= Solution
Take the sheet to lie in the $x_1x_2$ plane, with lengths $a,b$ in those directions. Its surface density is $M/(ab)$. Reflection symmetry about the coordinate axes makes the off-diagonal entries of the <inertia tensor> zero. Its <principal moments of inertia> are
$$
I_1=\frac M{ab}\int_{-a/2}^{a/2}\int_{-b/2}^{b/2}x_2^2\,dx_2dx_1=\frac{Mb^2}{12},
$$
and similarly $I_2=Ma^2/12$. Since $I_3$ integrates $x_1^2+x_2^2$, the <perpendicular axis theorem> gives
$$
\boxed{I_1=Mb^2/12,\quad I_2=Ma^2/12,\quad I_3=M(a^2+b^2)/12.}
$$
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