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: 13

Answer

$49$

Work Step by Step

We know that the line equation is defined as: $r(t)=r_0+kt$ Thus, $r_0+kt=(0,0,0)+t \lt 2, 3, -6 \gt=\lt 2t, 3t, -6t \gt$ So, $x=2t \implies dx= 2 dt$; $y=3t \implies dy=3 dt$ and $z=-6t \implies dz=-6dt$ We need to substitute all the values in the given integral. $\int_{0}^14t(2 dt)+6t (3 dt)-12 t(-6 dt)=49$
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