Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 900: 12

Answer

6

Work Step by Step

$\int^1_{-1} \int^1_0 \int^2_0 (x+y+z)$ dy dx dz =$\int^1_{-1} \int^1_0 =[xy+\frac{1}{2}y^2+zy]^2_0$ dx dz =$\int^1_{-1} \int^1_0 (2x+2+2z)$ dx dz =$\int^1_{-1}[x^2+2x+2zx]^1_0 dz $ =$\int^1_{-1}(3+2z)$ dz =$[3z+z^2]^1_{-1}$ =6
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