## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 16: Integrals and Vector Fields - Section 16.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises 16.3 - Page 966: 7

#### Answer

$x^2+\dfrac{3}{2}y^2+2z^2$

#### Work Step by Step

Given: $f(x,y,z)=x^2+g(y,z)$ Also, $\nabla f =F$ Here, $g_y(y,z)=3y$ and $g(y,z)=\dfrac{3}{2}y^2+h(z)$ and $h(z)=2z^2+C$ Then, $f(x,y,z)=x^2+\dfrac{3}{2}y^2+h(z)$ Thus, $f(x,y,z)=x^2+\dfrac{3}{2}y^2+2z^2$

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