Answer
$-\pi$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
Here, Here, $F \cdot \dfrac{dr}{dt}=-\sin^2 t+\cos t$
Then, we have
$\iint _S (\nabla \times F) \cdot n d\sigma=\int _{0}^{2 \pi} (\cos t-\sin^2 t) dt $
This implies that
$[ \sin t+\dfrac{\sin 2t}{4}-\dfrac{t}{2}]_{0}^{2 \pi}=-\pi$
