Answer
We prove:
${I_x} = \frac{1}{4}M{R^2} + \frac{1}{{12}}M{H^2}$
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 $y = r\sin \theta $, we evaluate the moment of inertia ${I_x}$ in cylindrical coordinates:
${I_x} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{y^2} + {z^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \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}{{\sin }^2}\theta } \right)r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ + \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} {z^2}r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
(1) ${\ \ \ \ }$ ${I_x} = \frac{M}{{\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\sin }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = - H/2}^{H/2} {\rm{d}}z} \right)$
$ + \frac{M}{{\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R r{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = - H/2}^{H/2} {z^2}{\rm{d}}z} \right)$
Consider the first integral on the right-hand side of (1). Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
$\mathop \smallint \limits_{\theta = 0}^{2\pi } {\sin ^2}\theta {\rm{d}}\theta = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {1 - \cos 2\theta } \right){\rm{d}}\theta $
$\mathop \smallint \limits_{\theta = 0}^{2\pi } {\sin ^2}\theta {\rm{d}}\theta = \frac{1}{2}\left( {\left( {\theta - \frac{1}{2}\sin 2\theta } \right)|_0^{2\pi }} \right) = \pi $
Thus, equation (1) becomes
${I_x} = \frac{M}{{\pi H{R^2}}}\left( \pi \right)\left( {\frac{1}{4}{R^4}} \right)\left( H \right) + \frac{M}{{\pi H{R^2}}}\left( {2\pi } \right)\left( {\frac{1}{2}{R^2}} \right)\left( {\frac{1}{3}{z^3}|_{ - H/2}^{H/2}} \right)$
$ = \frac{1}{4}M{R^2} + \frac{M}{{\pi H{R^2}}}\left( {2\pi } \right)\left( {\frac{1}{2}{R^2}} \right)\left( {\frac{2}{{24}}{H^3}} \right)$
$ = \frac{1}{4}M{R^2} + \frac{1}{{12}}M{H^2}$
Hence, ${I_x} = \frac{1}{4}M{R^2} + \frac{1}{{12}}M{H^2}$.
