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 1097: 33

Answer

$m=2k \pi$ and $(\overline {x}, \overline {y})=(\dfrac{4}{\pi},0)$

Work Step by Step

Here, $ds=\sqrt{(dx)^2+(dy)^2+(dz)^2}$ or, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$ Thus, $ds=\sqrt{(-2\sin t)^2+(2\cos t)^2}dt=2 dt$ Here, $m=\int_C k ds=2k\int_{-\pi/2}^{\pi/2} dt=2k \pi$ Now, $\overline {x}=\dfrac{1}{2k \pi}\int_{C} xk ds=\dfrac{1}{2 \pi}\int_{C} (2 \cos t) 2dt=\dfrac{4}{\pi}$ and $\overline {y}=\dfrac{1}{2k \pi}\int_{C} yk ds=\dfrac{1}{2 \pi}\int_{C} (2 \sin t) 2dt=0$ Hence, the mass of the wire is: $m=2k \pi$ and $(\overline {x}, \overline {y})=(\dfrac{4}{\pi},0)$
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