Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Section 15.6 - Triple Integrals - 15.6 Exercise - Page 1093: 23

Answer

$\dfrac{16}{3}$

Work Step by Step

Consider $V=\int_{0}^2 \int_{0}^{4-2x} \int_{0}^{4-2x-y} dz dy dx= \int_{0}^2 \int_{0}^{4-2x} [z]_{0}^{4-2x-y} dy dx$ or, $=\int_{0}^2 \int_{0}^{4-2x}(4-2x-y) dy dx$ or, $= \int_{0}^{2}(4y-2xy-\dfrac{y^2}{2}]_0^{4-2x} dx$ or, $=\int^{0}_{2} 4(4-2x)-2x(4-2x)-\dfrac{(4-2x)^2}{2} dx$ or, $=\int_0^2 2x^2-8x+8 dx$ or, $=[(2/3) x^3-4x^2+8x]_0^2$ or, $=\dfrac{16}{3}$
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