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: 14



Work Step by Step

Let us consider that $I=\iiint_E (x-y) dV$ $I= \int_{-1}^1 \int_{0}^{2} \int_{1-x^2}^{x^2-1} (x-y) dz dy dx$ Thus, $\int_{-1}^1 \int_{0}^{2} (x-y)[z]_{1-x^2}^{x^2-1} dy dx= \int_{-1}^1 \int_{0}^{2}[ (x-y)(x^2-1-1+x^2)] dy dx=2 \int_{-1}^1 \int_{0}^{2}[ (x-y)(x^2-1)] dy dx$ and $-2 \int_{-1}^1 [x(1-x^2)y-(1-x^2)\dfrac{y^2}{2}]_0^2dx=-4 \int^{-1}_{1}[x-x^3-1+x^{2}] dx$ Hence, $\iiint_E (x-y) dV=-4[\dfrac{1}{2}x^2-\dfrac{1}{4}x^2-x+\dfrac{1}{3}x^3]^{-1}_{1}=\dfrac{-16}{3}$
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