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: 16

Answer

$\frac{1}{12}$

Work Step by Step

$\int^1_0 \int^{1-x^2}_0 \int^{4-x^2-y}_3$ x dz dy dx =$\int^1_0 \int^{1-x^2}_0 x(1-x^2-y)dydx $ =$\int^1_0 x[(1-x^2)^2-\frac{1}{2}(1-x^2)]dx $ =$\int^1_0 \frac{1}{2} x (1-x^2)^2$ dx =$[\frac{-1}{12}(1-x^2)^3]^1_0$ =$\frac{1}{12}$
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