sin \phi \cos \theta $, ${\ \ \ }$ $y = \rho \sin \phi \sin \theta $, ${\ \ \ }$ $z = \rho \cos \phi $
Evaluate the average of the $x$-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}\left( {1 - \cos {\phi _0}} \right)}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{{\phi _0}} \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( {1 - \cos {\phi _0}} \right)}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{{\phi _0}} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^R {\rho ^3}{\rm{d}}\rho } \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta = \sin \theta |_0^{2\pi } = 0$, so $\bar x = 0$.
Evaluate the average of the $y$-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}\left( {1 - \cos {\phi _0}} \right)}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{{\phi _0}} \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( {1 - \cos {\phi _0}} \right)}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{{\phi _0}} {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^R {\rho ^3}{\rm{d}}\rho } \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = - \cos \theta |_0^{2\pi } = 0$, so $\bar y = 0$.
Evaluate the average of the $z$-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}\left( {1 - \cos {\phi _0}} \right)}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{{\phi _0}} \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( {1 - \cos {\phi _0}} \right)}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{{\phi _0}} \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( {1 - \cos {\phi _0}} \right)}}\left( {2\pi } \right)\left( {\frac{1}{2}\mathop \smallint \limits_{\phi = 0}^{{\phi _0}} \sin \left( {2\phi } \right){\rm{d}}\phi } \right)\left( {\frac{1}{4}{\rho ^4}|_0^R} \right)$
$ = \frac{3}{{2\pi {R^3}\left( {1 - \cos {\phi _0}} \right)}}\left( {2\pi } \right)\left( { - \frac{1}{4}\cos \left( {2\phi } \right)|_0^{{\phi _0}}} \right)\left( {\frac{1}{4}{R^4}} \right)$
$ = \frac{{3R}}{{16\left( {1 - \cos {\phi _0}} \right)}}\left( {1 - \cos \left( {2{\phi _0}} \right)} \right)$
$ = \frac{{3R\left( {1 - \cos \left( {2{\phi _0}} \right)} \right)}}{{16\left( {1 - \cos {\phi _0}} \right)}}$
So, the centroid of ${\cal W}$ is $\left( {0,0,\frac{{3R\left( {1 - \cos \left( {2{\phi _0}} \right)} \right)}}{{16\left( {1 - \cos {\phi _0}} \right)}}} \right)$.
