## Calculus 8th Edition

Published by Cengage

# Chapter 16 - Vector Calculus - 16.5 Curl and Divergence - 16.5 Exercises - Page 1149: 20

#### Answer

There does not exist such a vector field $G$.

#### Work Step by Step

Suppose, we have a vector field $G$ such that $div [curl (G)]=0$ when $F=A i+B j+C k$, then we have $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}$ Given: $curl G=xyz i-y^2 zj+yz^2 k$ $div[curl(G)]=div (xyz i-y^2 zj+yz^2 k)$ and $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}=\dfrac{\partial xyz}{\partial x}+\dfrac{\partial (-y^2z)}{\partial y}+\dfrac{\partial (yz^2)}{\partial z}$ $\implies div[curl(G)]=yz-2yz+2yz=yz$ Hence, we can conclude that there does not exist such a vector field $G$.

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