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 850: 23

Answer

$$-3$$

Work Step by Step

The line equation can be written as: $ r(t)=r_0+kt=(1,1,1)+t \space \lt 1,2,-2 \gt=\lt 1+t, 1+2t, 1-2t \gt $ or, $x=1+t$ or, $ dx= dt \\ y=1+2t $ or, $ dy= 2 dt \\ z= 1-2t$ or, $ dz= -2dt $ Now, we will substitute all the above values in the given integral. $ \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$ or, $= [2t^2 -5t])_0^1 $ or, $=-3$
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