Answer
$\dfrac{-\pi}{4}$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
$r_u=\lt \cos v,\sin v,1 \gt$ and $r_u=\lt -u\sin v,u\cos v,0 \gt$
Then, we have
$\iint _S (\nabla \times F) \cdot n d\sigma=\int _{0}^{2 \pi}\int_{0}^{1} (-2 u^4 \cos v \sin^3 v-u^3 \cos v) du dv $
This implies that
$\int _{0}^{2 \pi} [ -\dfrac{2}{5} \cos v \sin^3 v-\dfrac{1}{4} \cos^2 v]dv=\dfrac{-\pi}{4}$
