= 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 is
$$
\boxed{f(u,v)=(\cos v\cos u,\cos v\sin u,\sin v)}.
$$
On 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 are
$$
K=\frac{\cos v}{2+\cos v},
\qquad
dA=(2+\cos v)\,du\,dv.
$$
Therefore the positive outer and negative inner contributions cancel:
$$
\boxed{\int_{f^{-1}(U)}K\,dA
=\int_0^{2\pi}\int_0^\pi\cos v\,dv\,du=0}.
$$
Back to article page
