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 1126: 37


$I_x=k( \dfrac{\pi}{2}-\dfrac{4}{3}) $ and $I_y=k( \dfrac{\pi}{2}-\dfrac{2}{3}) $

Work Step by Step

$I_x=\int_C y^2 \rho(x,y) ds=\int_0^{\pi} \sin ^2 t[k(1-\sin t)] dt=k\int_0^{\pi} (\sin ^2 t-\sin^3 t) dt =k \times [\int_0^{\pi} (\dfrac{1-\cos 2t}{2})-(\dfrac{3 \sin t-\sin 3t}{4}) dt] =k( \dfrac{\pi}{2}-\dfrac{4}{3}) $ $I_y=\int_C x^2 \rho(x,y) ds=\int_0^{\pi} \cos ^2 t[k(1-\sin t)] dt=k\int_0^{\pi} (\cos^2 t-\cos^2 t \sin t) dt =k \times \int_0^{\pi} \dfrac{1+ \cos 2t}{2} dt+k[\dfrac{\cos^3 t}{3}]_0^{\pi}=k( \dfrac{\pi}{2}-\dfrac{2}{3}) $ Therefore, $I_x=k( \dfrac{\pi}{2}-\dfrac{4}{3}) $ and $I_y=k( \dfrac{\pi}{2}-\dfrac{2}{3})$
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