= Inertia tensor of a sheared rectangular lamina
{title2=$I=\frac M3\begin{pmatrix}b^2&-sb^2&0\\-sb^2&a^2+s^2b^2&0\\0&0&a^2+(1+s^2)b^2\end{pmatrix}$}
For a uniform planar lamina with $|x_1-sx_2|\leq a$ and $|x_2|\leq b$, put $u=x_1-sx_2$, $v=x_2$. The <Jacobian determinant> is one, the mass is $M=4ab\sigma$, and symmetry puts the centre of mass at zero. Uniform rectangular moments give $\langle u^2\rangle=a^2/3$, $\langle v^2\rangle=b^2/3$, $\langle uv\rangle=0$. Substitution into $I_{ij}=\int(|x|^2\delta_{ij}-x_ix_j)dm$ yields the displayed <inertia tensor>. The in-plane <principal moments of inertia> are $M[a^2+(1+s^2)b^2\pm\sqrt{(a^2+(s^2-1)b^2)^2+4s^2b^4}]/6$; the perpendicular moment is their sum. This separates the geometric shear from the elementary rectangular integrals.
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