Answer
The volume of the region is $\frac{4}{3}{\left( {{a^2} - {b^2}} \right)^{3/2}}\pi $.
Work Step by Step
Using ${r^2} = {x^2} + {y^2}$, we have the equation of a sphere of radius $a$ in cylindrical coordinates:
${r^2} + {z^2} = {a^2}$
And the equation of the central cylinder of radius $b$ in cylindrical coordinates:
$r=b$
Let ${\cal W}$ denote the region of the sphere of radius $a$ from which a central cylinder of radius $b$ has been removed, where $0 < b < a$.
We project ${\cal W}$ onto the $xy$-plane and obtain the domain ${\cal D}$. From the figure attached, we see that the description of ${\cal D}$ in polar coordinates is
${\cal D} = \left\{ {\left( {r,\theta } \right)|b \le r \le a,0 \le \theta \le 2\pi } \right\}$
From the equation ${r^2} + {z^2} = {a^2}$, we obtain the upper and lower boundaries of ${\cal W}$:
$z = \pm \sqrt {{a^2} - {r^2}} $
So, ${\cal W}$ is bounded below by $z = - \sqrt {{a^2} - {r^2}} $ and bounded above by $z = \sqrt {{a^2} - {r^2}} $.
Thus, the description of ${\cal W}$ is
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|b \le r \le a,0 \le \theta \le 2\pi , - \sqrt {{a^2} - {r^2}} \le z \le \sqrt {{a^2} - {r^2}} } \right\}$
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_{r = b}^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 = b}^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 = b}^a r\sqrt {{a^2} - {r^2}} {\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}}|_b^a} \right){\rm{d}}\theta $
$ = \frac{2}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\left( {{a^2} - {b^2}} \right)^{3/2}}{\rm{d}}\theta $
$ = \frac{2}{3}{\left( {{a^2} - {b^2}} \right)^{3/2}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta $
$ = \frac{4}{3}{\left( {{a^2} - {b^2}} \right)^{3/2}}\pi $
Thus, the volume of the region is $\frac{4}{3}{\left( {{a^2} - {b^2}} \right)^{3/2}}\pi $.
