Answer
The volume of the region is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V \simeq 21.766$.
Work Step by Step
Using ${\rho ^2} = {x^2} + {y^2} + {z^2}$, we have the equation of a sphere of radius $2$ in spherical coordinates:
${x^2} + {y^2} + {z^2} = {\rho ^2} = 4$
$\rho = 2$
Using $x = \rho \sin \varphi \cos \theta $ and $y = \rho \sin \varphi \sin \theta $, the equation of the central cylinder of radius $1$ in spherical coordinates:
${x^2} + {y^2} = {\rho ^2}{\sin ^2}\varphi {\cos ^2}\theta + {\rho ^2}{\sin ^2}\varphi {\sin ^2}\theta = 1$
${\rho ^2}{\sin ^2}\varphi = 1$
$\rho = \frac{1}{{\sin \varphi }}$
Let ${\cal W}$ denote the region of the sphere of radius $2$ from which a central cylinder of radius $1$ has been removed.
From the figure attached, we see that $\rho$ ranges from $\rho = \frac{1}{{\sin \varphi }}$ to $\rho = 2$.
Since the central cylinder is removed, each ray starts from the angle where the cylinder intersects the sphere at the top and ends at the angle where the cylinder intersects the sphere at the bottom. So, we need to find the angle $\varphi $. Using the radius of the cylinder $1$ and the radius of the sphere $2$, we get
$\sin \varphi = \frac{1}{2}$
So, ${\varphi _1} = \frac{\pi }{6}$, ${\varphi _2} = \frac{{5\pi }}{6}$. Thus, $\varphi $ ranges from ${\varphi _1} = \frac{\pi }{6}$ to ${\varphi _2} = \frac{{5\pi }}{6}$.
Thus, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|\frac{1}{{\sin \varphi }} \le \rho \le 2,\frac{\pi }{6} \le \varphi \le \frac{{5\pi }}{6},0 \le \theta \le 2\pi } \right\}$
Evaluate the volume of ${\cal W}$ using spherical coordinates:
$\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 = \pi /6}^{5\pi /6} \mathop \smallint \limits_{\rho = 1/\sin \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 = \pi /6}^{5\pi /6} \left( {{\rho ^3}|_{1/\sin \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 = \pi /6}^{5\pi /6} \left( {8 - \frac{1}{{{{\sin }^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 = \pi /6}^{5\pi /6} \left( {8\sin \varphi - {{\csc }^2}\varphi } \right){\rm{d}}\varphi {\rm{d}}\theta $
Recall from Eq. (16) of the Table of Trigonometric Integrals (Section 8.2):
$\smallint {\csc ^2}x{\rm{d}}x = - \cot x$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( { - 8\cos \varphi + \cot \varphi } \right)|_{\pi /6}^{5\pi /6}} \right){\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {4\sqrt 3 - \sqrt 3 + 4\sqrt 3 - \sqrt 3 } \right){\rm{d}}\theta $
$ = 2\sqrt 3 \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 4\sqrt 3 \pi \simeq 21.766$
Thus, the volume of the region is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V \simeq 21.766$.
