 Solution

For a closed surface, $\chi =V-E+F$ for any finite cell decomposition. Gauss–Bonnet for a geodesic polygon saysand for a closed smooth surface it says ${\int }_{S}KdA=2\pi \chi (S)$.

For a large radius, the graph surface bounded over that circle is flat near its boundary. The boundary geodesic-curvature integral is $2\pi$, so Gauss–Bonnet on the disc gives $\int K=0$. Since $K\ge 0$, continuity forces $K\equiv 0$.

For the torus,Thus $T_{+}$ has $\cos u>0$ and $T_{-}$ has $\cos u<0$. Cap the two boundary circles of the outer half $T_{+}$ by flat discs. The result is a sphere, and the caps contribute no Gaussian curvature, so Gauss–Bonnet gives ${\int }_{T_{+}}K=4\pi$. The full torus has Euler characteristic zero, hence total curvature zero and