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: 35

Answer

$4\pi $

Work Step by Step

$ V=2\int^2_{-2} \int^{\sqrt{4-x^2}/2}_0 \int^{x+2}_0 $ dz dy dx =$2\int^2_{-2} \int^{\sqrt{4-x^2}}_0 (x+2)$ dy dx =$\int^2_{-2} (x+2)\sqrt{4-x^2}dx $ =$\int^2_{-2}2 \sqrt{4-x^2}dx+\int^2_{-2}x\sqrt{4-x^2}$ dx =$[x\sqrt{4-x^2}+4sin^{-1}\frac{x}{2}]^{2}_{-2}+[-\frac{1}{3} (4-x^2)^{3/2}]^{2}_{-2}$ =$4\frac{\pi}{2}-4(-\frac{\pi}{2})$ =$4\pi $
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