Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Practice Exercises - Page 932: 6

Answer

$\frac{4}{35}$

Work Step by Step

$\int^1_0 \int^{\sqrt{y}}_y \sqrt{x} $ dx dy =$\int^1_0 [\frac{2}{3}x^{3/2}]^{\sqrt{y}}_y dy $ =$\frac{2}{3} \int^1_0 (y^{3/4}-y^{3/2})dy=\frac{2}{3} [\frac{4}{7}y^{7/4}-\frac{2}{5}y^{5/2}]^1_0$ =$\frac{2}{3}(\frac{4}{7}-\frac{2}{5})=\frac{4}{35}$ =$\int^1_0 \int^x_{x^2}\sqrt{x}$ dy dx =$\int^1_0 x^{1/2}(x-x^2)dx $ =$\int^1_0 (x^{3/2}-x^{5/2})dx $ =$[\frac{2}{5}x^{5/2}-\frac{2}{7}x^{7/2}]^1_0$ =$\frac{2}{5}-\frac{2}{7}=\frac{4}{35}$
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