Answer
$$\pi \sqrt {14}$$
Work Step by Step
We have $$A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2 } dA \\= \iint_{D} \sqrt {1+(-1/3)^2+(-2/3)^2 } dA \\=\dfrac{\sqrt {14}}{3} \iint_{D} dA $$
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $D$ is the area of the region inside the circle $x^2+y^2=3$ having radius $\sqrt 3$ and the area of the circle is: $3 \pi$
Therefore $\iint_{D} dA= 3 \pi$
The area of the plane in first quadrant is:
$$A(S)=\dfrac{\sqrt {14}}{3} \iint_{D} dA \\=3 \pi \times \dfrac{\sqrt {14}}{3} \\= \pi \sqrt {14}$$
