= Solution
For an oriented regular surface with unit normal $N$, the <Gauss map> sends each point $p$ to $N(p)\in S^2$. The outward normal to the catenoid is
$$
N(u,v)=(\operatorname{sech}u\cos v,
\operatorname{sech}u\sin v,-\tanh u).
$$
Its image is the open southern hemisphere, apart from its limiting equator and south pole. The Gauss map is one-to-one after taking $v$ modulo $2\pi$ and reverses orientation. Since the Jacobian of the Gauss map is the <Gaussian curvature>, its signed spherical area is the <total Gaussian curvature>:
$$
\boxed{\int_{S'}K\,dA=-\operatorname{area}(\text{hemisphere})=-2\pi}.
$$
Equivalently, $K\,dA=-\operatorname{sech}^2u\,du\,dv$, whose integral over $u>0$ and $0\leq v<2\pi$ is $-2\pi$.
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