Thomas' Calculus 13th Edition

Published by Pearson

Chapter 15: Multiple Integrals - Section 15.6 - Moments and Centers of Mass - Exercises 15.6 - Page 909: 27

Answer

$$M= 36$$

Work Step by Step

The plane $y+2 z=2$ is the top of the wedge \begin{align*} I_{L}&=\int_{-2}^{2} \int_{-2}^{4} \int_{-2}^{4} \int_{-1}^{(2-y) / 2}\left[(y-6)^{2}+z^{2}\right] d z d y d x\\ &=\int_{-2}^{2} \int_{-2}^{4}\left[\frac{(y-6)^{2}(4-y)}{2}+\frac{(2-y)^{3}}{24}+\frac{1}{3}\right] d y d x \end{align*} Let $$t=2-y,\ \ dt=-dy$$ Then \begin{align*} I_{L}&=4 \int_{-2}^{4}\left(\frac{133^{3}}{24}+5 t^{2}+16 t+\frac{49}{3}\right) d t\\ &=4 \left(\frac{133^{3}}{24}t+\frac{5}{3} t^{3}+8 t^2+\frac{49}{3}t\right)\bigg|_{-2}^{4} \\ &=1386 \end{align*} Then $$M=\frac{1}{2}(3)(6)(4)=36$$

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