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 - Practice Exercises - Page 1029: 33

Answer

$$2x+y^2+yz+z+C $$

Work Step by Step

A vector field is said to be conservative when $ curl F=\nabla \times F=0$ Now, $ f(x,y,z)=\int_0^x f_1 (t,0,0) dt+ \int_0^y f_2 (x,t,0) \space dt+\int_0^z f_3 (x,y,t) \space dt $ Further, $$ f(x,y,z)=\int_0^x (2) \space dt+ \int_0^y (2t+0) dt+\int_0^z (y+1) \space dt\\=[2t]_0^x+[t^2]_0^y \space [y+1](t)_0^z+C \\=2x+y^2+yz+z+C $$
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