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 901: 24

Answer

$\frac{2}{3}$

Work Step by Step

$ V=\int^1_0 \int^{1-x}_0 \int^{2-2z}_0$ dy dz dx =$\int^1_0 \int^{1-x}_0(2-2z)$ dz dx =$\int^1_0 [2z-z^2]^{1-x}_0$ dx =$\int^1_0 (1-x^2)dx $ =$[x-\frac{x^3}{3}]^1_0$ =$\frac{2}{3}$
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