Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 7 / a / Solution

For the stated square side length $2a$, the volume is $4a^{2}h$ and the uniform mass density is ${\rho }_1=m_1/(4a^{2}h)$. Direct integration about its central vertical axis gives the moment of inertia of a uniform square plate: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 $m_2={\rho }_2\pi a^{2}h$:For the sphere, use spherical coordinates and $m_3=4\pi {\rho }_3{a}^3/3$: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.