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 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 21



Work Step by Step

average height=$\frac{1}{4} \int^{2}_{0} \int^{2}_{0}(x^2+y^2)dydx$ =$\frac{1}{4}\int^{2}_{0}[x^2y+\frac{y^3}{3}]^2_0dx$ =$\frac{1}{4}\int^{2}_{0}(2x^2+\frac{8}{3})dx=\frac{1}{2}[\frac{x^3}{3}+\frac{4x}{3}]^2_0$ =$\frac{8}{3}$
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