Surface area of the part of a sphere above a hexagon 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?

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
$$\color{blue}{S_G=\left(-4+3 \sqrt{3}\right) \pi}$$
