## University Calculus: Early Transcendentals (3rd Edition)

$0$
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, $\dfrac{2}{3}(0-0)=0$