Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 6 / b

Integrating the Dirac delta distribution in $x_3$ reduces the inertia integral to the planar lamina. Use $u=x_1-x_2$, $v=x_2$, so $x_1=u+v$ and the parallelogram becomes the square $-D\le u,v\le D$, with unit Jacobian determinant. The mass is $M=4\sigma D^2$, and symmetry gives a centre of mass at the origin.

Odd rectangular integrals vanish. The required second moments areThe mixed moments involving $x_3$ vanish. Hence the inertia tensor isThe off-diagonal entries are minus the product moment, not the product moment itself. This is the unit-shear case of the inertia tensor of a sheared rectangular lamina.

Put $J_0=4\sigma D^4/3=M{D}^2/3$. The normal to the lamina is already an eigenvector, with moment $3J_0$. The planar block has characteristic polynomialso the principal moments of inertia areA corresponding orthonormal set of principal axes isThe sign of each axis vector is immaterial. The smaller-moment axis makes angle $\frac{1}{2}\arctan (2)$ with the positive $x_1$ direction, approximately $31.7^{\circ }$; the other planar axis is perpendicular. As a check, $I_{-}+I_{+}=3J_0=I_3$, the perpendicular axis theorem for a planar lamina.