University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 15 - Section 15.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises - Page 849: 9



Work Step by Step

Here, $f=xe^{y+2z}+g(y,z)$ and $g(y,z)=0$ or, $g(y,z)=h(z)=0$ Then $f=xe^{y+2z}+h(z)$ This implies that $f_z=2xe^{y+2z}+h'(z)=2xe^{y+2z}$ Hence, $f=xe^{y+2z}+C$
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