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: 54


$\dfrac{1748 \pi }{35}$

Work Step by Step

The volume of a region can be calculated as: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ where, $R_{outside}=x^3+2-(-2)=x^3+4 $ and $ R_{inside}=4-x^2-(-2)=6-x^2$ Now, $V=\pi \int_1^2 [(x^3+4)^2-(6-x^2)^2] \ dx =\pi \int_1^2 [x^6+8x^3+16-36+12x^2-x^4] \ dx \\=\pi \int_1^2 [x^6+8x^3+12x^2-x^4-20] \ dx \\=\pi [\dfrac{x^7}{7}+2x^4 +4x^3-\dfrac{x^5}{5}-20 x]_1^2 \\=\pi [\dfrac{1256}{35}+\dfrac{492}{35}] \\=\dfrac{1748 \pi }{35}$
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