= Solution
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 $-D\leq u,v\leq D$, 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 are
$$
\begin{aligned}
\int x_1^2dm&=\sigma\int_{-D}^D\!\int_{-D}^D(u+v)^2du\,dv=\frac83\sigma D^4,\\
\int x_2^2dm&=\sigma\int_{-D}^D\!\int_{-D}^D v^2du\,dv=\frac43\sigma D^4,\\
\int x_1x_2dm&=\sigma\int_{-D}^D\!\int_{-D}^D(u+v)v\,du\,dv=\frac43\sigma D^4.
\end{aligned}
$$
The mixed moments involving $x_3$ vanish. Hence the <inertia tensor> is
$$
\boxed{I=\frac{4\sigma D^4}{3}\begin{pmatrix}1&-1&0\\-1&2&0\\0&0&3\end{pmatrix}=\frac{MD^2}{3}\begin{pmatrix}1&-1&0\\-1&2&0\\0&0&3\end{pmatrix}.}
$$
The 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>.
Back to article page
