Find volume by these bounded regions $y=\dfrac{1}{x}, x=1, x=2, y= 0$ about $x= 3$ (shell method)
Not sure what is wrong with my integral here.
$$2\pi \int_1^2 \frac{(3-x}{x} \, \mathrm{d}x+ \frac{1}{2} \int_1^2 3-x \, \mathrm{d}x$$
It does not match up with the answer, $6\pi \ln 2 - 2\pi$
I split it up into two integral sincet here are two heights, one shape is like a rectangle another is a curvature.
If you used the disc method, you would need to break the region up into two pieces. However, applying the shell method allows you to use a single region here.
For a given shell, the radius of the annular region is $3-x$ and the height is $1/x$, so
$dV = h \cdot 2\pi r \cdot dr = \dfrac{1}{x} \cdot 2\pi (3-x) dx = 2\pi\ \Big(\dfrac{3}{x} - 1\Big) dx$
So
$I = 2\pi \displaystyle\int\limits_{1}^{2}{\Big(\dfrac{3}{x} - 1}\Big) dx = 2\pi \Big[3\ln{|x|}-x\Big]_{x=1}^{x=2} = 2\pi\ (3\ ln{\ 2}-1)$