Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 15 - Multiple Integrals - 15.3 Exercises - Page 1021: 59

Answer

$\dfrac{3}{4}$

Work Step by Step

Total area of the triangle: $\ Area; A = \dfrac{3}{2}$ Let us consider that $Volume =\int_0^1 \int_0^{3x} x \ y \ dy \ dx \\=\dfrac{1}{2} \times \int_0^1 x [y^2]_0^{3x} dx \\=\dfrac{9}{2} \times \int_0^1 x^3 dx \\=\dfrac{9}{8} [ x^3]_0^1 dx \\ =\dfrac{9}{8}$ Now, the average volume becomes: $V_{Av}= \dfrac{Total \ Volume}{Total \ Area}=\dfrac{9/8}{3/2} =\dfrac{3}{4}$
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