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: 10

Answer

$\frac{3}{2}$

Work Step by Step

$\int^1_0 \int^{3-3x}_0 \int^{3-3x-y}_0$ dz dy dx =$\int^1_0 \int^{3-3x}_0 (3-3x-y)$ dy dx =$\int^1_0[(3-3x)^2-\frac{1}{2}(3-3x)^2]dx $ =$\frac{9}{2} \int^1_0 (1-x)^2$ dx =$\frac{-3}{2}[(1-x)^3]^1_0$ =$\frac{3}{2}$
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