= Solution
Write $I_a=\bar I_{aa}$ and use the right-handed rotation
$$
L(t)=\begin{pmatrix}\cos\Omega t&-\sin\Omega t&0\\
\sin\Omega t&\cos\Omega t&0\\0&0&1\end{pmatrix}.
$$
In the integral defining the <second mass moment tensor>, change variables to $\mathbf y=L^{-1}\mathbf x$. The <determinant> is one and $x_i=L_{ik}y_k$, giving
$$
I_{ij}(t)=\int\rho(\mathbf y)L_{ik}y_kL_{jl}y_l\,d^3y
=L_{ik}L_{jl}\bar I_{kl},\qquad \boxed{I(t)=L(t)\bar I L(t)^T.}
$$
Set $C=(I_1+I_2)/2$ and $\Delta=I_1-I_2$. Multiplication gives
$$
\boxed{\begin{aligned}
I_{11}(t)&=C+\frac\Delta2\cos(2\Omega t),\\
I_{22}(t)&=C-\frac\Delta2\cos(2\Omega t),\\
I_{12}(t)=I_{21}(t)&=\frac\Delta2\sin(2\Omega t),\\
I_{33}(t)&=I_3,\qquad I_{13}(t)=I_{23}(t)=0.
\end{aligned}}
$$
In particular, $I_{kk}=I_1+I_2+I_3$ is constant. These expressions also show how the rotating anisotropy enters the <mass quadrupole moment>.
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