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: 42

Answer

$56 \pi$

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, $V=2\pi \int_{0}^{2} (x+3) (x^2+2) \ dx\\= 2 \pi \int_0^2 (x^3+2x+3x^2+6) \ dx \\= 2\pi [\dfrac{x^4}{4}+x^2+x^3+6x]_0^2 \\=2 \pi [\dfrac{16}{4}+4+8+12-0] \\=56 \pi$
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