Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1029: 32

Answer

Conservative

Work Step by Step

A vector field is said to be conservative when $ curl F=\nabla \times F=0$ Now, $$ curl F= \nabla \times F \\=(\dfrac{\partial z}{\partial y}-\dfrac{\partial y}{\partial z}) \space i+(\dfrac{\partial x}{\partial z}-\dfrac{\partial z}{\partial x}) \space j+(\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{\partial y}) \space k $$ From the given equation, we have $$ curl F=-yz \times (x+yz)^{-2}+(x+yz)^{-1}+yz \times (x+yz)^{-2}-(x+yz)^{-1}-y(x+yz)^{-2}+y (x+yz)^{-2}+z \times (x+yz)^{-2}-z (x+yz)^{-2} \\= 0$$ Thus, the given field $ F $ is conservative.
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