Answer
$$\dfrac{4}{15} [3^{5/2}-2^{7/2}+1]$$
Work Step by Step
$A(S)=\iint_{D} \sqrt {1+(z_x )^2+(z_y )^2 } dA= \iint_{D} \sqrt {1+(x^{1/2})^2+(y^{1/2})^2 } dA \\= \iint_{D} \sqrt {1+x+y} dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $$A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA \\=\int_0^1 \int_{0}^{1} \sqrt {1+x+y} dy dx\\= \dfrac{2}{3} \int_0^1 [(1+x+y)^{3/2} ]_{0}^1 dx \\=\dfrac{2}{3} \times \int_0^1(2+y)^{3/2} -(1+y)^{3/2} dy \\ =\dfrac{4}{15} \times [3^{5/2}-2^{5/2} -2^{5/2} +1] \\=\dfrac{4}{15} [3^{5/2}-2^{7/2}+1]$$
