= Solution
For the stated square side length $2a$, the volume is $4a^2h$ and the uniform <mass density> is $\rho_1=m_1/(4a^2h)$. Direct integration about its central vertical axis gives the <moment of inertia of a uniform square plate>:
$$
I_1=\rho_1h\int_{-a}^a\int_{-a}^a(x^2+y^2)\,dx\,dy
=\rho_1h\frac{8a^4}{3}
=\boxed{\frac23m_1a^2}.
$$
Thus the PDF's coefficient $1/6$ is inconsistent with its stated side length. The coefficient $1/6$ would apply to a square of side $a$.
For the disc, use <cylindrical coordinates> and $m_2=\rho_2\pi a^2h$:
$$
I_2=\rho_2h\int_0^{2\pi}\int_0^a r^2r\,dr\,d\theta
=\frac{\pi\rho_2ha^4}{2}=\boxed{\frac12m_2a^2}.
$$
For the sphere, use <spherical coordinates> and $m_3=4\pi\rho_3a^3/3$:
$$
I_3=\rho_3\int_0^a r^4dr\int_0^\pi\sin^3\theta\,d\theta\int_0^{2\pi}d\phi
=\frac{8\pi\rho_3a^5}{15}=\boxed{\frac25m_3a^2}.
$$
These agree with the <moment of inertia of a uniform disc> and the <moment of inertia of a uniform solid sphere>. The square's thickness does not enter its axial <moment of inertia> once the mass is fixed.
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