Chapter 15 - Section 15.7 - Stokes' Theorem - Exercises - Page 896: 10

$-\pi$

Applying Stoke's Theorem, we have $\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$ Here, $n=xi+yj+zk$ and $d \sigma=\dfrac{dA}{z}$ Then, we have $\iint _S (\nabla \times F) \cdot n d\sigma=\iint _{R} (-z)(\dfrac{dA}{z}) $ This implies that $-\iint _{R} dA=-\pi$