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 1086: 44


$\dfrac{m(b^2-a^2)}{2} $

Work Step by Step

The work done is given by $W=\int_C F\cdot dr=\int_C F\cdot r'(t) dt=\int_0^{\pi/2} (-ma \sin t i-mb \cos t j) \cdot (a \cos t i-b \sin t j+ck) dt$ or, $W=m (b^2-a^2) \int_0^{\pi/2} (\sin t) (\cos t) dt$ and $W=\dfrac{m(b^2-a^2)}{2} \int_0^{\pi/2} \sin 2 t dt =(m) \dfrac{(b^2-a^2)}{2} [\dfrac{-\cos 2t}{2}]_0^{\pi/2}$ Thus, we have: $W=(m) \dfrac{(b^2-a^2)}{2} [\dfrac{-\cos 2t}{2}]_0^{\pi/2}=(m) \dfrac{(b^2-a^2)}{2} [\dfrac{1-(-1)}{2}=\dfrac{m(b^2-a^2)}{2} $
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