Solution
= Solution
The relevant projection is the disk $x^2+y^2\le9$. The <surface area of a graph> is $\iint\sqrt{1+z_x^2+z_y^2}\,dx\,dy$. Here $z_x=-2x$ and $z_y=-2y$, so <polar coordinates> give
$$
A=2\pi\int_0^3r\sqrt{1+4r^2}\,dr
=\frac\pi6\left[(1+4r^2)^{3/2}\right]_0^3
=\boxed{\frac\pi6\bigl(37\sqrt{37}-1\bigr).}
$$
The boundary circle has zero surface area, so including the points at height zero does not change the result.