Chapter 15 - Section 15.7 - Stokes' Theorem - Exercises - Page 896: 8

$-4 \pi$

Applying Stoke's Theorem, we have $\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$ $curl F=\nabla \times F=-2j$ $\oint F \cdot dr=\iint _S (-2) dA$ This implies that $\iint _R (-2) dA=(-2) [\pi (1)(2)]=-4 \pi$