Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 13 - Multiple Integration - 13.4 Triple Integrals - 13.4 Exercises - Page 1002: 8

Answer

0

Work Step by Step

$$\eqalign{ & \int_{ - 1}^1 {\int_1^2 {\int_0^1 {6xyz\,dydxdz} } } \cr & {\text{Integrate with respect to }}y \cr & = \int_{ - 1}^1 {\int_1^2 {\left[ {3x{y^2}z} \right]_0^1dx} } dz \cr & = \int_{ - 1}^1 {\int_1^2 {\left( {3xz} \right)dx} } dz \cr & = 3\int_{ - 1}^1 {z\int_1^2 {xdx} } dz \cr & {\text{Integrate with respect to }}x \cr & = 3\int_{ - 1}^1 {z\left[ {\frac{{{x^2}}}{2}} \right]_1^2} dz \cr & = 3\int_{ - 1}^1 {z\left( {\frac{3}{2}} \right)} dz \cr & = \frac{9}{2}\int_{ - 1}^1 {zdz} \cr & {\text{Integrate with respect to }}z \cr & = \frac{9}{2}\left( {\frac{{{z^2}}}{2}} \right)_{ - 1}^1 \cr & = \frac{9}{2}\left( {\frac{1}{2} - \frac{1}{2}} \right) \cr & = 0 \cr} $$
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