Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - 15.6 Triple Integrals - 15.6 Exercises - Page 1077: 6


$\dfrac{\ln (2)}{3}$

Work Step by Step

Need to integrate the given integral first with respect to $x$, then $z$, and then $y$ $ \int_{0}^1 \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \dfrac{z}{y+1} dx dz dy= \int_{0}^1 \int_{0}^{1} [\dfrac{xz}{y+1}]_{0}^{\sqrt{1-z^2}} dz dy$ or, $=\int_{0}^1 \dfrac{1}{y+1} dy \int_{0}^{1} z\sqrt{1-z^2} dz$ Plug $k=1-z^2 \implies z dz=-\dfrac{dk}{2}$ or, $=[\ln |y+1|]_{0}^1 \int_{0}^{1} \sqrt k [-\dfrac{dk}{2}]$ or, $= (\ln 2)(\dfrac{-1}{2}) \int_0^1 [k^{1/2}] dk$ or, $=(ln 2) (\dfrac{1}{3})$ or, $=\dfrac{\ln (2)}{3}$
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