Answer
$-15 \pi$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
$r_{\phi}=\lt\sqrt 3 \cos \phi \cos \theta, \sqrt 3 \cos \phi \sin \theta,-\sqrt 3 \sin \phi \gt$ and $r_{\theta}=\lt -\sqrt 3 \sin \phi \sin \theta, \sqrt 3 \sin \phi \cos \theta,0 \gt$
Then, we have
$\iint _S (\nabla \times F) \cdot n d\sigma=\int _{0}^{\pi/2}\int_{0}^{2 \pi} (\dfrac{-15}{2} \sin 2 \phi )d\phi d\theta $
This implies that
$(-15 \pi) \int _{0}^{\pi/2} \sin 2 \phi d \phi=-15 \pi$
