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: 25



Work Step by Step

Here,we have $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$ or, $=\sqrt{(2t)^2+(3t^2)^2+(\dfrac{1}{2\sqrt t})^2}=\sqrt {4t^2+9t^4+(\dfrac{1}{4}) t} dt$ Now,we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_1^2 (t^2) \cdot (t^3) \tan^{-1} (\sqrt t) [\sin (t^3+t^4)] \cdot \sqrt {4t^2+9t^4+\dfrac{1}{4}t} dt$ Need to use calculator. $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=94.8231$
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