Answer
$\dfrac{2 \pi}{3}(2 \sqrt{2}-1)$
Work Step by Step
We have: $z=f(x,y)=xy$, with $
D=\left\{ (x, y) | x^{2}+y^{2} \leq 1\right\}$
and $f_{x}=y, \space f_{y}=x$
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{y^{2}+x^{2}+1} d A $
and, $\iint_{D} dA$ is the area of the region inside $D$.
We can use the polar co-ordinates because of the part $x^2+y^2+z^2$
Therefore, $A(S) =\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{r^{2}+1} r d r d \theta \\
=\int_{0}^{2 \pi}\left[\dfrac{1}{3}\left(r^{2}+1\right)^{3 / 2}\right]_{0}^{1} d \theta \\
=\int_{0}^{2 \pi} \dfrac{1}{3}(2 \sqrt{2}-1) d \theta \\
=\dfrac{2 \pi}{3}(2 \sqrt{2}-1)$
