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 909: 22


\begin{align*} I_{x} &=208 \\ I_{y}&= 280\\ I_{z}&= 360 \end{align*}

Work Step by Step

The plane $z=\frac{4-2 y}{3}$ is the top of the wedge, so \begin{align*} I_{x}&=\int_{-2}^{3} \int_{-2}^{4} \int_{-2}^{4} \int_{-4 / 3}^{(4-2 y) / 3}\left(y^{2}+z^{2}\right) d z d y d x \\ &=\int_{-3}^{3} \int_{-2}^{4}\left[\frac{8 y^{2}}{3}-\frac{2 y^{3}}{3}+\frac{8(2-y)^{3}}{81}+\frac{64}{81}\right] d y d x\\ &=\int_{-3}^{3} \frac{104}{3} d x=208 \\ I_{y}&=\int_{-3}^{3} \int_{-2}^{4} \int_{-4 / 3}^{(4-2 y) / 3}\left(x^{2}+z^{2}\right) d z d y d x\\ &=\int_{-3}^{3} \int_{-2}^{4}\left[\frac{(4-2 y)^{3}}{81}+\frac{x^{2}(4-2 y)}{3}+\frac{4 x^{2}}{3}+\frac{64}{81}\right] d y d x\\ &=\int_{-3}^{3}\left(12 x^{2}+\frac{32}{3}\right) d x\\ &=280\\ I_{z}&=\int_{-3}^{3} \int_{-2}^{4} \int_{-4 / 3}^{(4-2 y) / 3}\left(x^{2}+y^{2}\right) d z d y d x\\ &=\int_{-3}^{3} \int_{-2}^{4}\left(x^{2}+y^{2}\right)\left(\frac{8}{3}-\frac{2 y}{3}\right) d y d x\\ &=12 \int_{-3}^{3}\left(x^{2}+2\right) d x\\ &=360 \end{align*}
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