Answer
$\dfrac{2 \pi}{3}(2 \sqrt 2-1)$
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+(y)^2+(x)^2 } dA= \iint_{D} \sqrt {1+y^2+x^2} dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA=\iint_{D} \sqrt {1+y^2+x^2} dA =\int_0^{2 \pi} \int_{0}^{1} \sqrt {1+r^2} r dr d \theta $
Substitute $a=1+r^2 $ and $da= 2 r dr$
Now, $A(S)= \dfrac{1}{2} \int_0^{2 \pi} \int_{1}^{2} \sqrt {a} da d \theta=\dfrac{1}{2} \int_0^{2 \pi} [\dfrac{2a^{3/2}}{3}]_{1}^{2} d \theta$
This implies that,
$A(S)=\dfrac{2 \pi}{3}(2 \sqrt 2-1)$
