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: 14



Work Step by Step

Consider $I=\iiint_E (x-y) dV$ $I= \int_{-1}^1 \int_{0}^{2} \int_{1-x^2}^{x^2-1} (x-y) dz dy dx= \int_{-1}^1 \int_{0}^{2} (x-y)[z]_{1-x^2}^{x^2-1} dy dx$ or, $=\int_{-1}^1 \int_{0}^{2}(2) (x-y)(x^2-1) dy dx$ or, $=(-2) \int_{-1}^1 [x(1-x^2)y-(1-x^2)\dfrac{y^2}{2}]_0^2dx$ or, $= (-4) \int^{-1}_{1}[x-x^3-1+x^{2}] dx$ or, $= (-4)[\dfrac{x^2}{2}-\dfrac{x^2}{4}-x+\dfrac{x^3}{3}]^{-1}_{1}$ or, $=\dfrac{-16}{3}$
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