Answer
The volume of the region is $V = \frac{{2\pi {R^3}}}{3}\left( {1 - \cos {\varphi _0}} \right)$.
Work Step by Step
We have region ${\cal W}$, lying above the cone $\varphi = {\varphi _0}$ and below the sphere $\rho = R$.
Since ${\cal W}$ is bounded above by $\rho = R$, the range of $\rho$ is $0 \le \rho \le R$.
The cone $\varphi = {\varphi _0}$ implies that ${\cal W}$ is bounded by the angle ${\varphi _0}$ with respect to the $z$-axis. Therefore, $\varphi $ varies from $0$ to ${\varphi _0}$, that is, $0 \le \varphi \le {\varphi _0}$.
Thus, the description of ${\cal W}$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le R,0 \le \varphi \le {\varphi _0},0 \le \theta \le 2\pi } \right\}$
Using Eq. (8), the volume of ${\cal W}$ 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}^{{\varphi _0}} \mathop \smallint \limits_{\rho = 0}^R {\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}^{{\varphi _0}} \left( {{\rho ^3}|_0^R} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{{R^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{{\varphi _0}} \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = - \frac{{{R^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \varphi |_0^{{\varphi _0}}} \right){\rm{d}}\theta $
$ = - \frac{{{R^3}}}{3}\left( {\cos {\varphi _0} - 1} \right)\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta $
$ = \frac{{2\pi {R^3}}}{3}\left( {1 - \cos {\varphi _0}} \right)$
Thus, the volume of the region is $V = \frac{{2\pi {R^3}}}{3}\left( {1 - \cos {\varphi _0}} \right)$.
