Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1124: 16



Work Step by Step

$I=\int_{C_1}(y+z) dx+(x+z) dy+(x+y) dz+\int_{C_2}(y+z) dx+(x+z) dy+(x+y) dz$ $=\int_{0}^{1} (0+t) dt +(x+z)(0) +(t+0)dt+\int_{0}^{1} [t+(1+t)](-dt) +[(-t+1)+(1+t)] dt +[(-t+1)+t]dt$ $=\int_0^1 2t dt+\int_{0}^{1} (-2t-1) dt+2 dt+1 dt$ $=[t^2]_0^1+[-t^2+2t]_{0}^{1}$ $=(1)^2-0+1$ $=2$
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