Answer
$0$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
Here, $n=\dfrac{2}{7}i+\dfrac{6}{7}j-\dfrac{3}{7}k$ and $d \sigma=\dfrac{7dA}{3}$
Then, we have
$\iint _S (\nabla \times F) \cdot n d\sigma=\iint _{R} (\dfrac{6}{7} y)(\dfrac{7dA}{3}) $
This implies that
$\int_0^{2 \pi} \int_0^{1} 2r \sin \theta r dr d \theta=\int_0^{2 \pi} \dfrac{2}{3} \sin \theta d \theta $
Thus, $\dfrac{2}{3}(0-0)=0$