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: 19



Work Step by Step

Here, we have $V=\int_{0}^2 \int_{0}^{4-2x} \int_{0}^{4-2x-y} dz dy dx= \int_{0}^2 \int_{0}^{4-2x} [z]_{0}^{4-2x-y} dy dx$ and $\int_{0}^2 \int_{0}^{4-2x}4-2x-y dy dx= \int_{0}^{2}(4y-2xy-\dfrac{y^2}{2}]_0^{4-2x} dx$ $\implies \int^{0}_{2} [4(4-2x)-2x(4-2x)-\dfrac{1}{2}(4-2x)^2] dx$ $\implies \int_0^2 2x^2-8x+8 dx$ Hence, $V=[\dfrac{2 x^3}{3}-4x^2+8x]_0^2=[\dfrac{2 (2-0)^3}{3}-4(2-0)^2+8(2-0)]=\dfrac{16}{3}$
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