Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.5 Curl and Divergence - 16.5 Exercises - Page 1150: 32


$div F=\dfrac{3-p}{(x^2+y^2+z^2)^{p/2}}$ ; $div F=0$ when $p=3$

Work Step by Step

$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}=\dfrac{2x^2(-p/2)}{(x^2+y^2+z^2)^{(p/2)+1}}+\dfrac{1}{(x^2+y^2+z^2)^{(p/2)}}+\dfrac{2y^2(-p/2)}{(x^2+y^2+z^2)^{(p/2)+1}}+\dfrac{1}{(x^2+y^2+z^2)^{(p/2)}}+\dfrac{2z^2(-p/2)}{(x^2+y^2+z^2)^{(p/2)+1}}+\dfrac{1}{(x^2+y^2+z^2)^{(p/2)}}=\dfrac{-p(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{(p/2)+1}}+\dfrac{3}{(x^2+y^2+z^2)^{(p/2)}}$ This implies that $div F=\dfrac{3-p}{(x^2+y^2+z^2)^{p/2}}$ ; $div F=0$ when $p=3$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.