Answer
$\dfrac{4 \sqrt 2-2}{3}$
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, $\iint_S y dS =\int_{0}^{\pi} \int_{0}^{1} (u \sin v) \sqrt {1+u^2} dA$
$=\int_{0}^{\pi} \sin v dv \times \int_{0}^{1} u [ \sqrt {1+u^2}] du$
$=\int_{0}^{1} 2u \sqrt {1+u^2} du$
Plug $1+u^2=t \implies dt=2u du$
$=\int_1^2 \sqrt {t} dt$
$=\dfrac{2}{3}t^{3/2}$
$=\dfrac{4 \sqrt 2-2}{3}$
