Answer
${I_x} = \frac{1}{4}M{R^2}$
${I_y} = \frac{1}{4}M{R^2}$
Work Step by Step
We have ${\cal D}$, defined by the half-disk ${x^2} + {y^2} \le {R^2}$, $x \ge 0$ (in meters), with total mass $M$ kilograms and uniform mass density. So, the area of ${\cal D}$ is $A = \frac{1}{2}\pi {R^2}$.
Thus, the mass density is constant $\delta$ given by $\delta = \frac{M}{A} = \frac{{2M}}{{\pi {R^2}}}$.
We choose to compute ${I_x}$ and ${I_y}$ using polar coordinates. Since $x \ge 0$, so the description of ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le R, - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}} \right\}$
1. Evaluate the moment of inertia relative to the $x$-axis:
${I_x} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^2}\delta \left( {x,y} \right){\rm{d}}A = \frac{{2M}}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^R {\left( {r\sin \theta } \right)^2}r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{{2M}}{{\pi {R^2}}}\left( {\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {{\sin }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$
Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
${I_x} = \frac{M}{{\pi {R^2}}}\left( {\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {1 - \cos 2\theta } \right){\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$
${I_x} = \frac{M}{{\pi {R^2}}}\left( {\left( {\theta - \frac{1}{2}\sin 2\theta } \right)|_{ - \pi /2}^{\pi /2}} \right)\left( {\frac{1}{4}{r^4}|_0^R} \right)$
${I_x} = \frac{M}{{\pi {R^2}}}\left( \pi \right)\left( {\frac{1}{4}{R^4}} \right) = \frac{1}{4}M{R^2}$
2. Evaluate the moment of inertia relative to the $y$-axis:
${I_y} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}\delta \left( {x,y} \right){\rm{d}}A = \frac{{2M}}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^R {\left( {r\cos \theta } \right)^2}r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{{2M}}{{\pi {R^2}}}\left( {\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {{\cos }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$
Using the Double-angle formulas in Section 1.4:
${\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)$
we get
${I_y} = \frac{M}{{\pi {R^2}}}\left( {\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {1 + \cos 2\theta } \right){\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$
${I_y} = \frac{M}{{\pi {R^2}}}\left( {\left( {\theta + \frac{1}{2}\sin 2\theta } \right)|_{ - \pi /2}^{\pi /2}} \right)\left( {\frac{1}{4}{r^4}|_0^R} \right)$
${I_y} = \frac{M}{{\pi {R^2}}}\left( \pi \right)\left( {\frac{1}{4}{R^4}} \right) = \frac{1}{4}M{R^2}$
