Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.5 Curl and Divergence - 16.5 Exercises - Page 1150: 28

Answer

$div (\nabla f \times \nabla g)=0$

Work Step by Step

$div (F \times G)=\dfrac{\partial}{\partial x}[G_3F_2-G_2F_3]-\dfrac{\partial}{\partial y}[G_3F_1-G_1F_3]+\dfrac{\partial}{\partial z}[G_2F_1-G_1F_2]$ or, $=[G_1(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z})-G_2(\dfrac{\partial F_3}{\partial x}-\dfrac{\partial F_1}{\partial z})+G_3(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y})]-[F_1(\dfrac{\partial G_3}{\partial y}-\dfrac{\partial G_2}{\partial z})-F_2(\dfrac{\partial G_3}{\partial x}-\dfrac{\partial G_1}{\partial z})+F_3(\dfrac{\partial G_2}{\partial x}-\dfrac{\partial G_1}{\partial y})]$ or, $=(curl F) \cdot G-F \cdot (curl G)$ But $div (\nabla f \times \nabla g)=(curl (\nabla f)) \cdot (\nabla g)-(\nabla f) \cdot (curl (\nabla g))$ or, $=(0) \cdot (\nabla g)-(\nabla f) \cdot (0)=0$ Hence, $div (\nabla f \times \nabla g)=0$ (proved)
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