Answer
$\dfrac{\sqrt 2}{6}$
Work Step by Step
We have $Surface \ Area =\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 \times \iint_{D} dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $Surface \ Area =\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA \\ =\int_0^1 \int_{x^2}^x \sqrt 2 dy dx \\ = \sqrt 2 \times \int_0^1 [y]_{x^2}^x dx \\ \\ =\sqrt 2 \times \int_0^1 (x-x^2) dx \\ =\sqrt 2 \times (\dfrac{1}{2}-\dfrac{1}{3}) \\ =\dfrac{\sqrt 2}{6}$
