Answer
$\dfrac{2 \pi}{3}(2 \sqrt 2-1)$
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 = \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, $Surface \ Area ; 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} \times r \ dr \ d\theta $
Set $1+r^2 =u$ and $du= 2 r dr$
Now, $Surface \ Area; A(S)= \dfrac{1}{2} \int_0^{2 \pi} \int_{1}^{2} \sqrt {u} \ du \ d \theta=\dfrac{1}{2} \times \dfrac{2}{3} \int_0^{2 \pi}[u^{3/2}]_{1}^{2} \ d \theta$
Therefore , $Surface \ Area A(S)=\dfrac{2 \pi}{3}(2 \sqrt 2-1)$
