Answer
$\pi \sqrt 2$
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ denotes the unit vector.
and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
The flux through a surface can be defined only when the surface is orientable.
Consider $\iint_D f(r(u,v)) y dS =\iiint_{S}(4u^2v^2 +(u^2+v^2)^2] ( 4 \sqrt 2 (u^2+v^2) dA$
or, $=\int_{0}^{2 \pi} \int_0^1 (4r^4 \cos^2 \theta \sin^2 \theta +(r^2 \cos^2 \theta -r^2 \sin^2)^2] \times [4 \sqrt 2 r^3 dr d \theta]$
or, $= 4 \sqrt 2 \times \int_{0}^{2 \pi} \int_0^1 [r^4 \sin^2 (2 \theta) +(r^4 \cos^2 (2 \theta) r^3 dr d\theta$
or, $=4 \sqrt 2\times \int_{0}^{2 \pi} \int_0^1 r^7 dr d\theta$
or, $=\dfrac{\sqrt 2}{2} \times \int_0^{2 \pi} d \theta$
or, $=\dfrac{\sqrt 2}{2} \times [\theta]_0^{2 \pi}$
Thus, we have $\iint_D f[r(u,v)] y dS=\pi \sqrt 2$
