Answer
$\pi \sqrt 2$
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, $\iint_D f(r(u,v)) y dS =\iiint_{S}(4u^2v^2 +(u^2+v^2)^2] ( 4 \sqrt 2 (u^2+v^2) dA$
$=\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] 4 \sqrt 2 r^3 dr d \theta $
$= 4 \sqrt 2 \int_{0}^{2 \pi} \int_0^1 [r^4 \sin^2 (2 \theta) +(r^4 \cos^2 (2 \theta) r^3 dr d\theta$
$=4 \sqrt 2\int_{0}^{2 \pi} \int_0^1 r^7 dr d\theta$
$=(\sqrt 2/2) \int_0^{2 \pi} d \theta$
$=\pi \sqrt 2$
