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 900: 15

Answer

$=\frac{7}{6}$

Work Step by Step

$\int^1_0 \int^{2-x}_0 \int^{2-x-y}_0$ dz dy dx =$\int^1_0 \int^{2-x}_0 (2-x-y)$ dy dx =$\int^1_0 [(2-x)^2-\frac{1}{2}(2-x)^2]dx $ =$\frac{1}{2} \int^1_0 (2-x)^2$ dx =$[-\frac{1}{6}(2-x)^3]^1_0$ =$\frac{-1}{6} +\frac{8}{6}=\frac{7}{6}$
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