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: 34


The mass of the wire is: $m=\dfrac{ka^3}{2}$ and The center of mass is: $(\overline {x}, \overline {y})=(\dfrac{2a}{3},\dfrac{2a}{3})$

Work Step by Step

The mass of the wire is given by: $m=\int_C \rho(x,y) ds=(k a^3) \int_{0}^{\pi/2} \cos t \sin t dt=\dfrac{1}{2}(ka^3)$ Now, $\overline {x}=\dfrac{1}{m}\int_{C} y \rho(x,y) ds=\dfrac{2}{ka^3}(k a^3)\int_{0}^{\pi/2} (a \sin t)(\cos t \sin t ) dt$ or, $=2a \int_{0}^{\pi/2} \sin^2 t \cos t dt$ or, $=\dfrac{2a}{3}$ Since, the quarter-circle is symmenrtic about the line $y=x$. So, we have $\overline {x}=\overline {y}=\dfrac{2a}{3}$ Thus, the center of mass is: $(\overline {x}, \overline {y})=(\dfrac{2a}{3},\dfrac{2a}{3})$
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