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 1078: 21



Work Step by Step

Let us consider that $V=\int_{-1}^1 \int_{x^2}^{1} \int_{0}^{1-y} dz dy dx= \int_{-1}^1 \int_{-x^2}^{1} [z]_{0}^{1-y} dy dx $ This implies that $\int_{-1}^{1} \int_{x^2}^{1} (1-y) dy dx=\int_{-1}^1[y-(\dfrac{1}{2})y^2]_{x^2}^1 dx $ or, $\int_{-1}^{1} \dfrac{1}{2}-x^2+\dfrac{x^4}{2} dx=[\dfrac{1}{2}x-\dfrac{1}{3}x^3+\dfrac{1}{(2)(5)}x^5]_{-1}^1$ Thus, $V= \dfrac{8}{5}$
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