Answer
$\dfrac{\sqrt 2}{6}$
Work Step by Step
We have $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial 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 $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA=\int_0^1 \int_{x^2}^x \sqrt 2 dy dx$
or, $A(S)= \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}) $
This implies that , $A(S)=\dfrac{\sqrt 2}{6}$
