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 15.4 - Double Integrals in Polar Form - Exercises 15.4 - Page 893: 19

Answer

$\frac{\pi}{2}(2ln2-1)$

Work Step by Step

$\int^{ln2}_0 \int^{\sqrt{(ln2)^2-y^2}}e^{\sqrt{x^2+y^2}}dxdy $ =$\int^{\pi/2}_0 \int^{ln2}_0 re^r dr d\theta)$ =$\int^{\pi/2}_0(2ln2-1)d\theta $ =$\frac{\pi}{2}(2ln2-1)$
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