Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.5 - Curl and Divergence - 16.5 Exercise - Page 1109: 19


Does not exist.

Work Step by Step

Suppose, we have a vector field $G$ such that $div [curl (G)]=0$ Let us consider $F=ai+b j+c k$ $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$ Given: $curl G=\lt x \sin y, \cos y, z-xy \gt$ This implies that $div[curl(G)]=\sin y-\sin y+1 $ we can see that $div [curl (G)]=1 \ne 0$ Hence, we can conclude that there does not exist such a vector field $G$.
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