Gauss–Bonnet theorem
…characteristic is 1. On the left hand side of the theorem, we have K = 1 / R 2 and k g = 0 , because the boundary is the equator and the equator is a geodesic of the sphere. Then, since M has area 4 π R 2 2 , we obtain that ∫ M K d A = 2 π . On the other hand, suppose we flatten the hemisphere to make it into…