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