Answer
The volume of the region is $\frac{5}{3}\pi $.
Work Step by Step
We have the region ${\cal W}$ bounded below by the plane $z=1$ and above by the sphere ${x^2} + {y^2} + {z^2} = 4$.
Step 1. Find the range of $\rho$
We see from the figure attached, each ray begins at the plane $z=1$ and ends at the sphere. In spherical coordinates:
- the plane $z=1$ is given by $\rho \cos \varphi = 1$.
So, $\rho = \frac{1}{{\cos \varphi }}$.
- the sphere ${x^2} + {y^2} + {z^2} = 4$ has radius $2$, thus $\rho = 2$.
Thus, $\rho$ ranges from $\rho = \frac{1}{{\cos \varphi }}$ to $\rho = 2$.
Step 2. Find the range of $\varphi $
From the figure attached, we see that $\varphi $ varies from $\varphi = 0$ to the angle where the sphere intersects the plane $z = 1 = \rho \cos \varphi $, where $\rho = 2$. So
$\cos \varphi = \frac{1}{2}$, ${\ \ \ \ }$ $\varphi = \frac{\pi }{3}$
Thus, the region description of ${\cal W}$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|\frac{1}{{\cos \varphi }} \le \rho \le 2,0 \le \varphi \le \frac{\pi }{3},0 \le \theta \le 2\pi } \right\}$
Using Eq. (8) of Theorem 3. we evaluate the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \mathop \smallint \limits_{\rho = 1/\cos \varphi }^2 {\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 /3} \left( {{\rho ^3}|_{1/\cos \varphi }^2} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \left( {8 - \frac{1}{{{{\cos }^3}\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}^{\pi /3} \left( {8\sin \varphi - {{\cos }^{ - 3}}\varphi \sin \varphi } \right){\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta - \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} {\cos ^{ - 3}}\varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
Consider the inner integral of the second integral on the right-hand side:
$ - \frac{1}{3}\mathop \smallint \limits_{\varphi = 0}^{\pi /3} {\cos ^{ - 3}}\varphi \sin \varphi {\rm{d}}\varphi $
Write $u = \cos \varphi $. So, $du = - \sin \varphi d\varphi $. Thus,
$ - \frac{1}{3}\mathop \smallint \limits_{\varphi = 0}^{\pi /3} {\cos ^{ - 3}}\varphi \sin \varphi {\rm{d}}\varphi = \frac{1}{3}\mathop \smallint \limits_{u = 1}^{1/2} {u^{ - 3}}{\rm{d}}u$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta + \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{u = 1}^{1/2} {u^{ - 3}}{\rm{d}}u{\rm{d}}\theta $
$ = - \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \varphi |_0^{\pi /3}} \right){\rm{d}}\theta - \frac{1}{6}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{u^{ - 2}}|_1^{1/2}} \right){\rm{d}}\theta $
$ = - \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{2} - 1} \right){\rm{d}}\theta - \frac{1}{6}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {4 - 1} \right){\rm{d}}\theta $
$ = \frac{4}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta - \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta $
$ = \frac{8}{3}\pi - \pi = \frac{5}{3}\pi $
Thus, the volume of the region is $\frac{5}{3}\pi $.
