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: 38

Answer

$I_x=I_y= 4 \sqrt{13} k \pi (1+6 \pi^2); I_z=8 \sqrt{13} k \pi$

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, $ds=\sqrt{(2\cos t)^2+(-2\sin t)^2+3^2}dt=\sqrt{13} dt$ Now, $I_x=\int_{C} (y^2+z^2) \rho(x,y,z) ds=\int_{0}^{2\pi} (4 \cos^2 t+9t^2)(k) \sqrt{13} dt=\sqrt {13} k \int_{0}^{2\pi} (4 \cos^2 t+9t^2) dt=\sqrt{13} k \int_{0}^{2\pi} 2 (1+\cos 2t)+9t^2= 4 \sqrt{13} k \pi (1+6 \pi^2)$ $I_y=\int_{C} (x^2+z^2) \rho(x,y,z) ds=\int_{0}^{2\pi} (4 \sin^2 t+9t^2)(k) \sqrt{13} dt=\sqrt {13} k \int_{0}^{2\pi} (4 \sin^2 t+9t^2) dt=\sqrt{13} k \int_{0}^{2\pi} 2 (1-\cos 2t)+9t^2= 4 \sqrt{13} k \pi (1+6 \pi^2)$ and $I_z=\int_{C} (x^2+y^2) \rho(x,y,z) ds=\int_{0}^{2\pi} (4 \sin^2 t+4 \cos^2 t)(k) \sqrt{13} dt=\sqrt {13} k \int_{0}^{2\pi} (4) dt=8 \sqrt{13} k \pi$ Hence, $I_x=I_y= 4 \sqrt{13} k \pi (1+6 \pi^2); I_z=8 \sqrt{13} k \pi$
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