A disc of mass 'm' and radius R has a concentric hole of radius 'r'. Its M.I. about an axis through center and normal to its plane is

Area of disc $=\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right);$

Mass of disc=m

$\mathrm{Mass} \mathrm{per} \mathrm{unit} \mathrm{area}=\frac{\mathrm{m}}{\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)}$ 

Consider a concentric ring of thickness dx at a distance x, 

Area of elementary ring is given by $\mathrm{dA}=2\mathrm{\pi }\times \mathrm{dx}$

Mass of ring = Area x mass per unit area

$\mathrm{dm}=\frac{\mathrm{m}\times 2\mathrm{\pi }\times \mathrm{dx}}{\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)}$

$$\begin{gathered} \mathrm{Moment} \mathrm{of} \mathrm{inertia}=\int {\mathrm{dmx}}^2={\int }_{\mathrm{r}}^{\mathrm{R}}\frac{\mathrm{m}\times 2\mathrm{\pi }\times \mathrm{dx}}{\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)}{\mathrm{x}}^2 \\ =\frac{2\mathrm{m}}{\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)}{\int }_{\mathrm{r}}^{\mathrm{R}}{\mathrm{x}}^3\mathrm{dx} \\ \mathrm{I}=\frac{1}{2}\mathrm{m}\left({\mathrm{R}}^2+{\mathrm{r}}^2\right) \end{gathered}$$