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 902: 30

Answer

$\frac{128}{15}$

Work Step by Step

$\int^2_0 \int^{4-x^2}_0 \int^{4-x^2-y}_0 $ dz dy dx =$\int^2_0 \int^{4-x^2}_0 (4-x^2-y)$ dy dx =$\int^2_0 [(4-x^2)^2-\frac{1}{2}(4-x^2)^2]dx $ =$\frac{1}{2}\int^2_0 (4-x^2)^2$ dx =$\int^2_0 (8-4x^2+x^4/2)dx $ =$\frac{128}{15}$
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