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 - Section 16.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises 16.3 - Page 966: 5

Answer

Not Conservative

Work Step by Step

As we know that $\text{curl} F =(\dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z})i +(\dfrac{\partial P}{\partial z}-\dfrac{\partial R}{\partial x}) k+(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})k $ A vector field is conservative iff the $\text{curl} F =0$ Given: $F=(z+y) i+zj+(y+x) k$ Now, curl F$=(1-1) i+(1-1)j +(0-1) k$ and $-k \ne 0$ This shows that the vector field is Not Conservative.
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