Volume of solid of revolution about a line other than the axis - using Cylindrical Shells method I am struggling with a concept of finding the radius in the shell method of finding the volume of the solid when revolved around a line other than the axis.  Maybe the issue here is that I am looking for a "rule" instead of actually visualizing the problem.  It was my understanding that when revolving around a line other than the axis a formula could be used as follows: (this example uses revolution about a vertical line other than the $y$ axis)
the radius would be $|verticalvalue|$ $+/-$ $x$.  Use $+$ if the region of revolution is to the right of the line and use $-$ if the region is to the left.
I thought I had worked many problems using this concept and gotten the correct answer.  However, doesn't work in this problem:
Use the method of cylindrical shells to find the volume generated by rotating $y=x^2$, $y=0$, $x=1$, $x=2$ about the line $x=1$.
My integral, using my above "concept" was to form the following integral:
$V=2\pi\int(1+x)(x^2) dx$ evaluated from lower limit of 1 to upper limit of 2.  I used the absolute value of $1$ PLUS (since the region of revolution is to the right of the vertical line of revolution.
This does not produce the book answer and the book used $x-1$ for the radius instead of $x+1$, obviously producing very different results.
My question:  is there a rule for writing the radius?  Why didn't my rule work?
 A: You would only "use the absolute value of 1 plus the x" if the region of rotation was to the right of the vertical line of revolution AND the left boundary of the region of rotation were to the left of the y-axis. Perhaps most of the problems you did in the past were this case.

There are rules for these types of problems. If you are rotating about a vertical line, and the axis of rotation is the left boundary, you use $x-h$. But if you're rotating about the right boundary, you use $h-x$.
As for your problem, it is usually better to draw a picture. I don't know anyone who doesn't at least visualize these types of find-the-volume-of-rotation problems.

Since $1\le 1< 2$, use $\int_1^2 (x-1)f(x)dx$. You get
$$2\pi\int_1^2(x-1)x^2dx$$
However if you were to rotate it around the line h=2, not shown, you get
$$2\pi\int_1^2(2-x)x^2dx$$
Note that you would get a larger volume for the first rotation (h=1) because the taller part of it has to rotate more, covering more volume, but for the second rotation (h=2), the larger part is in the "center" and creates less volume.
A: " Maybe the issue here is that I am looking for a "rule" instead of actually visualizing the problem". I am not sure if that is a good idea. However, what you CAN do is, perform a horizontal translation (a shift of $1$ units to the left) on all functions in question. So "Use the method of cylindrical shells to find the volume generated by rotating $y=x^2$,$ y=0$, $x=1$, $x=2$ about the line $x=1$, now becomes "rotating the region formed by$y=(x+1)^2$, $y=0$, $x=0$, $x=1$ about $x=0$. This way you can now apply the "standard" formulas for cylindrical shells since the axis of rotation is now a coordinate axis. I would still make a graph of the situation, try to understand why this works
