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: 32

Answer

$12\pi $

Work Step by Step

$ V=\int^{2}_{-2} \int^{\sqrt{4-x^2}}_{-\sqrt{4-x^2}} $ dz dy dx =$\int^2_{-2} \int^{\sqrt{4-x^2}}_{-\sqrt{4-x^2}} (3-x)$ dy dx =$2\int^2_{-2}(3-x)\sqrt{4-x^2}$ dx =$3\int^2_{-2}2\sqrt{4-x^2} dx-2\int^2_{-2}x\sqrt{4-x^2}dx $ =$3[x\sqrt{4-x^2} +4sin^{-1}\frac{x}{2}]^2_{-2}$ =$12sin^{-2}1-12sin6{-1}$ =$12(-\frac{\pi}{2}-12(\frac{\pi}{2}))$ =$12\pi $
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