Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 16 - Vector Calculus - 16.2 Exercises - Page 1098: 41

Answer

$\dfrac{7}{3}$

Work Step by Step

Work done: $W=\int_C F\cdot dr=\int_0^{1} \lt2t-t^2,-t^2 +3t-1, 1-t-4t^2 \gt \cdot \lt 2,1,-1 \gt dt$ or, $=\int_0^{1} 2(2t-t^2)+(-t^2 +3t-1)-( 1-t-4t^2) dt$ or, $=\int_0^{1} 4t-2t^2-t^2 +3t-1- 1+t+4t^2 dt$ or, $=\int_0^{1} t^2+8t-2dt$ or, $=[\dfrac{t^3}{3}+4t^2-2t]_0^1$ Work done, $W=\dfrac{1}{3}+4-2=\dfrac{7}{3}$
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