Answer
$\approx 2.69588$
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}{1+y^2})^2+(-\dfrac{2y(1+x^2)^2}{(1+y^2)^4})^2} dA $
or, $=\iint_{D} \sqrt {1+\dfrac{4x^2}{1+y^2})^2+\dfrac{4y^2(1+x^2)^2}{(1+y^2)^4}} dA $
By using a calculator, we get
$A(S) = \int_{-1}^{1} \int_{|x|-1}^{1-|x|} \dfrac{1}{(1+y^2)^2} \sqrt {(1+y^2)^4+4x^2 (1+y^2)^2+4y^2 (1+x^2)^2} dA \approx 2.69588$
