= Solution
The <quadrupole formula> in units $G=c=1$ is
$$
P=\frac15\left\langle\dddot Q_{ij}\dddot Q_{ij}\right\rangle,
\qquad
Q_{ij}=I_{ij}-\frac13\delta_{ij}I_{kk}.
$$
Here $I_{kk}=ma^2/3$ is constant, so its trace subtraction has no third derivative. If $B=4ma^2\omega^3/3$, then
$$
\dddot I_{xx}=B\sin2\omega t,
\quad
\dddot I_{yy}=-B\sin2\omega t,
\quad
\dddot I_{xy}=-B\cos2\omega t.
$$
Thus $\dddot I_{ij}\dddot I_{ij}=2B^2$ and
$$
\boxed{\langle P\rangle=\frac{32}{45}m^2a^4\omega^6}.
$$
Restoring units multiplies this by $G/c^5$.
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