Area of a surface The question is; 
If the equation of a surface $S$ is $z = f(x, y)$, where $x^2 + y^2 \leq R^2$, and you know that $|f_x| \leq 1$ and $|f_y| \leq 1$, what can you say about $Area(S)$?
Options are: 
1) $Area(S) = π^2R$
2) $πR^2 ≤ Area(S) ≤ \sqrt3πR^2$
3) $Area(S) > 3πR^2$
4) $πR ≤ Area(S) ≤ πR^2$
5) $πR ≤ Area(S) ≤ 3πR$
My workings so far:
$Area(S)= \iint_D \sqrt{(\frac{df}{du})^2 + (\frac{df}{dv})^2 +1} dA$
Since $|f_x| \leq 1$ and $|f_y| \leq 1$,
$Area(S)= \iint_D \sqrt{(\frac{df}{du})^2 + (\frac{df}{dv})^2 +1}dA \leq \iint_D \sqrt3 dA$
where the domain is $ D = ${$(x,y): x^2 + y^2 \leq R^2$}.
And now I'm stuck, where do I go from here? 
If I had to guess, I would choose (option 2) $πR^2 ≤ Area(S) ≤ \sqrt3πR^2$

From your work you obtain
$$\iint_D \sqrt3 dA=\sqrt 3\pi R^2$$
