Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.4 The Method of Cylindrical Shells - Exercises - Page 312: 39


$\dfrac{1024 \pi}{15}$

Work Step by Step

The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by: $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$ Now, $V=2\pi \int_{2}^{6} (y+2) \sqrt {y-2} \ dy$ Let us use substitution method. So, substitute $a=y-2 \implies da= dy$ $V= 2 \pi \int_2^6 (a+4) a^{1/2} \ dy \\= 2\pi [\dfrac{2}{5}a^{5/2}+\dfrac{8}{3}a^{3/2}]_2^6 \\=2\pi [\dfrac{2}{5}(y-2)^{5/2}+\dfrac{8}{3}(y-2)^{3/2}]_2^6 \\=\dfrac{1024 \pi}{15}$
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