Answer
$\dfrac{\pi(25 \sqrt 5+1)}{120}$
Work Step by Step
When the surface is orientable then the flux through a surface is defined as: $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ is the unit vector and $dS=n \times dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
Since, $\iint_S z^2 dS =\iint_{R} (z^2) \times \sqrt {1+4z^2+4x^2} dA=\int_{0}^{2\pi} \int_0^1 \sin^2 \theta[ \sqrt {1+4r^2}](r) \times (r^2) dr d\theta$
and $\iint_S z^2 dS = \int_0^1 (r^3) \times [ \sqrt {1+4r^2}]dr \int_{0}^{2\pi} \sin^2 \theta$
Suppose $1+4r^2 =t; dt=8r dr$
$\iint_S z^2 dS =\int_{0}^{2 \pi} \dfrac{1}{2} -\dfrac{\cos 2 \theta}{2} \times (1/8) \times \int_1^{5} \dfrac{(t-1) \sqrt t}{4} dt d \theta=\dfrac{\pi}{32} \times [(2/5) t^{5/2} -(2/3) t^{3/2}|_1^{5} d\theta=\dfrac{\pi(25 \sqrt 5+1)}{120}$
