Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \rho {\rm{d}}V = \frac{{79\pi }}{{12}}$
Work Step by Step
We have $f\left( {x,y,z} \right) = \rho $ and the region ${\cal W}:{x^2} + {y^2} + {z^2} \le 4,z \le 1,x \ge 0$.
From the region description, we see that ${\cal W}$ is part of a sphere of radius $2$, bounded above by the plane $z=1$ and bounded below by the sphere $\rho = 2$. Since $x \ge 0$, ${\cal W}$ is located at the right-hand side of the $y$-axis. We notice from the figure attached, each ray begins at the origin and ends on the plane $z=1$ until it touches the intersection of the plane and the sphere. Then, it ends on the sphere until the angle of the ray is $\varphi = \pi $. Thus, we divide the region into two subregions ${{\cal W}_1}$ and ${{\cal W}_2}$, which are divided by the angle at the intersection ${\varphi _0}$ of the plane and the sphere. Thus, ${\cal W} = {{\cal W}_1}\bigcup {{\cal W}_2}$.
Next, we find the angle of the intersection ${\varphi _0}$ by solving:
$z = 2\cos {\varphi _0} = 1$
$\cos {\varphi _0} = \frac{1}{2}$, ${\ \ \ \ \ }$ ${\varphi _0} = \frac{\pi }{3}$
1. Subregion ${{\cal W}_1}$
Since $z = \rho \cos \varphi $, the plane $z=1$ in spherical coordinates is
$\rho \cos \varphi = 1$
$\rho = \frac{1}{{\cos \varphi }}$
Thus, $\rho$ ranges from $0$ to $\rho = \frac{1}{{\cos \varphi }}$.
In region ${{\cal W}_1}$, $\varphi $ ranges from $0$ to ${\varphi _0} = \frac{\pi }{3}$.
Since $x \ge 0$, the range of $\theta$ is $ - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}$.
Thus, the region description of ${{\cal W}_1}$ in spherical coordinates is given by
${{\cal W}_1} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le \frac{1}{{\cos \varphi }},0 \le \varphi \le \frac{\pi }{3}, - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}} \right\}$
Using Eq. (8), the triple integral of $f\left( {x,y,z} \right)$ over ${{\cal W}_1}$ in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal W}_1}}^{} \rho {\rm{d}}V = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \mathop \smallint \limits_{\rho = 0}^{1/\cos \varphi } {\rho ^3}\sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \left( {{\rho ^4}|_0^{1/\cos \varphi }} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{\varphi = 0}^{\pi /3} {\sec ^4}\varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
Since $\sec \varphi = \frac{1}{{\cos \varphi }}$, write ${\sec ^4}\varphi \sin \varphi = {\sec ^3}\varphi \tan \varphi $.
So, the integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal W}_1}}^{} \rho {\rm{d}}V = \frac{1}{4}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{\varphi = 0}^{\pi /3} {\sec ^3}\varphi \tan \varphi {\rm{d}}\varphi {\rm{d}}\theta $
Consider the inner integral of the integral on the right-hand side:
$\mathop \smallint \limits_{\varphi = 0}^{\pi /3} {\sec ^3}\varphi \tan \varphi {\rm{d}}\varphi $
Write $u = \sec \varphi $. So, $du = \sec \varphi \tan \varphi d\varphi $. Thus
$\mathop \smallint \limits_{\varphi = 0}^{\pi /3} {\sec ^3}\varphi \tan \varphi {\rm{d}}\varphi = \mathop \smallint \limits_{u = 1}^2 {u^2}{\rm{d}}u = \frac{1}{3}\left( {{u^3}|_1^2} \right) = \frac{7}{3}$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \rho {\rm{d}}V = \frac{1}{4}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {\mathop \smallint \limits_{\varphi = 0}^{\pi /3} {{\sec }^3}\varphi \tan \varphi {\rm{d}}\varphi } \right){\rm{d}}\theta $
$ = \frac{7}{{12}}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\rm{d}}\theta $
$ = \frac{{7\pi }}{{12}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal W}_1}}^{} \rho {\rm{d}}V = \frac{{7\pi }}{{12}}$.
2. Subregion ${{\cal W}_2}$
In region ${{\cal W}_2}$, $\varphi $ ranges from $\varphi = {\varphi _0} = \frac{\pi }{3}$ to $\varphi = \pi $.
Since $x \ge 0$, the range of $\theta$ is $ - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}$.
Thus, the region description of ${{\cal W}_2}$ in spherical coordinates is given by
${{\cal W}_2} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le 2,\frac{\pi }{3} \le \varphi \le \pi , - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}} \right\}$
Using Eq. (8), the triple integral of $f\left( {x,y,z} \right)$ over ${{\cal W}_2}$ in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal W}_2}}^{} \rho {\rm{d}}V = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{\varphi = \pi /3}^\pi \mathop \smallint \limits_{\rho = 0}^2 {\rho ^3}\sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{\varphi = \pi /3}^\pi \left( {{\rho ^4}|_0^2} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = 4\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{\varphi = \pi /3}^\pi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = - 4\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {\cos \varphi |_{\pi /3}^\pi } \right){\rm{d}}\theta $
$ = - 4\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( { - 1 - \frac{1}{2}} \right){\rm{d}}\theta $
$ = 6\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\rm{d}}\theta = 6\pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal W}_2}}^{} \rho {\rm{d}}V = 6\pi $.
Using the linearity property of the triple integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \rho {\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal W}_1}}^{} \rho {\rm{d}}V + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal W}_2}}^{} \rho {\rm{d}}V$
$ = \frac{{7\pi }}{{12}} + 6\pi = \frac{{79\pi }}{{12}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \rho {\rm{d}}V = \frac{{79\pi }}{{12}}$.
