University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises - Page 805: 48

Answer

$$\dfrac{3\pi-4}{6}$$

Work Step by Step

Our aim is to integrate the integral as follows: $$ Volume =\int^{\pi/2}_0 \int^{\cos\theta}_0 \int^{3\sqrt{1-r^2}}_0 \space dz \space r \space dr \space d\theta \\=\int^{\pi/2}_0 \int^{\cos\theta}_0 3 \times \space r \space \sqrt{1-r^2} \space dr \space d\theta \\=\int^{\pi/2}_0 [-(1-r^2)^{3/2}]^{cos\theta}_0 \space d\theta\\=\int^{\pi/2}_0[-(1-cos^2\theta)^{3/2}+1] \space d\theta \\=\dfrac{\pi}{2}+\dfrac{2}{3}[\cos\theta]^{\pi/2}_0 \\=\dfrac{3\pi-4}{6}$$
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