Answer
$\approx 2.695884$
Work Step by Step
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
Our aim is to calculate the area of the given surface.
$A(S)=\iint_{D} \sqrt{1+(\dfrac{2x}{1+y^2})^{2}+(\dfrac{-2y(1+x^2)}{1+y^2})^{2}} d A \\ =\iint_{D} \sqrt{1+\dfrac{4x^2}{(1+y^2)^2}+\dfrac{4y^2(1+x^2)^2}{(1+y^2)^4}} \space d A \\=\iint_{D} \dfrac{1}{(1+y^2)^2} \sqrt {(1+y^2)^4+4x^2 \times (1+y^2)^2+4y^2 \times (1+x^2)^2} \space dA $
Now, we will apply the polar co-ordinates on the part $x^2+y^2$
Next we will use calculator to compute the result as follows:
$A(S)=4 \times \int_{0}^{1} \int_{0}^{1-x}\dfrac{1}{(1+y^2)^2} \sqrt {(1+y^2)^4+4x^2 (1+y^2)^2+4y^2(1+x^2)^2} dy dx \approx 2.695884$
