My second take on the Volumes of Revolution formula

Another way of looking at the Volumes of Revolution formula is to divide the area up into a series of... cylinders. Yes. You know how when you were learning about how integration "worked" you had those investigations where you had to divide the area up into small rectangles and calculate the combined area and work out an estimate? Well, that's exactly what I'm doing here, except with rectangles rotated 360 degrees about one of the axes because that just totally makes them less obnoxious. (Not really.)

Here's a triangle that wants to become a cone.

Now, one way that you could work out the volume of the cone-to-be is by using the area-of-base-times-height-divided-by-3 formula. But let's make stuff more complicated, because that's so much fun! (Okay, I should probably stop with the sarcasm now.)

One place to start off would be to divide the area between x = 0 and x = 2 into four rectangles and calculate the volume of each corresponding cylinder and then add the volumes together. (The fourth rectangle in the diagram below is that purple line at the bottom that has 0 area. I really should have picked (y = x^2 +3) or something else that would have given me a better area- an actual area, in fact- for the first rectangle. Ah well.)

As can be seen from the above diagram, these rectangles don't give a very good approximation of the area. Look at all those triangle bits being left out! To counter this partially, you have to divide the area into smaller rectangles like so:

And so on, until the rectangles are all so small that all you see is this vague purple blur.

It's easier to work out the area of each cylinder (formed from each rectangle) if you make yourself a nice simple formula and then work from there. The volume of each cylinder is (pi)(r^2)(h), but that's pretty generic so let's look at how it applies to the above diagram.

For the above diagram with the 8 rectangles (or rather, 7 rectangles plus one pathetic excuse of a rectangle), each cylinder is 2/8 = 1/4 units wide. Therefore, h in this case is 1/4 units for every rectangle. r is the tricky one, because it is different for each rectangle. r is equal to the y value at each x-coordinate. For the first rectangle, r = 0, then for the second one r = 1/4, then for the third rectangle r = 1/2, and so on (it sure helps having y = x as your formula! If it was x^2, though, then the rectangles' r values would be 0, then 1/16 and then 1/4 and so on.) Finally, pi is a constant, which is nice.

Now, as I said before, you eventually want to get a whole load of infinitesimally small rectangles, and therefore infinitesimally small cylinders too. Let's say that you want n small rectangles in that space. Therefore, the width of each rectangle is 1/n units. That gives us (1/n)(pi)(r^2). Now we need to work out a general formula for r^2. For the first rectangle, r = f(0), then, in the second rectangle, r = f(1/n), then r = f (2/n) and so on. A general formula for r would be f((p-1)(1/n)), or f((p-1)/n), where p is what number rectangle you're looking at.

To sum up:

Volume of first cylinder = (1/n)(pi)(f(0/n)^2)
Volume of second cylinder: (1/n)(pi)(f(1/n)^2)
Volume of final (nth) cylinder: (1/n)(pi)(f((n-1)/n)^2)

Combined volume of cylinders: (1/n)(pi)(f(0/n)^2 + f(1/n)^2 + f(2/n)^2 ... f((n-1)/n)^2)

Which can further be summed up in the handy dandy summation formula that I just learned! (Too late for the previous investigation, unfortunately, but hey, better late than never!)

Now, to find the exact area under the curve, you want n to be as large as possible- infinitely large, in fact. That's where limits come in! Now, I'm not sure whether I've used limits correctly in the following equation, so as soon as I've published this I'll shoot off an email to my teachers to ask them. Of course, if you're a teacher yourself, or if you're just plain good at maths, then you can tell me whether I'm right or wrong in my notation.

Let's analyse this formula a bit more. f((p-1)/n) gives the y-value for various x-coordinates, since f at something gives the y-value of whatever x-coordinate "something" is. (f((p-1)/n))^2 is therefore where the y^2 part from the volumes of revolution formula comes from. Pi, our good friend, is there. Finally, the 1/n is the width of the tiny little rectangles/cylinders, made even tinier as n approaches infinity. (1/(a large number) = a very small number.) I believe this is where the "dx" part comes from in the volumes of revolution formula. The volumes of revolution formula also has the start and end point marked in, but I'm not sure how to include that in the summation formula above.

And that is a second explanation for the Volumes of Revolution formula, just because one explanation is simply not enough!