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