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: 12


$$$ f=\tan^{-1} (x \space y)+sin^{-1} (yz)+\ln |z|+C $$

Work Step by Step

We have $$$ f=\arctan (xy)+g(y,z) \\g(y,z)=\arcsin (yz)+h(z)\\=\tan^{-1} (xy)+sin^{-1} (yz)+h(z)$$ Now, $$ f_z=\dfrac{y}{\sqrt {1-y^2z^2}}+\dfrac{1}{z}$$ and $$ h(z)=\ln |z|+c $$ Therefore, $$ f=\tan^{-1} (x \space y)+\sin^{-1} (yz)+\ln |z|+C $$
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