## Calculus 8th Edition

Published by Cengage

# Chapter 15 - Multiple Integrals - 15.6 Triple Integrals - 15.6 Exercises - Page 1077: 3

#### Answer

$\dfrac{16}{15}$

#### 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, $=\dfrac{16}{15}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.