Answer
$\dfrac{\pi(391\sqrt{17}+1)}{60}$
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, $\iint_S y dS =\iint_{R} y \sqrt {1+4z^2+4x^2} dA$
$=\int_{0}^{2\pi} \int_0^2 r^3 \sqrt {1+4r^2} dr d\theta$
Plug $a=1+4r^2 \implies da=8r dr$
$=\dfrac{1}{8} \int_{0}^{2 \pi} \int_1^{17} a^{1/2} \times \dfrac{a-1}{4} du d \theta$
$=\dfrac{1}{32} \int_{0}^{2 \pi}(2/5) a^{5/2} -(2/3) a^{3/2}|_1^{17} d\theta$
$=\dfrac{1}{32} \int_{0}^{2 \pi} \dfrac{1564 \sqrt{17}+4}{15} d\theta$
$=\dfrac{\pi(391\sqrt{17}+1)}{60}$
