Answer
The volume of the cone is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{1}{3}\pi H{R^2}$.
Work Step by Step
Let ${\cal W}$ denote the cone as is shown in Figure 22.
From the figure we see that it is bounded above by the plane $z=H$. In spherical coordinates, the plane $z=H$ is expressed by
$z = \rho \cos \varphi = H$
$\rho = \frac{H}{{\cos \varphi }}$
Each ray starts from the origin and ends at the plane $\rho = \frac{H}{{\cos \varphi }}$. Thus, $\rho$ varies from $0$ to $\rho = \frac{H}{{\cos \varphi }}$, that is, $0 \le \rho \le \frac{H}{{\cos \varphi }}$.
Denote ${\varphi _0}$ the angle the cone makes with the $z$-axis. We have
$\tan {\varphi _0} = \frac{R}{H}$, ${\ \ \ \ \ }$ ${\varphi _0} = {\tan ^{ - 1}}\frac{R}{H}$
Therefore, $\varphi $ varies from $0$ to ${\varphi _0} = {\tan ^{ - 1}}\frac{R}{H}$, that is, $0 \le \varphi \le {\tan ^{ - 1}}\frac{R}{H}$.
Thus, the description of ${\cal W}$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le \frac{H}{{\cos \varphi }},0 \le \varphi \le {{\tan }^{ - 1}}\frac{R}{H},0 \le \theta \le 2\pi } \right\}$
Using Eq. (8), the volume of the cone 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}^{{{\tan }^{ - 1}}\left( {R/H} \right)} \mathop \smallint \limits_{\rho = 0}^{H/\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}^{{{\tan }^{ - 1}}\left( {R/H} \right)} \left( {{\rho ^3}|_0^{H/\cos \varphi }} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{{{\tan }^{ - 1}}\left( {R/H} \right)} \left( {\frac{{{H^3}}}{{{{\cos }^3}\varphi }}} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{{H^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{{{\tan }^{ - 1}}\left( {R/H} \right)} {\sec ^3}\varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{{H^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{{{\tan }^{ - 1}}\left( {R/H} \right)} {\sec ^2}\varphi \tan \varphi {\rm{d}}\varphi {\rm{d}}\theta $
Consider the inner integral above: $\mathop \smallint \limits_{\varphi = 0}^{{{\tan }^{ - 1}}\left( {R/H} \right)} {\sec ^2}\varphi \tan \varphi {\rm{d}}\varphi $
Write $u = \sec \varphi $. So, $du = \sec \varphi \tan \varphi d\varphi $.
Since $\tan {\varphi _0} = \frac{R}{H}$, so $\cos {\varphi _0} = \frac{H}{{\sqrt {{H^2} + {R^2}} }}$. We have $\sec {\varphi _0} = \frac{{\sqrt {{H^2} + {R^2}} }}{H}$.
Thus,
$\mathop \smallint \limits_{\varphi = 0}^{{{\tan }^{ - 1}}\left( {R/H} \right)} {\sec ^2}\varphi \tan \varphi {\rm{d}}\varphi = \mathop \smallint \limits_{u = 1}^{\sqrt {{H^2} + {R^2}} /H} u{\rm{d}}u$
$ = \frac{1}{2}\left( {{u^2}|_1^{\sqrt {{H^2} + {R^2}} /H}} \right) = \frac{1}{2}\left( {\frac{{{H^2} + {R^2}}}{{{H^2}}} - 1} \right)$
$ = \frac{{{R^2}}}{{2{H^2}}}$
Substituting this result back into the earlier integral gives
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{{{H^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{{{\sin }^{ - 1}}\left( {R/H} \right)} {\sec ^2}\varphi \tan \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{{H^3}}}{3}\frac{{{R^2}}}{{2{H^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta $
$ = \frac{1}{3}\pi H{R^2}$
Thus, the volume of the cone is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{1}{3}\pi H{R^2}$.
