Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.4 - Green's Theorem in the Plane - Exercises 16.4 - Page 978: 5

Answer

Outward flux $=2$ Counterclockwise Circulation =0

Work Step by Step

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 $
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