Answer
$0$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
$r_{\phi}=\lt 2 \cos \phi \cos \theta, 2 \cos \phi \sin \theta,-2\sin \phi \gt$ and $r_{\theta}=\lt -2 \sin \phi \sin \theta, 2 \sin \phi \cos \theta,0 \gt$
Then, we have
$\iint _S (\nabla \times F) \cdot n d\sigma=\int _{0}^{\pi/2}\int_{0}^{2 \pi} (-16 \sin^ 2 \phi \cos \phi (\cos \theta +\sin \theta) -2 \sin \theta +2 \cos 2 \phi \sin \theta )d\phi d\theta $
This implies that
$\int _{0}^{\pi/2}(-16 \sin^ 2 \phi \cos \phi (\cos \theta +\sin \theta) -2 \sin \theta +2 \cos 2 \phi \sin \theta )d\phi =0 $
