= Solution
Apply <differentiation under the integral sign> on the moving drop:
$$
\frac{dM}{dt}=R\,h(R,t)\dot R+\int_0^Rrh_t\,dr
=\left[rh^3h_r\right]_0^R.
$$
The moving-boundary term vanishes because $h(R,t)=0$. At the origin, symmetry and regularity give $h_r(0,t)=0$. At the front, the stated $h\propto(R-r)^{1/3}$ gives $h^3h_r=O((R-r)^{1/3})\to0$. Thus $\boxed{dM/dt=0}$, expressing <volume conservation> for this <axisymmetric viscous gravity current>.
With $\eta=r/t^\alpha$ and $R=\eta_0t^\alpha$, substitute the <self-similar solution> into the conserved integral:
$$
M=t^{2\alpha-1/4}\int_0^{\eta_0}\eta f(\eta)\,d\eta.
$$
For a nonzero finite drop volume, the power of $t$ must vanish, so $\boxed{\alpha=1/8}$. Balancing powers in the evolution equation gives the same result: its left side scales as $t^{-5/4}$ and its right side as $t^{-1-2\alpha}$. The front-edge exponent alone would not determine this spreading exponent without the conserved volume.
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