 Solution

For an oriented smooth embedded surface, the Gauss map sends each point to its chosen unit normal. This surface is the ring torus with major radius two and minor radius one. Its outward Gauss map isOn the $y=0$ cross-section, the normal points radially away from the center of each generating circle, which gives this formula for $u=0,\pi$ and then rotational symmetry gives it for all $u$.

The upper hemisphere condition is $\sin v>0$, hence $0<v<\pi$. For this torus, the Gaussian curvature and area element areTherefore the positive outer and negative inner contributions cancel: