= Solution
Centre the square on the $xy$ plane, with uniform surface <mass density> $m/a^2$. For an in-plane axis at angle $\alpha$ to the $x$ direction, the perpendicular distance of $(x,y)$ from the axis is $-x\sin\alpha+y\cos\alpha$. Therefore its <moment of inertia> is
$$
I_\alpha=\frac m{a^2}\int_{-a/2}^{a/2}\int_{-a/2}^{a/2}(-x\sin\alpha+y\cos\alpha)^2\,dx\,dy.
$$
The mixed integral vanishes by symmetry, while the $x^2$ and $y^2$ integrals are equal. Thus
$$
\boxed{I_\alpha=\frac{ma^2}{12}(\sin^2\alpha+\cos^2\alpha)=\frac{ma^2}{12}.}
$$
This is the in-plane isotropy of the square's <inertia tensor>: its two central in-plane principal moments coincide. It does not assert isotropy about arbitrary three-dimensional axes, nor about parallel axes that miss the centre.
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