Answer
$6\sqrt 2+\ln (2 \sqrt 2 +3)$
Work Step by Step
$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+(1)^2+(2z)^2} dA =2 \sqrt 2 \iint_{D} \sqrt {1+2z^2+17} \ dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $Surface \ Area; A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA \\=\int_0^{2} \int_{0}^{2} 2 \sqrt 2 \sqrt {1+2z^2+17} \ dy \ dz \\=2 \sqrt {2} \times \int_0^2 \sqrt {1+2z^2} \ dz$
Set $\sqrt 2 z = u $ and $\dfrac{dz}{ \sqrt 2} = du$
Now, $Surface \ Area; A(S)=2 \times \int_0^{2 \sqrt 2} \sqrt {1+u^2} du =6\sqrt 2+\ln (2 \sqrt 2 +3)$
