University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.5 - Triple Integrals in Rectangular Coordinates - Exercises - Page 785: 16

Answer

$$\dfrac{1}{12}$$

Work Step by Step

Our aim is to integrate the integral as follows: $$\int^1_0 \int^{1-x^2}_0 \int^{4-x^2-y}_3 (x) \space dz \space dy \space dx =\int^1_0 \int^{1-x^2}_0 (x) (1-x^2-y) \space dy \space dx \\= \int^1_0 x[(1-x^2)^2-\frac{1}{2}(1-x^2)] \space dx\\= \dfrac{1}{2} \times \int^1_0 x (1-x^2)^2 \space dx \\=[\dfrac{-1}{12} \times (1-x^2)^3]^1_0 \\=\dfrac{1}{12}$$
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