 Solution

In orthonormal Cartesian coordinates, a second-rank tensor has components transforming under a change of axes $Q$ as $M_{ij}^{\prime }=Q_{ik}{Q}_{j\ell }{M}_{k\ell }$, where $Q{Q}^T=I$. Its contraction is invariant:Thus $M_{ii}$ is a scalar. This is tensor contraction with the Euclidean metric. In general nonorthonormal coordinates the invariant expression is $g^{ij}{M}_{ij}$, not an unweighted sum of diagonal covariant components.

For the planar plate, every mass element has $z=0$. The inertia tensor is consequentlyIt follows directly that ${\mathbf{e}}_z$ is an eigenvector andThe real symmetric matrix has three real eigenvalues. Its trace is $2{\int }_D(x^2+y^2)\rho dS=2M_{\perp }$, so $M_1+M_2+M_{\perp }=2M_{\perp }$. Hence the perpendicular axis theorem gives $M_{\perp }=M_1+M_2$.

For a disc of constant areal density ${\rho }_0$, symmetry gives ${\int }_{D}xydS=0$, and polar integration gives ${\int }_D{x}^{2}dS={\int }_D{y}^{2}dS=\pi a^4/4$. Thereforewhere the scalar $M$ in the last expression denotes the total mass rather than the tensor.