= Solution
The rotation introduces <angular frequency> $\Omega$ and, after combining the two blades, harmonics such as $2\Omega$. The <acoustic compact-source approximation> requires the propagation time $a/c_0$ to be small compared with $\Omega^{-1}$. Thus
$$
\boxed{\mu=\Omega a/c_0\ll1.}
$$
It also bounds every blade element's <Mach number> by $\mu$. Consequently $|1-M_r|=1+O(\mu)$, and its leading value is one. The separate <acoustic far field> condition is $\Omega R/c_0\gg1$.
Choose the positive rotation sense so that a first blade at phase $\alpha$ has radial and tangential unit vectors
$$
e_r(\alpha)=\cos\alpha\,e_y+\sin\alpha\,e_z,\qquad
e_\phi(\alpha)=-\sin\alpha\,e_y+\cos\alpha\,e_z.
$$
Integrating the given line <force> from $r=0$ to $a$ yields $\mathcal F_1=F e_\phi(\Omega\tau)+D e_x$. Its axial component is constant, while $\dot{\mathcal F}_1=-F\Omega e_r$. Since $n=\cos\theta e_x+\sin\theta e_y$, the compact <acoustic dipole> sound from this blade is
$$
\boxed{\rho'_1=-\frac{F\Omega\sin\theta}{4\pi c_0^3R}\cos\big(\Omega(t-R/c_0)\big).}
$$
The other blade has phase $\alpha+\pi$ and contributes the opposite rotating <force>. At a common compact <retarded time>, their total <force> is $2D e_x$, so
$$
\boxed{\rho'_{\text{two blades}}=0\quad\text{at leading compact dipole order}.}
$$
This cancellation calls for the next source-delay correction; it does not mean that the complete moving-source field vanishes.
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