Solution
= Solution
The parametrisation $r(x,y)=(x,y,F(x,y))$ has area factor
$$
|r_x\times r_y|=\sqrt{1+F_x^2+F_y^2},
$$
so $\operatorname{area}(S)=\iint_D\sqrt{1+F_x^2+F_y^2}\,dx\,dy$.
For the unbounded saddle $F=(x^2-y^2)/2$, this factor is $\sqrt{1+x^2+y^2}$. Thus the requested <integral> is
$$
2\pi\int_0^\infty\frac{r\,dr}{(1+r^2)^{3/2}}=2\pi,
$$
which converges because its radial tail is $O(r^{-2})$.