Answer
$\dfrac{2 \pi}{3}(2 \sqrt{2}-1)$
Work Step by Step
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
We have: $z=f(x,y)=xy$ ; with $
D=\left\{ (x, y) | x^{2}+y^{2} \leq 1\right\}$
and $f_{x}=y , \space f_{y}=x$
Our aim is to calculate the area of the given surface. $A(S)=\iint_{D} \sqrt{y^{2}+x^{2}+1} d A $
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
Apply the polar co-ordinates because of the part $x^2+y^2+z^2$
Now, $A(S) =\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{r^{2}+1} \ r \ d r \ d \theta $
or,$=\int_{0}^{2 \pi}[\dfrac{1}{3}\left(r^{2}+1)^{3 / 2}\right]_{0}^{1} \ d \theta$
or, $=\int_{0}^{2 \pi} \dfrac{1}{3}(2 \sqrt{2}-1) \ d \theta$
or, $ =\dfrac{2 \pi}{3}(2 \sqrt{2}-1)$
