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 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 902: 34

Answer

$\frac{320}{3}$

Work Step by Step

$ V=\int^4_0 \int^8_z \int^{8-z}_z $ dx dy dz =$\int^4_0 \int^8_z(8-2z)dy $ dz =$\int^4_0 (8-2z)(8-z)dz $ =$\int^4_0 (64-24z+2z^2)dz $ =$[64z-12z^2+\frac{2}{3}z^3]^4_0$ $\frac{320}{3}$
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