Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/44

 gral curvature must vanish for $p = 0,$ the constant introduced by integration must be equal to the value of $\frac{\partial m}{\partial q}$  for $p = 0,$   i.e. , equal to unity. Thus we have

$$dq\left(1 - \frac{\partial m}{\partial p}\right){,}$$

where for $\frac{\partial m}{\partial p}$ must be taken the value corresponding to the end of this area on the line $CB.$  But on this line we have, by the preceding article,

$$\frac{\partial m}{\partial q}. dq = -d\theta,$$

whence our expression is changed into $dq + d\theta.$ Now by a second integration, taken from $q = 0$  to $q = A,$  we obtain for the integral curvature

$$A + \theta'- \theta^{\circ},$$

or

$$A + B + C - \pi.$$

The integral curvature is equal to the area of that part of the sphere which corresponds to the triangle, taken with the positive or negative sign according as the curved surface on which the triangle lies is concavo-concave or concavo-convex. For unit area will be taken the square whose side is equal to unity (the radius of the sphere), and then the whole surface of the sphere becomes equal to $4\pi.$ Thus the part of the surface of the sphere corresponding to the triangle is to the whole surface of the sphere as $\pm(A + B + C - \pi)$  is to $4\pi.$  This theorem, which, if we mistake not, ought to be counted among the most elegant in the theory of curved surfaces, may also be stated as follows:

The excess over $180^{\circ}$ of the sum of the angles of a triangle formed by shortest lines on a concavo-concave curved surface, or the deficit from $180^{\circ}$  of the sum of the angles of a triangle formed by shortest lines on a concavo-convex curved surface, is measured by the area of the part of the sphere which corresponds, through the directions of the normals, to that triangle, if the whole surface of the sphere is set equal to $720$  degrees.

More generally, in any polygon whatever of $n$  sides, each formed by a shortest line, the excess of the sum of the angles over $(2n - 4)$  right angles, or the deficit from $(2n - 4)$  right angles (according to the nature of the curved surface), is equal to the area of the corresponding polygon on the sphere, if the whole surface of the sphere is set equal to $720$  degrees. This follows at once from the preceding theorem by dividing the polygon into triangles.