Volumes of Revolution - Help!

[Arioch]

O great mathematicians of the Playground, hear my plea. I have homework involving volumes of revolution, and while doing them around the axes is easy, I can't work out how do do them around lines other than the axes. How does this affect the formula? Should I translate the whole thing so it's revolving around an axis and work from there?

I was on a school trip during the lesson they explained this, and I can't work it out. :smallfrown:

[Shishnarfne]

Actually, it's rather straightforward: simply change your coordinates so that you redefine your axes to be the lines around which you want to revolve.

(Also, that's a bit of a cheap trick to describe it that way.)

A more appropriate way to think about it is to re-think the problem in terms of distance from the axes of revolution rather than the standard axes.

For example, suppose you want to figure out the volume of a parabola (y=x^2) revolved about x=2 between y=0 and y=4.

Your distance from the line x=2 is given by:
distance=2-(sqrt(y))
Then you integrate pi*(distance^2) from y=0 to y=4.


(I could have found a simpler example, I think.)

[averagejoe]

The coordinate change is perfectly valid, and not at all cheap. You remember in algebra when you learned to shift and reflect functions? (i.e. y=x^2-4 is y=x^2 shifted to the down by four.) That's basically the technique you use. If you want to rotate around, say, the line y=4 then it's just like shifting your function down four and then rotating it around the y-axis.

However, make sure you don't screw up which way you have to shift it, which is an easy thing to do. An easy way to remember is to perform the same operation that will change your line of rotation into the origin. For example, if you're rotating around y=4 then 4-4=0, so you want to subtract four from your function. If you have a good geometric intuition then that's just as good. Ideally you'd want to do both; imagine what you have to do to the function, then prove it mathematically.

Rotating around lines parallel to the x-axis is just as easy, you just have to keep in mind that the way to shift left and right is sometimes counter intuitive. I generally find it easiest to solve for x, because then it becomes just like doing it with y. (For example, if you have y=f(x) then x=f^(-1)(y), where f^(-1) is f inverse. If you have the function in this form then shifting to the right and left is just like shifting y=f(x) up and down, for reasons that seem obvious enough to not need to explain, but that might be just because I have years of experience with this, so if you want one I can give it.)

(Incidentally, this is why shifting up and down seems backward. For example, f(x+4) is f(x) shifted left four spaces, when the intuitive thing is to shift up (since that's the positive direction.) However, if you take x=f^(-1)(y) (that is, solving for x) then it's easy to see that to shift to the left four you need to subtract four from the right side, since now it's exactly like shifting y=f(x) up and down. However, to get back to the form y=f(x) the first step you would have to do is add four to both sides; then you take the function of both sides and get y=f(x+4). This isn't really that important, but I've found it helps people "get it.")

Followup questions are welcome. I know I'm not always completely clear.