Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.6 - Moments and Centers of Mass - Exercises 15.6 - Page 908: 7


$$I_{o}=I_{x}+I_{y}=8 \pi$$

Work Step by Step

Since \begin{align*} I_{x}&=\int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} y^{2} d y d x\\ &=\int_{-2}^{2}\left[\frac{y^{3}}{3}\right]_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} d x\\ &=\frac{2}{3} \int_{-2}^{2}\left(4-x^{2}\right)^{3 / 2} d x\\ &=\frac{4}{3} \int_{0}^{2}\left(4-x^{2}\right)^{3 / 2} d x \\ & \text{Let } x=2\sin u: \\ &=\frac{8}{3} \int_{0}^{\pi/2}\left(4-4\sin^{2}u\right)^{3 / 2} \cos ud u\\ &=\frac{64}{3}\int_{0}^{\pi/2} \cos^4udu\\ &=4\pi \end{align*} and by symmetry $$ I_{y}=4 \pi,$$ then $$I_{o}=I_{x}+I_{y}=8 \pi$$
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