= Solution
Use the positive graph $z=g(x,y)=\sqrt{1+x^2+y^2}$. The printed PDF bounds are $\sqrt2\le z\le\sqrt5$, so its projection is the <annulus> $1\le r\le2$. The surface is the corresponding band of the upper sheet of a <two-sheeted hyperboloid>, bounded by circles of radii $1$ and $2$ at heights $\sqrt2$ and $\sqrt5$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-3-hyperboloid-annulus.png]
{title=The upper hyperboloid band between heights square root of two and square root of five, with annular projection}
For the parametrization $\mathbf R(x,y)=(x,y,g(x,y))$, its tangent <vectors> are $(1,0,x/z)$ and $(0,1,y/z)$. Their <cross product> gives the upward <vector surface element of a graph>:
$$
\boxed{d\mathbf S=\left(-\frac{x}{z},-\frac{y}{z},1\right)dx\,dy,
\qquad z=\sqrt{1+x^2+y^2}.}
$$
Its magnitude gives the scalar <surface area of a graph>:
$$
\boxed{dS=\sqrt{\frac{1+2(x^2+y^2)}{1+x^2+y^2}}\,dx\,dy.}
$$
Both formulas apply on $1\le\sqrt{x^2+y^2}\le2$. Reversing the orientation changes only the sign of the <vector> element.
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