Question

use the Shell Method to find the volume of the solid obtained by rotating region under the graph of f(x)=x^2+2 for 0≤x≤2 about x=2 .

Answer 1

32π/3 ≈ 33.51 cubic units

Explanation:

You want the volume of revolution of the region under f(x) = x² +2 between x=0 and x=2 around the line x=2.

Shell

The volume of a "shell" is the product of its circumference, height, and thickness:

dV = Ch·dx = (2π(2 -x))·(x² +2)·dx = 2π(-x³ +2x² -2x +4)dx

Volume

The volume is the integral of the differential volumes over the region of interest:

`\displaystyle V=2\pi\int_0^2{(-x^3+2x^2-2x+4)}\,dx=2\pi\left[-(x^4)/(4)+(2x^3)/(3)-x^2+4x\right]_0^2\\\\\\=(\pi)/(3)(-3(8)+8(4)-12(2)+48)\\\\\\\boxed{V=(32\pi)/(3)}`

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Additional comment

The circumference of the shell, C, is found the usual way: C = 2πr, where r is the radius. In terms of x, the radius of a shell is the distance from x to the center of revolution: r = 2 -x. Hence, C = 2π(2 -x).

The height of the shell is h = f(x).

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