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 1019: 1

Answer

$\int^{4}_{0}$ $\int^{\sqrt y}_{0}$ $xy^{2} dx dy = 32$

Work Step by Step

$\int^{4}_{0}$ $\int^{\sqrt y}_{0}$ $xy^{2} dx dy = $$\int^{4}_{0}[\frac{1}{2}x^{2}y^{2}]^{x=\sqrt y}_{x=0}$$ dy = $ $\int^{4}_{0}$$ \frac{1}{2} y^{2} [(\sqrt y)^2 - 0^2] dy = $ $\frac{1}{2}$$\int^{4}_{0} y^{3} dy = \frac{1}{2}[\frac{1}{4}y^{4}]^{4}_{0} = \frac{1}{2}(64-0) = 32$
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