Answer
$\dfrac{\pi(25 \sqrt 5+1)}{120}$
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ denotes the unit vector.
and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
The flux through a surface can be defined only when the surface is orientable.
Since, $\iint_S z^2 dS =\iint_{R} (z^2) \times \sqrt {1+4z^2+4x^2} dA$
$=\int_{0}^{2\pi} \int_0^1 r^2 \times \sin^2 \theta[ \sqrt {1+4r^2}]r dr d\theta$
$=\int_{0}^{2\pi} \sin^2 \theta \times \int_0^1 (r^3) \times [ \sqrt {1+4r^2}]dr$
Plug in $a=1+4r^2 \implies da=8r dr$
$=\dfrac{1}{8} \int_{0}^{2 \pi} (1/2) -(1/2) \cos 2 \theta \times \int_1^{5} a^{1/2} \times \dfrac{a-1}{4} du d \theta$
$=(\pi) \times \dfrac{1}{32} \times [(2/5) a^{5/2} -(2/3) a^{3/2}|_1^{5} d\theta$
$=\dfrac{\pi(25 \sqrt 5+1)}{120}$
