Answer
$$\approx 3.56180$$
Work Step by Step
We have $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dA$
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{2x}{x^2+y^2+2})^2+(\dfrac{2y}{(x^2+y^2+2)^2}} dA $
or, $=\iint_{D} \sqrt {1+\dfrac{4x^2+4y^2}{(x^2+y^2+2)^2}} dA $
By using a calculator, we get
$A(S) = \int_{0}^{2 \pi} \int_{0}^{1} r \sqrt {1+\dfrac{4r^2}{(r^2+2)^2}} dr \ d \theta= 2 \pi \int_{0}^{1} r \sqrt {1+\dfrac{4r^2}{(r^2+2)^2}} dr
\approx 3.56180$
