Did Archimedes Know Gauss-Bonnet?

One of the most beautiful propositions of mathematics is called the
Gauss-Bonnet formula, which states that the integral of the Gaussian
curvature over a region of a surface equals 2pi minus the integral 
of the geodesic curvature over the boundary of that region.  This 
was first stated and proved by Bonnet in 1848.  The name of Gauss 
is attached to the formula because Gauss published the special case
for geodesic triangles, i.e., triangles whose sides are geodesics 
of the surface.

John Stillwell notes that if we further specialize to a spherical
surface, Gauss' result was proved by Thomas Harriot in 1603.  Given
a spherical triangle with angles a,b,c, Harriot's proof was based on 
the fact that any two edges of the triangle, when extended, constitute 
two great circles that intersect at opposite poles, and the surface 
area between two such "longitude lines" is simply  2a r^2  where r 
is the radius of the sphere and "a" is the angle between the longitude
lines, i.e., one of the angles of the original triangle.  So, letting 
A denote the area of the spherical triangle, and letting A_a, A_b, 
A_c the areas of the three longitude slices less A, we have

     A + A_a = 2a r^2     A + A_b = 2b r^2     A + A_c = 2c r^2

Consequently

              3A + (A_a + A_b + A_c)  =  2r^2 (a+b+c)

Noting that

               2A + 2(A_a + A_b + A_c)  =  4pi r^2

because this covers the entire sphere, we can eliminate the sum in
parentheses on the left side to give the result

                 (1/r^2) A  =  (a + b + c - pi)

The Gaussian curvature of the sphere's surface is a constant 1/r^2, 
so the left side is the integral of the curvature over the surface
of the triangle, which equals the "angular excess", i.e., the amount
by which the sum of the three angles of the spherical triangle exceeds
pi.

It occurs to me that if we're willing to interpret early results in
such a charitable way, we could just as well make the case that
Archimedes proved a special case of the Gauss-Bonnet formula.  As
discussed in Archimedes on Spheres and Cylinders, Archimedes
proved that the surface area of a region of a sphere of radius R 
sliced off by a plane equals the area of a circle whose radius is 
the straight-line distance from the central point of that region to 
the perimeter.  Now, if we project the slice locus onto a plane 
tangent to the sphere at any point of the slice, it forms an ellipse
with major and minor radii Rsin(a) and Rsin(a)cos(a) respectively.
Hence the geodesic curvature of this boundary is

                      Rsin(a)cos(a)         1
              k   =  ---------------  =  --------
                      [R sin(a)]^2       R tan(a)

The circumference of that locus is 2pi R sin(a), so the integral of
the geodesic curvature around this locus is 

                 /
                 | k ds  =  2pi cos(a)
                 /
                 S

Archimedes proved that the surface area bounded by this locus equals

           /                           \
 A  =  pi ( [R-Rcos(a)]^2 + [Rsin(a)]^2 )  =  2pi R^2 [1-cos(a)]
           \                           /

The Gaussian curvature of the sphere is a constant 1/R^2, so the
integral of the curvature throughout the enclosed area is

      /            A
      | K dA  =   ---  =  2pi[1-cos(a)]  =  2pi - 2pi cos(a)
      /           R^2
      A

Thus we can express Archimedes' result in the form

                /                  /
                | K dA  =   2pi -  | k ds
                /                  /
                A                  S

which is the celebrated Gauss-Bonnet formula.  Admittedly Archimedes
was restricted to spherical surfaces, but unlike Harriot or Gauss
his perimeter was not a geodesic.