Answer
$$\dfrac{\sqrt 2}{6}$$
Work Step by Step
$$A(S)=\iint_{D} \sqrt {1+(z_x )^2+(z_y )^2 } dA \\= \iint_{D} \sqrt {1+(\dfrac{x}{\sqrt{x^2+y^2}})^2+(\dfrac{y}{\sqrt{x^2+y^2}})^2 } dA \\=\sqrt 2 \iint_{D} dA \\=\int_0^1 \int_{x^2}^x \sqrt 2 dy dx \\ = \sqrt 2 \int_0^1 [y]_{x^2}^x dx=\sqrt 2 \int_0^1 (x-x^2) dx \\=\sqrt 2 (\dfrac{1}{2}-\dfrac{1}{3}) \\=\dfrac{\sqrt 2}{6}$$
