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: 24

Answer

$\approx0.8527$

Work Step by Step

Here, $dr=(\cos t i-\sin t j+\sec^2 t k) dt$ and $F(r(t))=\sin te^{\sin t} i+\tan t \sin t e^{\cos t}j+\sin t\cos te^{\tan t} k$ $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{\pi/4} (\sin te^{\sin t} i+\tan t \sin t e^{\cos t}j+\sin t\cos te^{\tan t} k) \cdot (\cos t i-\sin t j+\sec^2 t k) dt$ or, $=\int_0^{\pi/4} \sin t \cos t e^{\sin t} -\tan t \sin^2 t e^{\cos t}+\tan t e^{\tan t} dt$ By using calculator, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}\approx0.8527$
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