Answer
The area of the surface $z$ is
$$
A(S) =\frac{4}{15}\left(3^{5 / 2}-2^{7 / 2}+1\right) .
$$
Work Step by Step
$$
z=\frac{2}{3}(x^{\frac{3}{2}}+y^{\frac{3}{2}})
$$
and
$$
D=\left\{ (x, y) | 0\leq x\leq 1 , \quad 0 \leq y \leq1\right\}.
$$
then
$$
f_{x}=x^{\frac{1}{2}}, \quad f_{y}=y^{\frac{1}{2}}
$$
and
$$
\begin{aligned} A(S) &=\iint_{D} \sqrt{(\sqrt{x})^{2}+(\sqrt{y})^{2}+1} d A \\
& =\int_{0}^{1} \int_{0}^{1} \sqrt{x+y+1} d y d x \\
&=\int_{0}^{1}\left[\frac{2}{3}(x+y+1)^{3 / 2}\right]_{y=0}^{y=1} d x \\ &=\frac{2}{3} \int_{0}^{1}\left[(x+2)^{3 / 2}-(x+1)^{3 / 2}\right] d x \\
&=\frac{2}{3}\left[\frac{2}{5}(x+2)^{5 / 2}-\frac{2}{5}(x+1)^{5 / 2}\right]_{0}^{1} \\
&=\frac{4}{15}\left(3^{5 / 2}-2^{5 / 2}-2^{5 / 2}+1\right) \\
&=\frac{4}{15}\left(3^{5 / 2}-2^{7 / 2}+1\right)
\end{aligned}
$$
Thus, the area of the surface $z$ is
$$
A(S) =\frac{4}{15}\left(3^{5 / 2}-2^{7 / 2}+1\right) .
$$
