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: 21


$\dfrac{6}{5}-\cos 1 -\sin 1$

Work Step by Step

Here, we have $ F(r(t)) =\sin t^3i +\cos t^2 j+t^4 k; dr=(3t^2i-2tj+k) dt$ $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{1} (3t^2 i -2tj+k) \cdot (\sin t^3i +\cos t^2 j+t^4 k) d t$ $= \int_0^{1} [3t^2 \sin t^3-2t \cos t^2+t^4] dt$ $=[-\cos (t)^3-\sin (t)^2+t^5/5]_0^1$ $=\dfrac{6}{5}-\cos (1) -\sin (1)$
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