Inertia tensor of a sheared rectangular lamina ($I=\frac{M}{3}\left(\begin{array}{ccc}{b}^2& -s{b}^2& 0\\ -s{b}^2& a^2+s^2{b}^2& 0\\ 0& 0& a^2+(1+s^2)b^2\end{array}\right)$)

For a uniform planar lamina with $|x_1-s{x}_2|\le a$ and $|x_2|\le b$, put $u=x_1-s{x}_2$, $v=x_2$. The Jacobian determinant is one, the mass is $M=4ab\sigma$, and symmetry puts the centre of mass at zero. Uniform rectangular moments give $⟨u^2⟩=a^2/3$, $⟨v^2⟩=b^2/3$, $⟨uv⟩=0$. Substitution into $I_{ij}=\int (|x{|}^2{\delta }_{ij}-x_i{x}_j)dm$ yields the displayed inertia tensor. The in-plane principal moments of inertia are $M[a^2+(1+s^2)b^2\pm \sqrt{(a^2+(s^2-1)b^2{)}^2+4s^2{b}^4}]/6$; the perpendicular moment is their sum. This separates the geometric shear from the elementary rectangular integrals.