Answer
$$\approx 2.695884$$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed as: $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dx \ dy$
and, $\iint_{D} dA$ represents the projection of the surface on the $xy$-plane.
The area of the given surface is: $A(S)=\iint_{D} \sqrt{1+(2x/1+y^2)^{2}+(-2y/-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}} d A $
and $A(S)=\iint_{D} \dfrac{1}{(1+y^2)^2} \sqrt {(1+y^2)^4+4x^2 (1+y^2)^2+4y^2(1+x^2)^2} dA $
Next, apply the polar co-ordinates because of the part $x^2+y^2$
Now, we will use a calculator to find the area of the given surface.
$$A(S)=4 \int_{0}^{1} \int_{0}^{1-x}\dfrac{1}{(1+y^2)^2} \sqrt {(1+y^2)^4+4x^2 (1+y^2)^2+4y^2(1+x^2)^2} dy dx \approx 2.695884$$
