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: 37

Answer

$\frac{31}{3}$

Work Step by Step

$ average=\frac{1}{8} \int^2_0 \int^2_0 \int^2_0 (x^2+p)$ dz dy dx =$\frac{1}{8} \int^2_0 \int^2_0 (2x^+18)$ dy dx =$\frac{1}{8} \int^2_0 (4x^2+36)dx $ =$\frac{31}{3}$
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