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 - Practice Exercises - Page 1028: 7


$$8 \pi \sin (1)$$

Work Step by Step

We calculate the line integral by using substitution as follows: $$ x=2 \cos t \\ \implies dx=-2 \sin t dt \\ y= 2 \sin t \\ \implies dy=2\cos t \space dt \\ z=-1 \\ \implies dz=(0) \space dt $$ We need to plug the above values in the given integral. $$\int_{2 \pi}^{0}2 \sin (1) \sin t-2 \sin 1 \space \cos t +4 \cos 1 \sin t \cos t = -4 \sin (1) \space \int_{2 \pi}^{0} (sin^2 t+\cos^2 t) \\=-4 \sin (1)\int_{2 \pi}^{0} dt \\=8 \pi \sin (1)$$
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