Answer
$\dfrac{4 \sqrt 2-2}{3}$
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.
Since, we have $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f[\overline{x}, \overline{y}, \overline{z}] AS_i$
Consider $\iint_S y dS =\int_{0}^{\pi} \sin v dv \times \int_{0}^{1} u [ \sqrt {1+u^2}] du=\int_{0}^{1} 2u \times \sqrt {1+u^2} du$
Suppose $a=1+u^2; da=2u du$
$\iint_S y dS=\int_1^2 \sqrt {a} da=(\dfrac{2}{3}) (a^{3/2})$
Hence,$\iint_S y dS=\dfrac{4 \sqrt 2-2}{3}$
