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The relevant projection is the disk $x^2+y^2\le 9$. The surface area of a graph is $\iint \sqrt{1+z_x^2+z_y^2}dxdy$. Here $z_x=-2x$ and $z_y=-2y$, so polar coordinates giveThe boundary circle has zero surface area, so including the points at height zero does not change the result.

Use the positive graph $z=g(x,y)=\sqrt{1+x^2+y^2}$. The printed PDF bounds are $\sqrt{2}\le z\le \sqrt{5}$, so its projection is the annulus $1\le r\le 2$. 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 $\sqrt{2}$ and $\sqrt{5}$.

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:Its magnitude gives the scalar surface area of a graph:Both formulas apply on $1\le \sqrt{x^2+y^2}\le 2$. Reversing the orientation changes only the sign of the vector element.

Parametrize the function graph by $\mathbf{R}(x,y)=(x,y,f(x,y))$. Its two tangent vectors are ${\mathbf{R}}_x=(1,0,f_x)$ and ${\mathbf{R}}_y=(0,1,f_y)$. A small parameter rectangle maps, to first order, to a tangent parallelogram with area $|{\mathbf{R}}_x\times {\mathbf{R}}_y|dxdy$. Sincethe surface area of a graph isFor example, $f\in C^1$ ensures a regular surface chart; more general domains can be handled by the corresponding convergent improper integral.