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: 1



Work Step by Step

We integrate the triple integral as follows: $\int^{1}_0 \int^{1-x}_0 \int^1_{x+z} F(x,y,z)dydzdx $ =$\int^{1}_0 \int^{1-x}_0 \int^{1}_{x+z} du dz dx $ =$\int^{1}_0 \int^{1-x}_0(1-x-z)dzdx $ =$\int^1_0[(1-x)-x(1-x)-\frac{(1-x)^2}{2}]dx $ =$\int^1_0\frac{(1-x)^2}{2}dx $ =$[-\frac{(1-x)^3}{6}]^1_0$ We plug in $1$ and $0$ and subtract to obtain: =$\frac{1}{6}$
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