It appears to me that after repeated applications of Girard's theorem on the area of spherical triangles that we can obtain the surface area of a spherical polygon with interior angles $\theta_{1},...,\theta_{n}$ on a sphere of radius $R$ is $$\text{area}(\text{spolygon}(\theta_{1},...,\theta_{n}))=R^2 \left(\sum_{i=1}^{n} \theta_{i} - (n-2)\pi\right).$$ Does this make sense to anyone else? I wanted to check before I used the result in a paper I am writing. Even if this just holds for spherical squares, it is all I need.
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asked Apr 10, 2015 at 18:13
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1$\begingroup$ This is essentially the Gauss-Bonnet theorem for geodesic polygon on a sphere. $\endgroup$– achille huiApr 10, 2015 at 18:43
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Yeah, you can always break a spherical n-gon in n−2 spherical triangles... You know, an interesting fact about higher dimensional spherical and euclidean simplices is that not all things generalize. For instance there is no relation between angle sums and volume in spherical simplices in the n-dimensional sphere for odd n (see here), and the fact that the sum of the angles is π for euclidean triangles does not generalize (see here) even in 3 dimensions...
answered Apr 10, 2015 at 18:22