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: 1

Answer

$\dfrac{27}{4}$

Work Step by Step

Let us consider that $I=\int \int \int xyz^{2} dV$ Here, we have $B = (x,y,z) | 0 \leq x \leq 1, -1 \leq y \leq\ 2, 0 \leq z \leq 3$ $I= \int^{1}_{0} \int^{3}_{0} \int^{2}_{-1} xyz^{2} dy dz dx=(\dfrac{1}{2}) \int^{1}_{0} \int^{3}_{0} xz^2|y^{2}|^{2}_{-1}dz dx$ or, $I=(\dfrac{3}{2}) \int^{1}_{0} \int^{3}_{0} xz^{2}dzdx$ or, $I=(\dfrac{1}{2})\int^{1}_{0} x|z^{3}|^{3}_{0}dx$ Hence, $I=(\dfrac{27}{2}) \int^{1}_{0} xdx =(\dfrac{27}{2})|x^{2}|^{1}_{0} = \dfrac{27}{4}(1-0)=\dfrac{27}{4}$
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