Answer
Using both spherical and cylindrical coordinates, we obtain the volume of the sphere:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{4}{3}\pi {a^3}$
Work Step by Step
Case 1. Spherical coordinates
The sphere ${x^2} + {y^2} + {z^2} = {a^2}$ in spherical coordinates is $\rho = a$.
Since the ranges are $0 \le \varphi \le \pi $ and $0 \le \theta \le 2\pi $, the description of ${\cal W}$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le a,0 \le \varphi \le \pi ,0 \le \theta \le 2\pi } \right\}$
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}^\pi \mathop \smallint \limits_{\rho = 0}^a {\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}^\pi \left( {{\rho ^3}|_0^a} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{{a^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = - \frac{{{a^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \varphi |_0^\pi } \right){\rm{d}}\theta $
$ = - \frac{{{a^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - 1 - 1} \right){\rm{d}}\theta $
$ = \frac{{2{a^3}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{4}{3}\pi {a^3}$
Thus, we obtain the well-known formula for the volume of the sphere of radius $a$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{4}{3}\pi {a^3}$
Case 2. Cylindrical coordinates
The sphere ${x^2} + {y^2} + {z^2} = {a^2}$ in cylindrical coordinates:
${r^2} + {z^2} = {a^2}$
$z = \pm \sqrt {{a^2} - {r^2}} $
In this case, the sphere is bounded above by $z = \sqrt {{a^2} - {r^2}} $ and bounded below by $z = - \sqrt {{a^2} - {r^2}} $.
Since $\theta$ ranges from $\theta = 0$ to $\theta = 2\pi $, we have $0 \le \theta \le 2\pi $.
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le a,0 \le \theta \le 2\pi , - \sqrt {{a^2} - {r^2}} \le z \le \sqrt {{a^2} - {r^2}} } \right\}$
Using Eq. (5), the volume of the sphere 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}^a \mathop \smallint \limits_{z = - \sqrt {{a^2} - {r^2}} }^{\sqrt {{a^2} - {r^2}} } r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^a r\left( {z|_{ - \sqrt {{a^2} - {r^2}} }^{\sqrt {{a^2} - {r^2}} }} \right){\rm{d}}r{\rm{d}}\theta $
$ = 2\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^a r\left( {\sqrt {{a^2} - {r^2}} } \right){\rm{d}}r{\rm{d}}\theta $
$ = - \frac{2}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{{\left( {{a^2} - {r^2}} \right)}^{3/2}}|_0^a} \right){\rm{d}}\theta $
$ = - \frac{2}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - {a^3}} \right){\rm{d}}\theta $
$ = \frac{2}{3}{a^3}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{4}{3}\pi {a^3}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{4}{3}\pi {a^3}$.
