= Solution
{c}
The <gravity-capillary drop with a meridional curvature approximation> uses the stipulated $\kappa=-h''$ for this calculation. At equilibrium the <thin-film equation> gives
$$
rh^3(\rho gh'-\gamma h''')=C.
$$
Regularity at $r=0$ forces $C=0$. In the wetted interior $h>0$, the equilibrium <pressure gradient> therefore vanishes, giving
$$
h'''-\frac{h'}{l^2}=0,\qquad
\boxed{l=\sqrt{\frac{\gamma}{\rho g}}.}
$$
This is the <capillary length>. Integrating the constant-coefficient <ordinary differential equation> gives
$$
h=C_0+C_1\cosh(r/l)+C_2\sinh(r/l).
$$
The central <boundary condition> $h'(0)=0$ sets $C_2=0$, and contact at $r=R$ gives $C_0=-C_1\cosh(R/l)$. Write $C_1=-\alpha l$. The prescribed <contact angle> fixes
$$
h'(R)=-\alpha\sinh(R/l)=-\tan\theta_c,
\qquad
\boxed{\alpha=\frac{\tan\theta_c}{\sinh(R/l)},\quad
h(r)=\alpha l[\cosh(R/l)-\cosh(r/l)].}
$$
Here $\alpha$ is the dimensionless coefficient of this drop solution, unrelated to the hinge angle in question 1.
The axisymmetric <volume integral> is $V=2\pi\int_0^Rrh(r)\,dr$. Substituting $\xi=r/l$ gives
$$
\boxed{1=\frac{2\pi l^3\alpha}{V}\int_0^{R/l}[\cosh(R/l)-\cosh\xi]\xi\,d\xi.}
$$
For numerical evaluation, with $s=R/l$, the integral is
$$
I(s)=\frac{s^2}{2}\cosh s-s\sinh s+\cosh s-1.
$$
Together with $\alpha=\tan\theta_c/\sinh s$, this determines $R$ for fixed $V$ and $\theta_c$.
The <Bond number> $R^2/l^2=\rho gR^2/\gamma$ distinguishes the dominant balances. For $R\ll l$, <surface tension> dominates <gravity> and the surface is nearly a small-slope constant-<curvature> cap:
$$
h(r)\approx\frac{\tan\theta_c}{2R}(R^2-r^2),\qquad
V\approx\frac{\pi\tan\theta_c}{4}R^3.
$$
For $R\gg l$, <gravity> flattens the interior, while <surface tension> bends the edge over the <capillary length>. The stipulated model gives
$$
h(0)\approx l\tan\theta_c,\qquad
V\approx\pi R^2l\tan\theta_c,
$$
up to relative corrections of order $l/R$. The <contact angle> fixes the edge slope and therefore the height scale; <conservation of mass> then fixes the lateral extent. Consistency of <lubrication theory> requires small slopes, in particular $\tan\theta_c\ll1$.
\b[The stated approximation differs from <axisymmetric free-surface curvature>.] For an axisymmetric graph with the same sign convention, the sum of principal <curvatures> is
$$
\kappa=-\frac1r\frac{d}{dr}\left(\frac{rh'}{\sqrt{1+h'^2}}\right)
\approx-\left(h''+\frac{h'}r\right).
$$
The omitted $h'/r$ term is generally of the same order as $h''$. For example, $h=H(1-r^2/R^2)$ gives stipulated $\kappa=2H/R^2$ but axisymmetric small-slope $\kappa=4H/R^2$. Thus the required hyperbolic-cosine profile solves the explicitly prescribed reduced model, not the full axisymmetric capillary equilibrium. Keeping both small-slope terms instead gives
$$
h(r)=\frac{l\tan\theta_c}{I_1(R/l)}[I_0(R/l)-I_0(r/l)],
$$
where $I_0,I_1$ are <modified Bessel functions>. The qualitative <capillary length> and <Bond number> balances remain valid.
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