## Thomas' Calculus 13th Edition

Outward flux $=2$ Counterclockwise Circulation =0
The Normal form for Green's Theorem - Outward Flux $=\oint_C F \cdot n ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy$ Now, Outward Flux $=\oint_C F \cdot n ds= \iint_{R} (\dfrac{\partial (x-y)}{\partial x}-\dfrac{\partial (y-x)}{\partial y} dx dy =2 \iint_{R} dx dy$ So, Outward flux $=2$ The tangential form for Green Theorem - Counterclockwise Circulation $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy =\oint_C F \cdot n ds= \iint_{R} \dfrac{\partial (y-x)}{\partial x}-\dfrac{\partial (x-y)}{\partial y} dx dy = \iint_{R} -1 -(-1) dx dy=0$