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 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 900: 18

Answer

$\frac{2-\sqrt{e}}{6}$

Work Step by Step

$\int^1_0 \int^{\sqrt{e}}_1 \int^e_1 se^s ln r \frac{(\ln t)^2}{t}$ dt dr ds =$\int^1_0 \int^{\sqrt{e}}_1 (se^s \ln r)[\frac{1}{3}(\ln t)^3]^e_1$ dr ds =$\int^1_0 \int^{\sqrt{e}}_1 \frac{se^s}{3} \ln r $ dr ds =$\int^1_0 se^s/3 [r\ln r -r]^\sqrt{e}_1$ ds =$\frac{2-\sqrt{e}}{6} \int^1_0 se^s ds $ =$\frac{2-\sqrt{e}}{6} [se^s-e^s]^1_0$ =$\frac{2-\sqrt{e}}{6}$
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