= Solution
For a uniform thin disc, the <surface mass density> is $\sigma=m/(\pi a^2)$. An annulus of radius $r$ and thickness $dr$ has <mass> $dm=2\pi\sigma r\,dr$, all at distance $r$ from the normal central axis. Its contribution to the <moment of inertia> is $r^2dm$, so
$$
I=\int_0^a r^2(2\pi\sigma r\,dr)
=\frac{2m}{a^2}\left[\frac{r^4}{4}\right]_0^a,
\qquad\boxed{I=\frac12ma^2.}
$$
This establishes the <moment of inertia of a uniform disc>; the normal axis and uniform <surface mass density> are part of the result.
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