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 1110: 27


$div (F \times G)=G \cdot curl F-F \cdot curl G$

Work Step by Step

$div (F \times G)=\nabla \cdot (F \times G)$ This gives: $div (F \times G)=\dfrac{\partial}{\partial x}[G_3F_2-G_2F_3]-\dfrac{\partial}{\partial y}[G_3F_1-G_1F_3]+\dfrac{\partial}{\partial z}[G_2F_1-G_1F_2]$ 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, $=G \cdot curl F-F \cdot curl G$ This implies that $div (F \times G)=G \cdot curl F-F \cdot curl G$
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