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