Answer
$\dfrac{4(-2+4\sqrt {2}+9\sqrt 3)}{105}$
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.
$\iint_S y dS =\iiint_{D}y \sqrt {(x^{1/2})^2+(\sqrt y)^2+1^2} dA= \iint_D y \times \sqrt {x+y+1} dA=(\dfrac{2}{3}) \int_{0}^{1} [y(y+2)^{3/2}-y(y+1)^{3/2}] dy=(\dfrac{2}{3}) \times \int_{0}^{1}y(y+2)^{3/2} dy - (\dfrac{2}{3}) \times \int_0^1 y(y+1)^{3/2} dy$
Need to put the values of $y$, that is, $y=m^2-2$ in the first integral and $y=n^2-1$ in the second integral.
$\iint_S y =\int_{\sqrt 2}^{\sqrt 3} (m^2-2) m^3 (2m) dm \times (\dfrac{2}{3})- \int_{1}^{\sqrt 2} (n^2-1) n^3 (2n) dn \times (\dfrac{2}{3})=\int_{\sqrt 2}^{\sqrt 3} m^6-2m^4 dm \times (\dfrac{4}{3})- (4/3) \int_{1}^{\sqrt 2} (n^6-n^4) dn=\dfrac{4(-2+4\sqrt {2}+9\sqrt 3)}{105}$
