Given a polygon $P$, with geodesic edges, on the surface of the unit sphere in $\mathbb R^3$, what is the integral of the unit normal vector $\hat n$ over the polygon's area? (The normal vector is simply the position vector, or its opposite.)
$$\iint_P \hat n\,dA$$
I know that any polygon can be broken up into triangles, and any triangle can be split into 2 right triangles. So it suffices to integrate the normal vector over an arbitrary right triangle.
I've calculated the integral over a half-lune $H$, which is a triangle with 2 right angles. If the sphere is parametrized by $\vec x = (\cos\theta\sin\phi)\hat e_1 + (\sin\theta\sin\phi)\hat e_2 + (\cos\phi)\hat e_3$, and two vertices are at the equator ($\phi = \frac\pi 2$) and the other at the north pole ($\phi = 0$), then
$$\iint_H \hat n\,dA = \frac\pi 4(\sin\theta_2 - \sin\theta_1)\hat e_1 - \frac\pi 4(\cos\theta_2 - \cos\theta_1)\hat e_2 + \frac12(\theta_2 - \theta_1)\hat e_3$$
I'm still trying to figure out the geometric meaning of this, and I don't know what to do when the triangle has only 1 right angle.
Aha! I see that the half-lune's integral is half the sum of products of edge lengths with the unit vector that is tangent to the sphere and normal to the edge, pointing into the triangle.
$$\iint_H \hat n\,dA = \frac12(L_1\hat N_1 + L_2\hat N_2 + L_3\hat N_3)$$
I don't know if this generalizes, but it looks like it should.