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: 21


The vector field $F$ is irrotational.

Work Step by Step

The vector field $F$ will be irrotational when $curl F=0$ Let us consider that $F=P i+Q j+R k$ Then, we have $curl F=[R_y-Q_z]i+[P_z-R_z]j+[Q_x-P_y]k$ As we are given that $F(x,y,z)=f(x) i+g(y) j+h(z) k$ Here, we have $curl F= \nabla \times F=[R_y-Q_z]i+[P_z-R_z]j+[Q_x-P_y]k$ or, $=(\dfrac{\partial h(z)}{\partial y}-\dfrac{\partial g(y)}{\partial z})i+(\dfrac{\partial f(x)}{\partial z}-\dfrac{\partial h(z)}{\partial x})j+(\dfrac{\partial g(y)}{\partial x}-\dfrac{\partial f(x)}{\partial y})k=0$ Hence, we have: the vector field $F$ is irrotational.
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