Chapter 16: Integrals and Vector Fields - Section 16.4 - Green's Theorem in the Plane - Exercises 16.4 - Page 979: 23

$-16 \pi$

The tangential form for Green Theorem -- Counterclockwise Circulation $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $ Now, $\oint_C F \cdot n ds= \iint_{R} \dfrac{\partial (y+2x)}{\partial x}-\dfrac{\partial (6y+x)}{\partial y} dx dy $ or, $= -4 \iint_{R} dx dy$ or, $= -4 \times$ Area of circle with radius 2 or, $= -4 \times \pi \times (2)^2$ or, $=-16 \pi$