Answer
$ \approx 3.6258$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ is the projection of the surface on the xy-plane.
Now, the area of the given surface is: $A(S)=\iint_{D} \sqrt {1+(\dfrac{-2x}{(1+x^2+y^2)^2})^2+[\dfrac{-2y}{(1+x^2+y^2)^2}]^2} dA \\ =\int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} \sqrt {1+\dfrac{4x^2}{(1+x^2+y^2)^4}+(\dfrac{4y^2}{(1+x^2+y^2)^4}} dx \ dy$
By using a calculator, we have:
$A(S)=\int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} \sqrt {1+\dfrac{4x^2}{(1+x^2+y^2)^4}+(\dfrac{4y^2}{(1+x^2+y^2)^4}} dx \ dy \approx 3.6258$
