Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 305: 37


$\dfrac{32 \pi}{3} $

Work Step by Step

The volume of a revolution can be calculated as: $V=\pi \int_{m}^{n} (R^2_{outside}-r^2_{inside}) \ dy$ where, $R_{outside}=2-0=2$ and $ R_{inside}=2-\sqrt{y-2}$ Now, $V=\pi \int_0^2 (2)^2 \ dy +\pi \int_2^6 [(2)^2-( \sqrt {y-2})^2 ] \ dy \\=\pi \int_0^2 4 \ dy + \pi \int_2^6 (4-4\sqrt {y-2} +y-2) \ dy \\=\pi [4y]_0^2 +\pi [\dfrac{y^2}{2}+2y-\dfrac{8(y-2)^{3/2}}{3}]_2^6 \\=8 \pi +\pi (\dfrac{26}{3}-6) \\=\dfrac{32 \pi}{3} $
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