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: 11


$$ F=x \ln x-x+\tan (x+y)+ \dfrac{\ln{(y^2+z^2)}}{2}+C $$

Work Step by Step

We have $$ F=[ \ln x +\sec^2 \space (x+y)]i +(\sec^2\space (x+y) +\dfrac{y}{y^2+z^2}) j+(\dfrac{z}{y^2+z^2}) \space k \\ \implies F=[x \ln x-x+\tan (x+y)]i +(\tan (x+y) +\dfrac{\ln{y^2+z^2}}{2}) \space j+(\dfrac{\ln{y^2+z^2}}{2}) \space k \\ =x \ln x-x+\tan (x+y)+ \dfrac{\ln{(y^2+z^2)}}{2}+C $$
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