Answer
$-18\pi$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
$r_u=\lt \cos v,\sin v,-2u \gt$ 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}^{3} (-2 u^2 \cos v-4u^2 \sin v-2u) du dv $
This implies that
$\int _{0}^{2 \pi} [ -18\cos v-36 \sin v-9]dv=-18\pi$
