Solid of revolution with function $1/x$ bound on $[1, 2]$ rotated about $x=0$ through the “shells method”

Question

I am trying to find the volume of the solid of revolution that that has bounds $x = 1, 2$ and enclosed by the functions $y= 1/x$ and $y=0$ and is rotated about the $x$ axis. I was able to obtain the answer of $\pi/2$ very simply through the “slicing” method but I cannot get the same results trying to do the function through the “shells” method of finding volumes. I can do the arithmetic, I just need an expression to show me the way forward. I have been working on this problem for several hours without stop now so any help would be very much appreciated

Answer 1

It is more natural to solve this problem using the slicing method rather than the shell method. But, if you want to use the shell method, then you're after the volume of the region obtained by rotating around the $x$-axis the region between the functions $f,g\colon[0,1]\longrightarrow\Bbb R$, with $f(y)=1$ and$$g(y)=\begin{cases}\frac1y&\text{ if }y\in\left[\frac12,1\right]\\2&\text{ if }y\in\left[0,\frac12\right]\end{cases}$$(see the picture below). So, the answer is$$\int_0^12\pi y\bigl(g(y)-f(y)\bigr)\,\mathrm dy=2\pi\left(\int_0^{1/2}y\,\mathrm dy+\int_{1/2}^1y\left(\frac1y-1\right)\,\mathrm dy\right).$$