Answer
$\dfrac{13 \sqrt 2}{12}$
Work Step by Step
The flux through a surface can be defined only when the surface is orientable.
We know that $\iint_S F \cdot dS=\iint_S F \cdot n dS$
Here, $n$ denotes the unit vector.
Since, $\iint_S z dS =\iint_{D} z \sqrt {2+16z^2} dydz$
$=\int_{0}^{1} z \sqrt {2+16z^2} dz$
Plug $a=2+16z^2 \implies da=32 z dz$
$=\dfrac{1}{32} \int_{2}^{18} a^{1/2} da$
$=\dfrac{1}{32} [\dfrac{2}{3} a^{3/2}]_2^{18}$
$=\dfrac{27\sqrt 2-\sqrt 2}{24}$
$=\dfrac{13 \sqrt 2}{12}$
