Area of a surface

Question

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) $πR^2 ≤ Area(S) ≤ \sqrt3πR^2$

Answer 1

From your work you obtain

$$\iint_D \sqrt3 dA=\sqrt 3\pi R^2$$

Answer 2

You have it right here $$Area(S)= \iint_D \sqrt{(\frac{df}{du})^2 + (\frac{df}{dv})^2 +1}dA \leq \iint_D \sqrt3 dA= \sqrt 3\pi R^2 $$