We assume that spherical geometry takes place on the unit sphere, that is, the sphere of radius 1 unit centered at the origin. If we later want to assume that the sphere is our planet, we just set 1 unit=6378.1 km. We will thus assume that the units of measurement (of lengths and areas) are chosen so that the radius of the sphere itself is 1 unit. For more background on spherical geometry, see Wikipedia's Spherical Geometry page [10], Wolfram MathWorld's Spherical Geometry page [11], or John Polking's Spherical Geometry Page [12].

When we measure the distance between points \(A\) and \(B\) in spherical geometry, we measure the shortest distance between the points *as measured on the surface of the sphere.* It is important to notice that, although the distance from New York to Melbourne through the interior of the earth is shorter than the distance along the surface, beings living in spherical geometry, like humans living on earth, cannot travel through the interior. When we speak of a circle of radius \(r\) in spherical geometry, we refer to the set of points on the sphere that are of distance \(r\) from a given point (also on the sphere), as measured on the surface of the sphere.

In spherical geometry, we do not calculate the area of a circle using the method of Subsection 2.1, since we cannot lay our slices out flat. We can, however, use variants of the other two methods.

The circumference of a circle begins at 0 and grows until the radius is \(\pi/2\), when the circumference is \(2\pi\). After this it shrinks until the radius is \(\pi\) (as large as possible), when the circumference is 0. Area of a circle also begins at 0, increases to \(2\pi\) (half the sphere) when \(r=\pi/2\), and further increases to \(4\pi\) (the whole sphere) when \(r=\pi\). It is difficult to find a "nice" formula for area that fits these criteria without thinking a bit further.

Rotate the sphere so that the center of the circle in question is at the point \((1,0,0)\)* *. The circle then must lie inside the Euclidean plane \(x=x_0\)* * for some appropriate \(x_0\), so the projection of the circle onto the \(xz\)-plane looks like the vertical line in Figure 4 (see also Figure 5 for the three dimensional picture). The vertical line also represents a diameter for the Euclidean circle corresponding to the spherical circle. Note that since we are working with a sphere of radius 1, the radius \(r\) of the spherical circle is equal to the angle measure \(r\) (measured in radians) shown in Figure 4, so that the radius of the Euclidean circle in the plane \(x=x_0\) (\(=\cos r\)) is \(\sin r\). Then it is straightforward to calculate the circumference of the circle: \(C=2\pi\sin r\).

Note that for the same reason as in Subsection 2.1, we can say that \(dA/dr\) is the length of the circumference. However, the circumference is now \(2\pi\sin r\). Hence we obtain \(dA/dr=2\pi\sin r\). Antidifferentiation gives us \(A=-2\pi\cos r+K\). Inspection of the known values of \(A\) gives us \(K=2\pi\), so that \(A=2\pi-2\pi\cos r\) (this is equivalent to \(4\pi\sin^2(r/2)\), given by some texts).

From our own experience, we know that when the radius of a circle in spherical geometry is small, the actual area is very nearly the Euclidean area \(\pi r^2\) and the actual circumference is very nearly the Euclidean circumference \(2\pi r\). Why? Because we live on a sphere, and a small radius on the unit sphere corresponds to a radius here on earth that is small compared to the radius of the earth. When we measure circles, we usually don't notice the curvature of the earth, obtaining (incorrectly) the Euclidean formulas. However, the Euclidean formulas and the (correct) spherical formulas are so close that we cannot tell the difference. In other words, we expect that the spherical circumference and area formulas are asymptotic to the Euclidean formulas as \(r \rightarrow 0\). We can check that this is true of our formulas by examining their MacLaurin series:

\[C=2\pi\sin r=\frac{2\pi}{1!}r-\frac{2\pi}{3!}r^3+\frac{2\pi}{5!}r^5-\dots\approx2\pi r\]

\[A=2\pi-2\pi\cos r=\frac{2\pi}{2!}r^2-\frac{2\pi}{4!}r^4+\frac{2\pi}{6!}r^6-\dots\approx\pi r^2\]

the latter approximations true when \(r\) is small. This verifies that our calculated formulas are reasonable. Note that the former implies

\[\lim_{r \rightarrow 0}\left(\frac{\sin r}{r} \right)=1\]

while the latter implies

\[\lim_{r \rightarrow 0}\left(\frac{1-\cos r}{r^2} \right)=\frac12\]

both familiar formulas from differential calculus.

We can calculate area of a spherical circle by using integral calculus as well. We assume that our circle is centered at the point \((1,0,0)\), and that the circle lies in the plane \(x=x_0\), parallel to the \(yz\)-plane, as in Figure 5. Then the interior of the circle is a surface of rotation, its area given by the integral

\[A=2\pi\int_{x_0}^1 y\left(1+\left[\frac{dy}{dx} \right]^2 \right)^{1/2} dx\]

Substituting \(y=(1-x^2)^{1/2}\) into this equation gives us a particularly simple integral, evaluating to

(1)\[A=2\pi(1-x_0)\]

Since we are interested in the relationship between area and radius, we now calculate the radius of the circle in terms of \(x_0\). Recall, radius in spherical geometry is as measured on the surface of the sphere. In other words, the radius can be seen as the arclength of the curve \(y=(1-x^2)^{1/2}\) from \(x=x_0\) to 1. Then we obtain the integral

\[r=\int_{x_0}^1\left(1+\left[\frac{dy}{dx} \right]^2 \right)^{1/2}dx\]

Making the appropriate substitutions yields \(r=\pi/2-\arcsin x_0\), or in reverse, \(x_0=\sin(\pi/2-r)=\cos r\). Note that substitution of this value into Equation (1) gives us

(2)\[A=2\pi-2\pi\cos r\]

just as in Subsection 3.1. From here we can differentiate to verify our formula for circumference.

For a spherical right triangle, we again denote the lengths of the sides \(a\) and \(b\), and the hypotenuse, that across from "the" right angle, we denote \(c\). If our spherical triangle has more than one right angle, we will have to agree on which right angle we are considering "the" right angle when we examine the "Spherical Pythagorean Theorem." Note that \(0\leq a\leq \pi\), and similarly for \(b\) and \(c\).

It is unlikely that students will discover the Spherical Pythagorean Theorem by trial and error as they might by simply examining the areas of circles constructed on each side as in Subsection 2.3. In Geometry Playground [1], we choose spherical geometry and construct a right triangle and its edge-circles as in Figure 1b, or simply open Geometry Playground with a pre-constructed spherical right triangle, and then construct the edge-circles on that right triangle as in Figure 1b.

Measuring the area \(A\) of each circle, and calculating \(\cos r=(2\pi-A)/(2\pi)\) from Equation (2), it shouldn't take long for the relationship \(\cos a\cdot \cos b = \cos c\) to stand out. In fact, this is the "Spherical Pythagorean Theorem."

However, we live on the surface of a sphere, and usually do not calculate the length of the hypotenuse of a right triangle using this formula; we use the Pythagorean Theorem (we rarely need the Spherical version, as the sides of our right triangles are usually quite small in comparison to unit length, the radius of the sphere!). This tells us that as \(a\), \(b\), and \(c\) decrease, we should expect the Spherical Pythagorean Theorem to approach the Euclidean one. This can again be verified via MacLaurin expansion. Substituting MacLauren series \(\cos r=1-r^2/2!+r^4/4!-\dots\) for \(\cos a\), \(\cos b\), and \(\cos c\), we obtain

\[\left(1-\frac1{2!}a^2+\frac1{4!}a^4-\dots \right)\left(1-\frac1{2!}b^2+\frac1{4!}b^4-\dots \right)=1-\frac1{2!}c^2+\frac1{4!}c^4-\dots\]

Distributing the left, we arrive at \(1-\frac1{2!}a^2-\frac1{2!}b^2+\dots=1-\frac1{2!}c^2+\dots\)*,* with all remaining terms of degree 4 or higher. Since we are assuming \(a\), \(b\), and \(c\) are all small, we ignore the higher order terms. Canceling the constant terms and multiplying by \(-2\), we obtain the Euclidean Pythagorean Theorem.