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: 44

Answer

$\frac{sin^2 4}{2}$

Work Step by Step

$\int^2_0 \int^{4-x^2}_0 \int^x_0 \frac{sin2z}{4-z}$ dy dz dx =$\int^2_0 \int^{4-x^2}_0 \frac{x \sin 2z}{4z}$ dz dx =$\int^4_0 \int^{\sqrt{4-z}}_0 (\frac{sin 2z}{4-z})dx $ dz =$\int^4_0 (\frac{sin2z}{4-z})\frac{1}{2} (4-z)dz $ =$[-\frac{1}{4}cos2z]^4_0 $ =$[-\frac{1}{4}+\frac{1}{2}sin^2z]^4_0$ =$\frac{sin^2 4}{2}$
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