Answer
$\dfrac{4}{15}(3^{5 / 2}-2^{7 / 2}+1) .
$
Work Step by Step
We have: $
D=\left\{ (x, y) | 0\leq x\leq 1, \quad 0 \leq y \leq1\right\}
$
and $f_{x}=\sqrt {x}, \ f_{y}=\sqrt {y}$
The surface area of the part $z=f(x,y)$ can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ is the projection of the surface on the xy-plane.
Now, the area of the given surface is: $A(S)=\iint_{D} \sqrt{(\sqrt{x})^{2}+(\sqrt{y})^{2}+1} d A$
and, $\iint_{D} dA$ is the area of the region inside $D$.
$A(S) =\int_{0}^{1} \int_{0}^{1} \sqrt{x+y+1} d y d x \\
=\int_{0}^{1}\left[\dfrac{2}{3}(x+y+1)^{3 / 2}\right]_{y=0}^{y=1} d x \\ =\frac{2}{3} \int_{0}^{1}\left[(x+2)^{3 / 2}-(x+1)^{3 / 2}\right] d x \\
=\frac{2}{3}\left[\dfrac{2}{5}(x+2)^{5 / 2}-\frac{2}{5}(x+1)^{5 / 2}\right]_{0}^{1} \\
=\dfrac{4}{15}(3^{5 / 2}-2^{7 / 2}+1) .
$
