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: 3



Work Step by Step

Need to integrate the given integral first with respect to $x$, then $y$, and then $z$ $ \int^{2}_{0} \int_{0}^{z^2} \int_{0}^{y-z} (2x-y) dx dy dz= \int^{2}_{0} \int_{0}^{z^2} (-yz+z^2)dy dz$ or, $=\int^{2}_{0}[\dfrac{-y^2z}{2}+yz^2]_0^{z^{2}} dz$ or, $=\int^{2}_{0}[\dfrac{-z^5}{2}+z^4] dz$ or, $=[\dfrac{-z^{6}}{12}+\dfrac{z^5}{5}]^{2}_{0}$ or, $=[\dfrac{-2^{6}}{12}+\dfrac{2^5}{5}]^{2}_{0}$ or, $=\dfrac{16}{15}$
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