Answer
The $z$-coordinate of the center of mass:
${z_{CM}} \simeq 0.3395$
Work Step by Step
Let ${\cal W}$ be the region of the first octant of the unit sphere.
Using spherical coordinates, the description of ${\cal W}$ is
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|0 \le \rho \le 1,0 \le \phi \le \frac{\pi }{2},0 \le \theta \le \frac{\pi }{2}} \right\}$
The mass density $\delta \left( {x,y,z} \right) = y$ in spherical coordinates:
$\delta \left( {\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi } \right) = \rho \sin \phi \sin \theta $
Step 1. Evaluate the mass of ${\cal W}$:
$M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \delta \left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^1 {\rho ^3}{\sin ^2}\phi \sin \theta {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{\pi /2} \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}^1 {\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
$M = \left( {\mathop \smallint \limits_{\theta = 0}^{\pi /2} \sin \theta {\rm{d}}\theta } \right)\left( {\frac{1}{2}\mathop \smallint \limits_{\phi = 0}^{\pi /2} \left( {1 - \cos 2\phi } \right){\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^1 {\rho ^3}{\rm{d}}\rho } \right)$
$ = - \left( {\cos \theta |_0^{\pi /2}} \right)\left( {\frac{1}{2}\left( {\phi - \frac{1}{2}\sin 2\phi } \right)|_0^{\pi /2}} \right)\left( {\frac{1}{4}{\rho ^4}|_0^1} \right)$
$ = - \frac{1}{8}\left( { - 1} \right)\left( {\frac{\pi }{2}} \right) = \frac{\pi }{{16}}$
Step 2. Evaluate the $z$-coordinate of the center of mass:
In spherical coordinates: $z = \rho \cos \phi $.
${z_{CM}} = \frac{{{M_{xy}}}}{M} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{{16}}{\pi }\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^1 {\rho ^4}\cos \phi {\sin ^2}\phi \sin \theta {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{{16}}{\pi }\left( {\mathop \smallint \limits_{\theta = 0}^{\pi /2} \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /2} \cos \phi {{\sin }^2}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^1 {\rho ^4}{\rm{d}}\rho } \right)$
Write $u = \sin \phi $. So, $du = \cos \phi d\phi $. Thus,
${z_{CM}} = - \frac{{16}}{\pi }\left( {\cos \theta |_0^{\pi /2}} \right)\left( {\mathop \smallint \limits_{u = 0}^1 {u^2}{\rm{d}}u} \right)\left( {\frac{1}{5}{\rho ^5}|_0^1} \right)$
$ = - \frac{{16}}{\pi }\left( { - 1} \right)\left( {\frac{1}{3}\left( {{u^3}|_0^1} \right)} \right)\left( {\frac{1}{5}} \right) = \frac{{16}}{{15\pi }} \simeq 0.3395$
So, the $z$-coordinate of the center of mass: ${z_{CM}} \simeq 0.3395$.
