Surface area of the part of a sphere above a hexagon

Question

I want to calculate the surface area of the part of a half-sphere, which lies above a regular 6-gon. (Radius $r=1$)

More formally, Let $G$ be the region on the $XY$-Plane, bounded by the points $\{P_k=(\cos(\frac{2\pi k}{6}), \sin(\frac{2\pi k}{6}))\}$ for $k=1, ..., 6$. It's just the hexagon whose nodes are on the unit circle.

I want to calculate the area of $S_G = \{(x,y,z)\in \mathbb{R}^3 \ |\ x^2+y^2+z^2 =1, \ z>0, \ and \ (x,y)\in G \}$.

What I did:

Similar to how one computes the sphere's area, I took $F(x,y,z)=x^2+y^2+z^2-1=0$, and eventually got to $Area(S_G)=\int_G{\frac{1}{\sqrt{1-x^2-y^2}}dx dy}$.

Seems kind of nasty since $G$ is not a nice domain to work on with polar coordinates (that would have helped if my domain was "round" in some way, because then it reduces to $\int{\frac{1}{\sqrt{1-s}}ds}$ ).

Any ideas?

Answer 1

The union of $S_{G}$ and its reflection across the $(x, y)$-plane is a unit sphere with six circular caps removed. Each cap is a zone cut by planes separated by $1 - \frac{\sqrt{3}}{2}$, and so has area $\pi(2 - \sqrt{3})$ by Archimedes' theorem. That is, the area of $S_{G}$ is $$ \tfrac{1}{2}\bigl[4\pi - 6\pi(2 - \sqrt{3})\bigr] = \pi\bigl[2 - 3(2 - \sqrt{3})\bigr] = \pi [3\sqrt{3} - 4] \approx 1.196152423\pi \approx 3.757823664. $$

Answer 2

Hint:

By simmetry, the area of the surface of the half sphere which lies above the equilateral triangle $OP_6P_1$ is $\frac16S_G$.

Now, $\triangle OP_6P_1$ is the set bounded by the $x$ axis, and the lines $y=\sqrt{3}x$ and $y=\sqrt{3}(1-x)$. In polar coordinates this region is $$\left\{(\theta,r)\in\mathbb{R}^2:0\leq\theta\leq\frac{\pi}{3},0\le r\le\frac{\sqrt{3}}{\sqrt{3}\cos\theta+\sin\theta}\right\}$$

Then, \begin{align*} \frac{1}{6}S_G&=\int_0^{\frac{\pi}{3}}\int_0^{\frac{\sqrt{3}}{\sqrt{3}\cos \theta+\sin \theta}}\frac{r}{\sqrt{1-r^2}}\,drd\theta\\[3pt] &=\int_0^{\frac{\pi}3}\left.-\sqrt{1-r^2}\right|_0^{\frac{\sqrt{3}}{\sqrt{3}\cos \theta+\sin \theta}}\,d\theta\\ &=\int_0^{\frac{\pi}{3}}\left(1-\sqrt{1-\frac{3}{4\sin^2\left(\theta+\frac{\pi}{3}\right)}}\right)\,d\theta \end{align*}

Last integral is hard to evaluate by hand, but Wolfram Mathematica give us $\frac{1}{6} \left(-4+3 \sqrt{3}\right) \pi $, then $$\color{blue}{S_G=\left(-4+3 \sqrt{3}\right) \pi}$$

Answer 3

The standard formula for area of a spherical segment is $ 2\pi R *$ Axial length.

Can be derived more easily as $ \int 2 \pi y \, ds = \int_{R-h}^R 2 \pi y \, \sqrt {1+y{\prime}^2} dx = 2 \pi R h $

For R =1,what is left after subtracting from hemi-sphere area is:

$$ 2 \pi R^2/3 - R ( 1 - \cos (\pi/6) ) \cdot 2\pi R \cdot * 6/2 $$

$$ \approx 3.75782 $$