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: 14



Work Step by Step

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