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: 38

Answer

$0$

Work Step by Step

$ average=\frac{1}{2}\int^1_0 \int^1_0 \int^2_0 (x+y-z)$ dz dy dx =$\frac{1}{2}\int^1_0 \int^1_0 (2x+2y-2)dy $ dx =$\frac{1}{2} \int^1_0(2x-1)dx $ =0
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