Answer
$4$
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.
Here, $\iint_S F \cdot n dS=\iint_D 2 (u^2-v^2) dA$
$=(2) \int_{0}^{1} \int_0^{2} 2 (u^2-v^2) du dv$
$=(2) \int_{0}^{1} [\dfrac{u^3}{3}-uv^2]_0^2 dv$
$=(2) \int_{0}^{1} \dfrac{2^3}{3}-(2)v^2 dv$
$=2[\dfrac{8}{3}-\dfrac{2}{3}]$
Hence, $Flux=4$
