Answer
$\pi \sqrt {14}$
Work Step by Step
We have $Surface \ Area=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA$
and $Surface \ Area = \iint_{D} \sqrt {1+(-1/3)^2+(-2/3)^2 } dA=\dfrac{\sqrt {14}}{3} \iint_{D} dA.....(a)$
and $\iint_{D} dA$ is the area of the region $D$
$D$ defines the area of the region inside circle $x^2+y^2=3$ with radius $\sqrt 3$ and area of a circle $3 \pi$ .
Therefore , $\iint_{D} dA= 3 \pi ....(b)$
From above equation (a) and (b) we can conclude that the area of the plane in first quadrant becomes: $\dfrac{\sqrt {14}}{3} \times \iint_{D} dA=\dfrac{\sqrt {14}}{3} \times 3 \pi$
or, $Surface \ Area= \pi \sqrt {14}$
