I am interested in calculating the least area of a surface spanning the boundary of an octant on the unit sphere; and short of precise values I am looking for upper bounds for this area.
In $\mathbf{S}^2 \subset \mathbf{R}^3$ consider the boundary of the first octant, namely $R = \partial \{ (x,y,z) \in \mathbf{S}^2 \mid x, y ,z \geq 0 \} \subset \mathbf{S}^2$. This is the set of points $(x,y,z) \in \mathbf{S}^2$ where $x,y,z \geq 0$ and at least one is zero. There exists a smooth surface $S$ with $\partial S = R$ with least area; moreover this is minimal. Comparing it to the octant one finds $\mathrm{Area}(S) < \frac{1}{8} \mathrm{Area}(\mathbf{S}^2) = \frac{\pi}{2}$, and comparing it to the three planes obtained by setting $x = 0, y = 0, z = 0$ respectively one finds $\mathrm{Area}(S) < \frac{3}{4} \mathrm{Area}(D_1) = \frac{3\pi}{4}$.
Question 1: What is the (approximate) value of $\mathrm{Area}(S)$? Are there simple improvements to the upper bound of $\frac{\pi}{2}$?
More generally, let $(r,\theta,z)$ define cylindrical polar coordinates on $\mathbf{R}^3$, and for $\varphi \in (0,\pi)$ let $R_{\varphi} \subset \mathbf{S}^2$ be $\partial \{ (r,\theta,z) \in \mathbf{S}^2 \mid 0 < \theta < \varphi, z > 0 \}$. This is the set of points in $\mathbf{S}^2$ with $z \geq 0$, $0 \leq \theta \leq \varphi$ and equality in one of the two. Let again $S_{\varphi}$ be the surface of least area with $\partial S_{\varphi} = R_{\varphi}$. The observations above yield $\mathrm{Area}(S_{\varphi}) < \varphi$ and $\mathrm{Area}(S_\varphi) < \frac{1}{2} (\mathrm{Area}(D_1) + \varphi) = \frac{1}{2}(\pi + \varphi)$. Note moreover that for all $\lambda \in [0,1]$, $A(\varphi) \leq A(\lambda \varphi) + A((1 - \lambda) \varphi)$.
Question 2: What is the (approximate) value of $\mathrm{Area}(S_\varphi)$? Can the upper bounds above be improved, for example when $\varphi$ is away from $0$ and $\pi$?
