Finding the radius of cylindrical shells when rotating two functions that make a shape about an axis of rotation (the shell method) I have been constantly writing and re-writing the procedure to find the radius in my notes. I know it is a simple concept but everywhere I look doesn't go over that concept- maybe I should know it already but I must have forgot. I know what the radius is by definition but I am still having trouble. 
At first I took the line perpendicular to height and usually came up with the function that the line touched. 
Then after getting it wrong I would subtract/add the difference of the axis of rotation and the start of the function except the radius always wanted to be $x$ or $y$. 
Then I came up with the idea that radius was the difference between $f(y)$, $f(x)$ and the axis of revolution. So the radius would be $x$ or $y$ plus/minus the difference from the axis of revolution. 
The problem is that there is not enough examples or questions with answers in my book to know if I understand it correctly.
 A: The key idea is that the radius $r$ is a variable which we create to integrate over. Let's look at an example: finding the volume of the region between the curves $f(x)=-(x-3)^2+5$ and $g(x)=x$ when it is rotated around the $y$-axis, using the method of cylindrical shells. Note that the two curves intersect at $x=1$ and at $x=4$.
Here is the figure:

Now, to get an idea of how the cylindrical shells work, imagine cutting a thin vertical sliver out of our picture and rotating just that around the $y$ axis - getting something that looks like a cylindrical shell.

Now we want to compute the volume of this shell: The volume of the shell is approximately area of the sliver multiplied by $2\pi r$ where $r$ is the distance from the axis of rotation. Why? Because for every point on the line of our sliver we are making a circle or radius $r$. We can compute the area of the sliver by approximating it as a rectangle with some very small base $dr$ and height $f(r)-g(r)$. Hence the volume of the shell is approximately
$$2\pi r(f(r)-g(r))dr$$
 
Now imagine dividing our figure entirely into thin slivers and rotating them. By taking an integral over all possible $r$, we can essentially sum up all of the volumes of these cylindrical shells and get the final volume. 
$$V=\int\limits_a^b2\pi r(f(r)-g(r))\,dr$$
For the sake of completing our example, notice that in our case $a=1$, $b=4$, $f(r)=-(r-3)^2+5$, and $g(r)=r$. Hence for our example we have
$$\begin{align}
V&=\int\limits_1^42\pi r(-(r-3)^2+5-r)\,dr
\\&=2\pi\int\limits_1^4 r(-r^2+5r-4)\,dr
\\&=2\pi\int\limits_1^4 -r^3+5r^2-4r\,dr
\\&=2\pi\left(-\frac{1}{4}r^4+\frac{5}{3}r^3-2r^2\right)\bigg|_{r=1}^{r=4}
\\&=2\pi\left(-64+\frac{5}{3}\times 64-32+\frac{1}{4}-\frac{5}{3}+2\right)
\\&=2\pi\left(\frac{32}{3}+\frac{7}{12}\right)
\\&=2\pi\left(\frac{135}{12}\right)
\\&=\frac{45\pi}{2}
\end{align}$$
