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