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Using the formula for the area of a spherical triangle, find and prove a formula relating the angle sum of a spherical polygon to its area


Thought:

Area (spherical triangle) $=\mathbb{R}^2(\alpha+\beta+\gamma-\pi)$ where $\alpha, \beta, \gamma$ are the interior angles

Let $n$ be the number of vertices/sides of a polygon.

Consider a quadrilateral $n=4$ with interior angle $(\alpha,\beta,\gamma,\delta)$

Area (spherical quadrilateral) $=\mathbb{R}^2(\alpha+\beta+\gamma+\delta-2\pi)$

then how do I induce to infinite $n$???

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  • $\begingroup$ Don't you mean: "how do I induce to any finite n" ? $\endgroup$ – Jean-Claude Arbaut Mar 12 '13 at 14:26
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The natural thing seems to be the area is the sum of the angles minus $\pi$ times two less than the number of sides. You should be able to prove this by cutting the polygon into triangles.This means that in Euclidean space all polygons have zero area.

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  • $\begingroup$ "This means that in Euclidean space all polygons have zero area." Of course this is wrong. In spherical geometry, angle excess is proportional to area. This does not happen in euclidian geometry. $\endgroup$ – Jean-Claude Arbaut Mar 12 '13 at 7:11
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    $\begingroup$ @arbautjc: true, but it is not as silly as it sounds. Euclidean geometry is like spherical geometry on an infinite radius sphere. All finite polygons have zero area relative to such a sphere. $\endgroup$ – Ross Millikan Mar 12 '13 at 13:02
  • $\begingroup$ There is an r^2 factor which tends toward infinity, that's why the limit is not zero. $\endgroup$ – Jean-Claude Arbaut Mar 12 '13 at 14:21
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A polygon with an infinite number of vertices/sides is a circle. So, what is the area of a circle in spherical geometry? It is the surface area covered by the region on $\mathbb{S}^2$ bounded by the curve resulting from intersecting $\mathbb{S}^2$ with a plane.

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    $\begingroup$ "A polygon with an infinite number of vertices/sides is a circle". Only if it's a regular polygon. However, the formula to be proved is true for any polygon on the sphere. $\endgroup$ – Jean-Claude Arbaut Mar 12 '13 at 14:23
  • $\begingroup$ @arbautjc: Can you please give me an example of a non-regular polygon with an infinite number of sides that has a finite area? :) $\endgroup$ – Samuel Reid Mar 12 '13 at 22:25
  • $\begingroup$ Given that "infinite number of side" is meaningless, I can't give you a correct answer - not even in the circle case. But if you take a limit, you can get any rectifiable curve (on the sphere, of course). A circle is only one such curve. You can get an idea here: books.google.fr/… For an example with finite area, let's take the Viviani curve: en.wikipedia.org/wiki/Viviani%27s_curve . Of course, remember that on the sphere you must take spherical polygons, not straight lines. $\endgroup$ – Jean-Claude Arbaut Mar 13 '13 at 4:38
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It is obvious that typically if the no. of sides, of a regular polygon having a finite circumscribed radius say $r$, tends to infinity i.e. $n \to \infty \space$ then the regular polygon becomes a circle with a radius $r$.

Now, let $r$ be the radius of circle on a spherical surface with a radius $R$ then the solid angle ($\omega$) subtended by the circle at the center of the sphere is given as $$\omega=2\pi \left(1-\sqrt{1-\left(\frac{r}{R}\right)^2}\right)$$
Hence, the area covered by a circle (i.e. a polygon having infinite no. of sides) on the sphere is given as $$Area=(solid \space angle) \cdot ({(radius)}^2)=\omega R^2=2\pi R^2\left(1-\sqrt{1-\left(\frac{r}{R}\right)^2}\right)$$

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