Mass per unit area of a circular disc of radius a depends

(1) Given,

mass per unit area of circular disc, $\sigma=A+B r$

Area of the ring $=2 \pi \mathrm{r} d r$

Mass of the ring, $d m=\sigma 2 \pi r d r$

Moment of inertia,

$I=\int d m r^{2}=\int \sigma 2 \pi r d r \cdot r^{2}$

$\Rightarrow \quad I=2 \pi \int_{0}^{a}(A+B r) r^{3} d r=2 \pi\left[\frac{A a^{4}}{4}+\frac{B a^{5}}{5}\right]$

$\Rightarrow I=2 \pi a^{4}\left[\frac{A}{4}+\frac{B a}{5}\right]$