 Solution

For a positively oriented curvilinear polygon bounding a disk $D$ in one coordinate patch, with exterior turning angles ${\theta }_i$, the Gauss-Bonnet theorem statesFor a closed compact surface the global formula isTriangulate the surface with each triangle contained in a coordinate patch. For each face, writing its exterior angles as $\pi$ minus its interior angles gives $\int K+\int {\kappa }_g=\sum \text{interior angles}-\pi$. On summing, the two contributions on each internal edge cancel, and all angles at each vertex sum to $2\pi$. Thus ${\int }_{S}K=2\pi V-\pi F$. Since $3F=2E$, this equals $2\pi (V-E+F)$, as required. For a nonorientable surface apply the orientable version to its orientation double cover: both the curvature integral and Euler characteristic double. With boundary, the same summation retains the boundary geodesic-curvature integral and the exterior angles at boundary corners, giving the corresponding global boundary formula with right side $2\pi \chi (S)$.