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: 13



Work Step by Step

We know that the line equation can be defined as: $$ r(t)=r_0+kt =(0,0,0)+t \lt 2, 3, -6 \gt=\lt 2t, 3t, -6t \gt $$ So, $$ x=2t \\ dx= 2 dt \\y=3t \\ dy=3 dt $$ and $$ z=-6t \implies dz=-6 \space dt $$ Substitute all the above values in the given integral as follows: $$\int_{0}^14t(2 dt)+6t (3 dt)-12 t(-6 dt)=49$$
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