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 311: 5


$18 \pi (2 \sqrt 2-1)$

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: $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$ Now, $V=2\pi \int_{0}^{3} (x)(\sqrt {x^2+9}) \ dx$ Let us use the substitution method. So, substitute $a=(x^2+9) \implies da=2x dx$ Now, $V= \pi \int_{9}^{18} a^{1/2} da \\=2 \pi [\dfrac{a^{3/2}}{3}]_{9}^{18} \\= 18 \pi (2 \sqrt 2-1)$
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