Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = - \frac{\pi }{{16}}$
Work Step by Step
We have $f\left( {x,y,z} \right) = y$ and the region
${\cal W}:{x^2} + {y^2} + {z^2} \le 1,x,y,z \le 0$
In spherical coordinates:
$f\left( {\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi } \right) = \rho \sin \varphi \sin \theta $
The region ${\cal W}$ is part of the sphere of radius $1$. Since $x,y,z \le 0$, so the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le 1,\frac{\pi }{2} \le \varphi \le \pi ,\pi \le \theta \le \frac{{3\pi }}{2}} \right\}$
Using Eq. (8), the triple integral of $f\left( {x,y,z} \right)$ over ${\cal W}$ in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = \mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \mathop \smallint \limits_{\varphi = \pi /2}^\pi \mathop \smallint \limits_{\rho = 0}^1 {\rho ^3}{\sin ^2}\varphi \sin \theta {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \mathop \smallint \limits_{\varphi = \pi /2}^\pi \left( {{\rho ^4}|_0^1} \right){\sin ^2}\varphi \sin \theta {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \mathop \smallint \limits_{\varphi = \pi /2}^\pi {\sin ^2}\varphi \sin \theta {\rm{d}}\varphi {\rm{d}}\theta $
Using Eq. (1) in Example 3 of Section 8.2:
$\smallint {\sin ^2}x{\rm{d}}x = \frac{x}{2} - \frac{{\sin 2x}}{4} + C$
we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = \frac{1}{4}\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \left( {\left( {\frac{\theta }{2} - \frac{{\sin 2\theta }}{4}} \right)|_{\pi /2}^\pi } \right)\sin \theta {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \left( {\frac{\pi }{2} - \frac{\pi }{4}} \right)\sin \theta {\rm{d}}\theta $
$ = \frac{\pi }{{16}}\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \sin \theta {\rm{d}}\theta $
$ = - \frac{\pi }{{16}}\left( {\cos \theta |_\pi ^{3\pi /2}} \right) = - \frac{\pi }{{16}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = - \frac{\pi }{{16}}$.
