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 967: 23



Work Step by Step

We know that the line equation can be expressed as: $$ r(t)=r_0+kt=(1,1,1)+t \space \lt 1,2,-2 \gt=\lt 1+t, 1+2t, 1-2t \gt \\ x=1+t \implies dx= dt \\ y=1+2t \implies dy= 2 dt \\ z= 1-2t \implies dz= -2dt $$ Substitute all the above values in the given integral as follows: $ \int_{0,1,1}^{2,3,-1} \ y dx +x \ dy +4 \ dz=\int_0^1 (1+2t) dt +(1+t) 2dt -8 dt=\int_0^1 (4t-5) dt \\= [2t^2 -5t])_0^1 \\=-3$
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