Attn! Ignore the post about rotational volume!

I bet you already did. But anyway, here's the reason why you should, assuming that you cared.



The last entry I posted about rotational volume was incorrect.

Mr. Cole wasn't lying when he said that if you make up a rule about integrals, it will be wrong. But anyway, here's why it was wrong. The mean value cuts through the top of an integral, and divides it so that the total area above the mean value equals the total area "missing" below the mean value. If you're not sure what I'm talking about, go to the graph in my previous entry and observe how some of the area from the integral goes above the mean value and some goes below. These areas should be equal if it is a mean value.

Which is fine and dandy, but now consider this. A square unit that is right on the x axis as it rotates will form a cylinder with a radius of x and a height of x. Let's say that we move the square two units from the x axis. Now it will form a cylinder with a hole. These volumes are not equal because the square that was further away from the axis travelled further and thus took up more volume.

Because of this simple fact, the rotational volume of the area above the mean value does not equal the corresponding rotational volume of the area below the mean value. Therefore, the mean value cylinder incorrectly represents the rotational volume of the 2 dimensional object.

So in order to correctly find the rotational volume, this is what you would do.

You would find the sum of an infinite amount of cylinder volumes, much as you would find the sum of the area of an infinite amount of rectangles. Since an integral can be broken up into a bunch of little rectangles, each of those rectangles can form little cylinders that we can find the area of. So let's take the Riemann Sum formula and modify it.

Σ(f((k(B-A)/n)),k,1,n) as n→∞

In order to modify it, we will use V=πr²h to change each rectangle into a cylinder. In this equation, r is represented by the height of each rectangle, or f((k(B-A)/n)). Height is Δx because the height is ever decreasing.

So, plugging into the equation, we get:

Σ(πΔx*(f((k(B-A)/n)))²,k,1,n) as n→∞

Which can be rewritten as:

π*Σ((f((k(B-A)/n)))²,k,1,n)Δx as n→∞

Notice how the function inside the sum is squared. If we think in terms of rectangles for a minute, squaring the function basically means that we're squaring the height of the rectangles, which basically means that we're squaring all of the function values. Therefore, what we're really finding is the integral of the squared function.

Thus, the equation in terms of integrals can be written as such:

V = π*∫(f(x))² dx [a,b]

There you go, and sorry for the inconvenience. I know that you all flipped out when you saw my fallacious entry.