Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 14


$\dfrac{\pi}{3} [1-\dfrac{1}{(1+H)^3}]$

Work Step by Step

The area $(A)$ of each horizontal cross-section is given by: $A(y)=\pi r^2 =\pi [(1+y)^{-2}]^2=\pi (1+y)^{-4}$ Thus, the volume of the solid is: $V=\pi \int_{0}^{H} (1+y)^{-4} \ dy \\=-\dfrac{\pi}{3} [(1+y)^{-3}]_0^{H}\\=\dfrac{\pi}{3} [1-(1+H)^{-3}] \\=\dfrac{\pi}{3} [1-\dfrac{1}{(1+H)^3}]$
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