University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.3 - Area by Double Integration - Exercises - Page 772: 21



Work Step by Step

The average height is equal to: $\dfrac{1}{4} \iint_{R} f dA=\dfrac{1}{4} \times \int^{2}_{0} \int^{2}_{0}(x^2+y^2) \space dy \space dx $ or, $=\dfrac{1}{4} \times \int^{2}_{0}[x^2y+\dfrac{y^3}{3}]^2_0dx $ or, $=\dfrac{1}{4} \times \int^{2}_{0}(2x^2+\dfrac{8}{3}) \space dx $ or, $=\dfrac{1}{2}[\dfrac{1}{3} x^3+\dfrac{4}{3}x]^2_0$ or, $=\dfrac{8}{3}$
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