Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.2 - Line Integrals - 16.2 Exercise - Page 1085: 26



Work Step by Step

Here, $ds=\sqrt{(dx)^2+(dy)^2+(dz)^2}$ or, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$ Thus, $ds=\sqrt{(3)^2+(2t)^2+(4t^3)^2}dt=\sqrt {9+4t^2+16t^{16}} dt$ Now, $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_{-1}^1 t^4 \ln [(1+3t)+(2+t^2)] (\sqrt {9+4t^2+16t^{16}} dt)$ By using the calculator, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=1.72599$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.