Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.6 - Triple Integrals - 15.6 Exercise - Page 1037: 4



Work Step by Step

Integrate the given integral first with respect to $z$, then $x$, and then $y$ $ \int_{0}^1 \int_{y}^{2y} \int_{0}^{x+y} 6xy dz dx dy= \int_{0}^1 \int_{y}^{2y} |6xyz|_{0}^{x+y} dx dy=\int_{0}^1\int_y^{2y} 6xy(x+y) dx dy$ This implies that $\int_0^1 \int_y^{2y} [6x^2y+6xy^2]dx dy=(6) \int^{0}_{1}[\dfrac{1}{3} 8y^4+\dfrac{4y^4}{2}] dy$ Hence, $ \int_{0}^1 \int_{y}^{2y} \int_{0}^{x+y} 6xy dz dx dy=6 [\dfrac{8y^5}{15}+\dfrac{2y^5}{5}]_{0}^{1}=\dfrac{23}{5}$
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