Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = 8\pi $
Work Step by Step
We have $f\left( {x,y,z} \right) = 1$ and the region
${\cal W}:{x^2} + {y^2} + {z^2} \le 4z,z \ge \sqrt {{x^2} + {y^2}} $
Since ${x^2} + {y^2} + {z^2} = {\rho ^2}$ and $z = \rho \cos \varphi $, in spherical coordinates the inequalities above become
${\rho ^2} \le 4\rho \cos \varphi $, ${\ \ \ }$ $\rho \cos \varphi \ge \rho \sin \varphi $
(1) ${\ \ \ \ }$ $\rho \le 4\cos \varphi $, ${\ \ \ }$ $\tan \varphi \le 1$
Since $\rho \ge 0$, the first inequality in equation (1) gives the range $0 \le \rho \le 4\cos \varphi $.
From the second inequality in equation (1) we obtain $0 \le \varphi \le \frac{\pi }{4}$.
Thus, the region description of ${\cal W}$ in spherical coordinates is given by
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le 4\cos \varphi ,0 \le \varphi \le \frac{\pi }{4},0 \le \theta \le 2\pi } \right\}$
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}^{} 1{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /4} \mathop \smallint \limits_{\rho = 0}^{4\cos \varphi } {\rho ^2}\sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /4} \left( {{\rho ^3}|_0^{4\cos \varphi }} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{64}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /4} {\cos ^3}\varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
Consider the inner integral: $\mathop \smallint \limits_{\varphi = 0}^{\pi /4} {\cos ^3}\varphi \sin \varphi {\rm{d}}\varphi $
Write $u = \cos \varphi $. So, $du = - \sin \varphi d\varphi $. Thus
$\mathop \smallint \limits_{\varphi = 0}^{\pi /4} {\cos ^3}\varphi \sin \varphi {\rm{d}}\varphi = - \mathop \smallint \limits_{u = 1}^{1/\sqrt 2 } {u^3}{\rm{d}}u$
$ = - \frac{1}{4}\left( {{u^4}|_1^{1/\sqrt 2 }} \right) = - \frac{1}{4}\left( {\frac{1}{4} - 1} \right) = \frac{3}{{16}}$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{{64}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /4} {\cos ^3}\varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{64}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{3}{{16}}} \right){\rm{d}}\theta $
$ = 4\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 8\pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = 8\pi $, which is the volume of ${\cal W}$.
