= Solution
Interpret the surface delta as $\delta_s=|\nabla F|\delta(F)$, so integration against it gives a surface integral. If $F$ is signed distance, this equals the notation $\delta(F)$ directly. This convention is needed because an arbitrary rescaling of the level-set function must not change the physical source.
Define the volume flux $\mathcal V(t)=\int_S V_n\,dS$ and total <pressure> <force> $f(t)=\int_S pn\,dS$. Here $\mathcal V$ is the rate of change of body volume, not the volume itself. The surface <acoustic monopole> contributes, in the <acoustic compact-source approximation>,
$$
\rho'_M=\partial_t\frac{\rho_0\mathcal V(\tau)}{4\pi c_0^2r}
=\frac{\rho_0\dot{\mathcal V}(\tau)}{4\pi c_0^2r}.
$$
The surface <acoustic dipole> contributes
$$
\rho'_D=-\nabla_x\cdot\frac{f(\tau)}{4\pi c_0^2r}
=\frac{n\cdot\dot f(\tau)}{4\pi c_0^3r}+\frac{n\cdot f(\tau)}{4\pi c_0^2r^2}.
$$
The second term is a nonradiating near-field term and is smaller in the <acoustic far field>. Writing $\mathcal F(t)=x\cdot f(t)$ with the observation point fixed, the leading augmentation is therefore
$$
\boxed{\rho'_{\rm surf}\sim\frac{\rho_0\dot{\mathcal V}(\tau)}{4\pi rc_0^2}
+\frac{\dot{\mathcal F}(\tau)}{4\pi r^2c_0^3}}.
$$
\b[The printed dipole coefficient is missing a factor $1/c_0$.] The extra inverse speed follows from differentiating $\tau=t-r/c_0$, and is required dimensionally: the printed expression with $c_0^2$ has the units of density times speed. This corrected <acoustic loading noise> term also agrees with the <far-field acoustic force and stress moments>. The monopole coefficient is unchanged.
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