Moment of inertia of a uniform disc ($I=M{a}^2/2$)

For a uniform disc of radius $a$ and mass $M$, annuli of radius $r$ and thickness $dr$ have mass $(M/\pi a^2)2\pi rdr$. Their axial moment of inertia sums to $I={\int }_0^a{r}^{2}dm=M{a}^2/2$.