Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 16 - Vector Calculus - 16.5 Exercises - Page 1121: 20

Answer

No

Work Step by Step

Let us consider a vector field $G$ such that $div [curl (G)]=0$ consider $F=A i+B j+C k$ $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}$ We are given that $curl G=xyz i-y^2 zj+yz^2 k$ $div[curl(G)]=div (xyz i-y^2 zj+yz^2 k) $ and $div F=\dfrac{\partial xyz}{\partial x}+\dfrac{\partial (-y^2z)}{\partial y}+\dfrac{\partial (yz^2)}{\partial z}$ $div[curl(G)]=yz-2yz+2yz=yz$ Thus, there does not exist such a vector field $G$.
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