A thin uniform disc has a mass $9M$ and radius $R$. A disc of radius $\frac{R}{3}$ is cut off as shown in figure. Find the moment of inertia of the remaining disc about an axis passing through $O$ and perpendicular to the plane of disc.

As the mass is uniformly distributed on the disc,

so mass density (per unit area) =$\frac{9M}{{\mathrm{\pi R}}^2}$. Mass of removed portion = $\frac{9M\mathrm{\pi }}{{\mathrm{\pi R}}^2}\times {\left[\frac{R}{3}\right]}^2=M$

So the moment of inertia of the removed portion about the stated axis by theorem of parallel axis is: $I_1=\frac{M}{2}{\left[\frac{R}{3}\right]}^2+M{\left[\frac{2R}{3}\right]}^2=\frac{M{R}^2}{2}$

The moment of inertia of the original complete disc about the stated axis is $I_2$ then

$I_2=9M\frac{R^2}{2}$

So the moment of inertia of the left over disc shown in figure is $I_2-I_1$.

i.e., $I_2-I_1=4M{R}^2$.