Answer
$25 \pi$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
$r_u=\lt \cos v,\sin v,1 \gt$ 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}^{5} (u \cos v+u \sin v+u) du dv $
This implies that
$\int _{0}^{2 \pi}(\dfrac{25}{2}) [ \cos v+\sin v+1]dv=25 \pi$
