Answer
The equation of the right-circular cone in Figure 22 in cylindrical coordinates is $z = \frac{{rH}}{R}$.
The volume of the region is $\frac{1}{3}\pi H{R^2}$.
Work Step by Step
From Figure 22 we see that the right-circular cone has axial symmetry with respect to the $z$-axis. Since, the right-circular cone is the surface consisting of all rays passing through the origin and a point on the cone, we see that the angle the rays make with the $z$-axis is given by
$\tan \theta = \frac{{\sqrt {{x^2} + {y^2}} }}{z} = \frac{r}{z}$
$z = \frac{r}{{\tan \theta }}$
But we also have $\tan \theta = \frac{R}{H}$. So, $z = \frac{{rH}}{R}$.
Thus, the equation of the right-circular cone in Figure 22 in cylindrical coordinates is $z = \frac{{rH}}{R}$.
Let ${\cal W}$ be the region of the right-circular cone as is shown in Figure 22. To find the volume of ${\cal W}$, we project it onto the $xy$-plane and obtain the domain ${\cal D}$. We see from Figure 22 and the figure attached, ${\cal D}$ is a disk of radius $R$, that is, ${x^2} + {y^2} \le {R^2}$. So, the description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le R,0 \le \theta \le 2\pi } \right\}$
Since ${\cal W}$ is bounded below by the cone $z = \frac{{rH}}{R}$ and bounded above by the plane $z=H$, the range of $z$ is from $z = \frac{{rH}}{R}$ to $z=H$. Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le R,0 \le \theta \le 2\pi ,\frac{{rH}}{R} \le z \le H} \right\}$
Evaluate the volume of ${\cal W}$ in cylindrical 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_{r = 0}^R \mathop \smallint \limits_{z = rH/R}^H r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {z|_{rH/R}^H} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {H - \frac{{rH}}{R}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {Hr - \frac{{{r^2}H}}{R}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{1}{2}H{r^2} - \frac{1}{3}\frac{H}{R}{r^3}} \right)|_0^R} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{2}H{R^2} - \frac{1}{3}H{R^2}} \right){\rm{d}}\theta $
$ = \frac{1}{6}H{R^2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{1}{3}\pi H{R^2}$
Thus, the volume of ${\cal W}$ is $\frac{1}{3}\pi H{R^2}$.
