University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 15 - Practice Exercises - Page 910: 32

Answer

Conservative

Work Step by Step

A vector field will be conservative when $curl F=\nabla \times F=0$ $curl F= \nabla \times F=(\dfrac{\partial z}{\partial y}-\dfrac{\partial y}{\partial z})i+(\dfrac{\partial x}{\partial z}-\dfrac{\partial z}{\partial x})j+(\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{\partial y})k$ This implies that $curl F=-yz(x+yz)^{-2}+(x+yz)^{-1}+yz(x+yz)^{-2}-(x+yz)^{-1}-y(x+yz)^{-2}+y (x+yz)^{-2}+z (x+yz)^{-2}-z (x+yz)^{-2}= 0$ Therefore, the given field is conservative.
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