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: 53


$\dfrac{563 \pi }{30}$

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, $V=2\pi \int_{1}^{2} (4-x)(x^3+2-4+x^2) \ dx \\ = 2\pi \int_1^2 (4-x) (x^3+x^2-2] \ dx \\= 2 \pi \int_1^2 (3x^3+4x^2-8-x^4+2x) \ dx \\= 2\pi [ \dfrac{ 3x^{4}}{4}+\dfrac{4x^{3}}{3}-8x-\dfrac{x^5}{5}+x^2]_1^2 \\=2 \pi ( \dfrac{64}{15}+\dfrac{ 307}{60}) \\= \dfrac{563 \pi }{30}$
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