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: 9


$$ x \space e^{y+2z}+c $$

Work Step by Step

We have $$ f=x \space e^{y+2z}+g(y,z) \\g(y,z)=0 \implies g(y,z)=h(z)=0$$ This yiedls: $$ f=xe^{y+2z}+h(z)$$ Now, $$ f_z=2xe^{y+2z}+h'(z)=2xe^{y+2z}$$ So, $$ f=x \space e^{y+2z}+c $$
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