= Solution
{c}
Integrating the radial <velocity> over the thickness gives the radial <volume flux per unit width>:
$$
q_r=\int_0^h v_r\,dz
=\frac{\rho gh_r+\gamma\kappa_r}{2\mu}
\left[\frac{h^3}{3}-h^3\right]
=-\frac{h^3}{3\mu}\left(\rho gh_r+\gamma\kappa_r\right).
$$
The full <volumetric flow rate> across a cylindrical surface, positive outward, is
$$
\boxed{Q(r,t)=2\pi r q_r=-\frac{2\pi r h^3}{3\mu}\left(\rho gh_r+\gamma\kappa_r\right).}
$$
An annulus of radial thickness $dr$ contains <volume> $2\pi rh\,dr$ to leading order. Its rate of gain equals incoming minus outgoing <volume flux>, so
$$
2\pi r h_t\,dr=Q(r,t)-Q(r+dr,t)=-Q_r\,dr+o(dr).
$$
Dividing and taking $dr\to0$ yields the axisymmetric <thin-film equation>:
$$
\boxed{h_t=\frac1r\partial_r\left[\frac{rh^3}{3\mu}\left(\rho gh_r+\gamma\kappa_r\right)\right].}
$$
The equation assumes no evaporation or injection through the boundaries. Integrating it over the whole drop also verifies <conservation of mass> when the edge carries zero <volume flux>.
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