University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 15 - Section 15.8 - The Divergence Theorem and a Unified Theory - Exercises - Page 906: 1



Work Step by Step

As we know that $div F=\dfrac{\partial A}{\partial x}i+\dfrac{\partial B}{\partial y}j$ Now, we have $div F=\dfrac{\partial}{\partial x}[\dfrac{-y}{(x^2+y^2)^{1/2}}]+\dfrac{\partial}{\partial y} [\dfrac{x}{(x^2+y^2)^{1/2}}]$ This implies that $\dfrac{2xy}{2}(x^2+y^2)^{-3/2}-\dfrac{2yx}{2}(x^2+y^2)^{-3/2}=0$
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