Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \sqrt {{x^2} + {y^2} + {z^2}} {\rm{d}}V = \frac{{8\pi }}{5}$
Work Step by Step
We have $f\left( {x,y,z} \right) = \sqrt {{x^2} + {y^2} + {z^2}} $ and the region ${\cal W}:{x^2} + {y^2} + {z^2} \le 2z$.
Since ${x^2} + {y^2} + {z^2} \ge 0$, we can write the region as ${\cal W}:0 \le {x^2} + {y^2} + {z^2} \le 2z$. This implies that $z \ge 0$.
Notice that the region is a sphere of radius $1$ centered at $\left( {0,0,1} \right)$. We verify this statement in the following. Write
${x^2} + {y^2} + {z^2} = 2z$
${x^2} + {y^2} + {z^2} - 2z = 0$
${x^2} + {y^2} + {\left( {z - 1} \right)^2} - 1 = 0$
${x^2} + {y^2} + {\left( {z - 1} \right)^2} = 1$
The last equation is a sphere of radius $1$ centered at $\left( {0,0,1} \right)$. Please see the figure attached.
Since $x = \rho \sin \varphi \cos \theta $, $y = \rho \sin \varphi \sin \theta $ and $z = \rho \cos \varphi $, in spherical coordinates:
$f\left( {\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi } \right) = \rho $
The inequalities ${x^2} + {y^2} + {z^2} \le 2z$ becomes
${\rho ^2} \le 2\rho \cos \varphi $
$\rho \le 2\cos \varphi $
Since $\rho \ge 0$, we obtain the range $0 \le \rho \le 2\cos \varphi $.
Since $z \ge 0$, $\varphi $ ranges from $\varphi = 0$ to $\varphi = \frac{\pi }{2}$. Thus, the region description of ${\cal W}$ in spherical coordinates is given by
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le 2\cos \varphi ,0 \le \varphi \le \frac{\pi }{2},0 \le \theta \le 2\pi } \right\}$
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}^{} \sqrt {{x^2} + {y^2} + {z^2}} {\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^{2\cos \varphi } {\rho ^3}\sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /2} \left( {{\rho ^4}|_0^{2\cos \varphi }} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = 4\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /2} {\cos ^4}\varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
Consider the inner integral: $\mathop \smallint \limits_{\varphi = 0}^{\pi /2} {\cos ^4}\varphi \sin \varphi {\rm{d}}\varphi $
Write $u = \cos \varphi $. So, $du = - \sin \varphi d\varphi $. Thus
$\mathop \smallint \limits_{\varphi = 0}^{\pi /2} {\cos ^4}\varphi \sin \varphi {\rm{d}}\varphi = - \mathop \smallint \limits_{u = 1}^0 {u^4}{\rm{d}}u$
$ = - \frac{1}{5}\left( {{u^5}|_1^0} \right) = \frac{1}{5}$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \sqrt {{x^2} + {y^2} + {z^2}} {\rm{d}}V = 4\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /2} {\cos ^4}\varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{4}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{8\pi }}{5}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \sqrt {{x^2} + {y^2} + {z^2}} {\rm{d}}V = \frac{{8\pi }}{5}$.
