Answer
$Surface \ Area ; A(S) \leq \pi \sqrt 3 R^2$
Work Step by Step
Since, $|f_x| \leq 1 $ or, $ (f_x)^2 \leq 1$
and $|f_x| \leq 1$ or, $(f_y)^2 \leq 1$
Now, $Surface \ Area ; A(S)=\iint_{D} \sqrt {1+(f_x )^2+(f_y)^2 } dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $Surface \ Area ; =\sqrt {1+(f_x )^2+(f_y)^2 } \leq \sqrt {1+1+1} $
or, $Surface \ Area ; =\sqrt {1+(f_x )^2+(f_y)^2 } \leq \sqrt 3$
Now, $Surface \ Area ; A(S) \leq \sqrt 3 \iint_{D} dA$
Since, $x^2+y^2 \leq R^2$ and area of the region $D$ is $\pi R^2$
Thus, $ \iint_{D} dA= \pi R^2$
Now, $Surface \ Area ; A(S) \leq \pi \sqrt 3 R^2$
