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 901: 19

Answer

$\frac{1}{2}-\frac{\pi}{8}$

Work Step by Step

$\int^{\pi/4}_0 \int^{\ln \sec v}_0 \int^{2t}_{-\infty} e^x $ dx dt dv =$\int^{\pi/4}_0 \int^{\ln \sec v}_0 \lim\limits_{b \to -\infty}(e^{2t}-e^b)$ dt dv =$\int^{\pi/4}_0 \int^{\ln \sec v}_0 e^{2t} dt $ dv =$\int^{\pi/4}_0 (\frac{1}{2}e^{2 \ln \sec v}-\frac{1}{2})dv $ =$\int^{\pi/4}_0 (\frac{sec^2}{v}-\frac{1}{2})dv $ =$[\frac{tan v}{2}-\frac{v}{2}]^{\pi/4}_0$ =$\frac{1}{2}-\frac{\pi}{8}$
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