University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 14 - Section 14.4 - Double Integrals in Polar Form - Exercises - Page 778: 40

Answer

$\dfrac{6\sqrt{2} \pi+40\sqrt{2}-64}{9}$

Work Step by Step

Our aim is to integrate the integral as follows: $ Volume=4\int^{\pi/4}_0 \int^{\sqrt{2 \cos 2\theta}} r\sqrt{2-r^2} \space dr \space d\theta $ or, $=-\dfrac{4}{3} \times \int^{\pi/4}_0 [(2-2 \cos2\theta)^{3/2}-2^{3/2}] \space d\theta $ or, $=\dfrac{2\pi\sqrt{2}}{3} -\dfrac{32}{3} \times \int^{\pi/4}_0 (1-\cos^2\theta) \space \sin\theta \space d\theta $ or, $=\dfrac{2\pi\sqrt{2}}{3}-\dfrac{32}{3} [\dfrac{\cos^3\theta}{4}-\cos\theta]^{\pi/4}_0$ or, $=\dfrac{6\sqrt{2} \pi+40\sqrt{2}-64}{9}$
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