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# Basis of the Shell Method

MathJax TeX Test Page A shell is a hollow cylinder, and the volume of that is the volume of the outside cylinder - the volume of the inside cylinder. If the outer radius is $p + \dfrac{w}{2}$, and the inner radius is $p - \dfrac{w}{2}$. The volume of the shell is $\pi\left(p + \dfrac{w}{2}\right)^2*h - \pi\left(p - \frac{w}{2}\right)^2*h = 2\pi{}phw = 2\pi$*radius * height *thickness.
In order to find the volume of that curve rotated around the y-axis, you need the radius $(x)$, the height $(f(x))$, and the thickness is the change in x, or $\Delta$x.
So, the volume of the whole thing is:
MathJax TeX Test Page $$2\pi\int_a^b f(x)\, \mathrm{d}x$$

# Example Problem

Rotate the given region around x = 1

MathJax TeX Test Page So, the problem is to rotate the given region around $x = 1$.  This would be slightly more annoying to do using disks, so we can use shells instead.  So, the volume is $$2\pi\int_{-1}^{1}\text{radius * height}\, \mathrm{d}x$$ We want to write the radius and the height separately in order to figure out the integral.
Radius = $1 - x$
This is because the radius is how far away the $x$ is from $x = 1$, so that's $1 - x$.
Height = $f(x)$, or $1 - x^2$
This is pretty straightforward, at any given shell, the height is $f(x)$. So the v = $$2\pi\int_{-1}^{1}(1-x)(1-x^2)\, \mathrm{d}x$$ You can expand, then you get $$\boxed{v = \dfrac{8\pi}{3}} = 8.3776$$
David Witten