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: 27

Answer

1

Work Step by Step

$ V=\int^1_0 \int^{2-2x}_0 \int^{3-3x-3y/2}_0$ dz dy dx =$\int^1_0 \int^{2-2x}_0 (3-3x-\frac{3}{2}y)dy $ dx =$\int^1_0 [6(1-x)^2-\frac{3}{4}.4(1-x)^2] dx $ =$\int^1_0 3(1-x)^2$ dx =$[-(1-x)^3]^1_0$ =1
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