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