Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.1 Vector Fields - Exercises - Page 919: 39

Answer

$$\phi (x,y) = \frac{1}{2}x^2+ \frac{1}{2}y^2 +C$$

Work Step by Step

Given $$\mathbf{F}=\langle x, y\rangle$$ We need to find $\phi (x,y)$ such that $$ \frac{\partial \phi }{\partial x}=x,\ \ \ \frac{\partial \phi }{\partial y}=y$$ By integration, we get \begin{align*} \phi (x,y)&= \frac{1}{2}x^2+C_1\\ \phi (x,y)&= \frac{1}{2}y^2+C_1\\ \end{align*} Then we can choose $$\phi (x,y) = \frac{1}{2}x^2+ \frac{1}{2}y^2 +C$$
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