The surface density (mass/area) of a circular disc having a radius...

Consider an elemental ring of radius $r$ and thickness $dr$ in the circular disc.

The area of this elemental ring is $dA = 2\pi r dr$.

The mass of the elemental ring is $dm = \rho(r) dA = (A + Br) 2\pi r dr$.

The moment of inertia of this elemental ring about the axis passing through the centre and perpendicular to the plane is $dI = r^2 dm$.

Substituting the expression for $dm$:

$dI = r^2 (A + Br) 2\pi r dr$

$dI = 2\pi (A r^3 + B r^4) dr$

The total moment of inertia of the disc is obtained by integrating $dI$ from $r = 0$ to $r = a$:

$I = \int_{0}^{a} 2\pi (A r^3 + B r^4) dr$

$I = 2\pi \left[ \dfrac{A r^4}{4} + \dfrac{B r^5}{5} \right]_{0}^{a}$

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