Answer
We prove: ${I_z} = \frac{1}{2}M{R^2}$
Work Step by Step
Let ${\cal W}$ be the region shown in Figure 18. The volume of ${\cal W}$ is $V = \pi {R^2}H$.
Let $M$ be the total mass of ${\cal W}$ and the mass density be uniform. Thus, the mass density is given by
$\delta = \frac{M}{V} = \frac{M}{{\pi H{R^2}}}$
We describe ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le R,0 \le \theta \le 2\pi , - \frac{H}{2} \le z \le \frac{H}{2}} \right\}$
Using ${x^2} + {y^2} = {r^2}$, we evaluate the moment of inertia ${I_z}$ in cylindrical coordinates:
${I_z} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{M}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = - H/2}^{H/2} \left( {{r^2}} \right)r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{M}{{\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = - H/2}^{H/2} {\rm{d}}z} \right)$
$ = \frac{M}{{\pi H{R^2}}}\left( {2\pi } \right)\left( {\frac{1}{4}{R^4}} \right)\left( H \right) = \frac{1}{2}M{R^2}$
Hence, ${I_z} = \frac{1}{2}M{R^2}$.
