how to find the surface area of a unit disk

Question

Suppose $z$ is a function for which $ \lvert \partial z$/ $\partial x \rvert$ and $\lvert \partial z/ \partial y \rvert$ are both at most $1$. How large could the surface area of the graph of $z$ be above the unit disk?

Answer 1

Note that the surface area is given,

$$\int_{r\le 1} \sqrt{1+(z_x')^2 + (z_y')^2} ~dS \le \int_{r\le 1} \sqrt3~ dS = \sqrt3 \pi$$

Thus, the largest surface area is $\sqrt3\pi $.