## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 15: Multiple Integrals - Section 15.6 - Moments and Centers of Mass - Exercises 15.6 - Page 909: 28

#### Answer

$$36$$

#### 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}^{(2-y) / 2}\left[(x-4)^{2}+y^{2}\right] d z d y d x\\ &= \frac{1}{2} \int_{-2}^{2} \int_{-2}^{4}\left(x^{2}-8 x+16+y^{2}\right)(4-y) d y d x\\ &=\int_{-2}^{2}\left(9 x^{2}-72 x+162\right) d x\\ &=\left(3 x^{3}-36 x^2+162x\right)\bigg|\\ &=696 \end{align*} Then $$M=\frac{1}{2}(3)(6)(4)=36$$

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