Answer
$$-2 \pi$$
Work Step by Step
Since, $F \cdot n =-8u^3+2u$
Now, $$\iint_{S} F \cdot n \ dS=\int_0^{2 \pi} \int_{0}^{1} (-8u^3+2u) \ du \ dv \\=\int_0^{2 \pi} [-2u^4+u^2]_{0}^{1} \ d v \\= -\int_0^{2 \pi} \ dv \\=-[v]_0^{2 \pi} \\=-(2 \pi- 0) \\=-2 \pi$$
