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: 15



Work Step by Step

Our aim is to integrate the integral as follows: $$\int^1_0 \int^{2-x}_0 \int^{2-x-y}_0 dz \space dy \space dx \\=\int^1_0 \int^{2-x}_0 (2-x-y) \space dy \space dx \\= \int^1_0 [(2-x)^2-\frac{1}{2}(2-x)^2]\space dx\\=\dfrac{1}{2} \times \int^1_0 (2-x)^2 dx\\= -\dfrac{1}{6} \times [(2-x)^3]^1_0 \\=-\dfrac{1}{6} +\dfrac{8}{6}\\=\dfrac{7}{6}$$
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