## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 6 - Applications of the Integral - 6.4 The Method of Cylindrical Shells - Exercises - Page 312: 45

#### Answer

$\dfrac{776\pi}{15}$

#### 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, $V=2\pi \int_{0}^{2} (x^2+4)^2 \ dx\\= 2 \pi \int_0^2 (x^4+16+8x^2-4) \ dx \\= \pi [\dfrac{x^5}{5}+12x+\dfrac{8x^3}{3}]_0^2 \\=2 \pi [\dfrac{32}{5}+24+\dfrac{63}{3}] \\=\dfrac{776\pi}{15}$

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