I would imagine that regardless of the method used, the volume should be the same, but in a particular problem they seem to produce different answers:
Rotate the area bounded by $xy=2$, $x=2$, $x=3$ and the x-axis.
With disk I get $\frac{2\pi}{3}$: $$\int_2^3{\pi\left(\frac{2}{x}\right)^2}dx=4\pi\int_2^3{\frac{1}{x^2}}dx=4\pi\left[-\frac{1}{x}\right]_2^3=4\pi\left(-\frac{1}{3}+\frac{1}{2}\right)=\frac{2\pi}{3}$$
With shell I get $\frac{4\pi}{3}$:$$\int_{\frac{2}{3}}^12\pi y\left(\frac{2}{y}\right)dy=\int_{\frac{2}{3}}^14\pi dy=4\pi\left[y\right]_{\frac{2}{3}}^1=4\pi\left[1-\frac{2}{3}\right]=\frac{4\pi}{3}$$
Any hints as to what mistake I've made, or an explanation as to how this is possible would be greatly appreciated!
There are two issues in what you have done, that both affect the volume you have calculated. Consider the following cross-sectional diagram:
The volume we want is that found by rotating the blue region, which you do successfully with the disk method. The volume you calculate with the second integral is that found by rotating the orange region. To calculate the blue region's volume, you have to include the shells with radius smaller than $2/3$, and not include the values of $x$ smaller than $2$.
So, to calculate using the shell method, we need $2<x<2/y$, so the width of the shell is $2/y-2$, and also include the shells of width $3-2=1$ for $0<y<2/3$. So we find the volume is $$ \int_{2/3}^1 2\pi y\left( \frac{2}{y}-2 \right) \, dy + \int_0^{2/3} (2\pi y)2 \, dy = \frac{2\pi}{9} + \frac{4\pi}{9} = \frac{2\pi}{3}, $$ agreeing with the disk method.
In general, the methods must give the same answer if applied correctly, since they are supposed to calculate the same volume!