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 1125: 26

Answer

$1.72599$

Work Step by Step

Here, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$ This implies that $ds=\sqrt{(3)^2+(2t)^2+(4t^3)^2}dt=\sqrt {9+4t^2+16t^{16}} dt$ Further, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_{-1}^1 (t^4) \cdot \ln [(1+3t)+(2+t^2)] \cdot (\sqrt {9+4t^2+16t^{16}} dt)$ Hence, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=1.72599$ (Using calculator)
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