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: 41

Answer

2 sin 4

Work Step by Step

$\int^4_0 \int^1_0 \int^{2}_{2y} \frac{4cos(x^2)}{2\sqrt{z}} dx $ dy dz =$\int^4_0 \int^1_0 \int^{x/2}_0 \frac{4cos(x^2)}{2\sqrt{z}}dy $ dx dz =$\int^4_0 \int^1_0 \frac{xcos(x^2)}{\sqrt{z}}$ dx dz =$\int^4_0 (\frac{sin 4}{2})z^{-1/2}$ dz =$[(sin 4)z^{1/2}]^4_0$ =2sin 4
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