[Flumper Barflusterbutt]

Anyone want a quick lesson in how calculus works? I thought so. Pull up a chair and I'll explain how to calculate the area of a circle if you don't know the forumula for it.


Imagine that we have a "clothesline" that is "C" feet long. It starts at point A and goes to point B.

[Flumper Barflusterbutt]

Now, I am going to take that "clothesline" and form a perfect circle with it. The resulting circle will have a perimeter of "C" feet and a radius = r

we know that the C = 2*PI*r

[Flumper Barflusterbutt]

Now, if I didn't know how to calculate the area of that circle, I could begin to cut it up into triangles that all meet at the Center of the circle.

I know that each one of these triangles has a side equal to r. But the round end makes it difficult to exactly calculate the area of them.

However, if I made them infinitely small, then the round end would be negligible.

[Flumper Barflusterbutt]

Imagine that I just kept cutting that circle up into tinier and tinier pie pieces. Each one is exactly the same width. We will call that width "Delta C" or ^c or dc.

Basically those triangles would look like this:

[Flumper Barflusterbutt]

And we know the area of a triangle is (1/2)*b*h.

So, the area of that little triangle is (1/2)*dc*r

Now, if we unfold our "clothesline" we see that we have an infinite number of these little triangles all exactly the same shape and size. They are all "r" long and dc wide.

Now, all we have to do is add them up. and to do that we use "integration".

[Flumper Barflusterbutt]

All we need now is to "integrate" the forumula:

(1/2)*r*dc from one end of that clothesline to the other. From 0 ---> C

Written as:

[Flumper Barflusterbutt]

Now, the integral of dc is C. You Multiply dc by C and divide by the resulting power. In this case the power of C is 1. So dividing C by 1 gives you C.

So that forumula, after it is integrated is (1/2)*r * C

Now, if we substitute the value of C (C=2*PI*r) then we get:

(1/2) * r * 2 * pi * r

which can be reduced to: Area = Pi * r^2

And there you have it.

[leeman]

Now, the integral of dc is C. You Multiply dc by C and divide by the resulting power. In this case the power of C is 1. So dividing C by 1 gives you C.

So that forumula, after it is integrated is (1/2)*r * C

Now, if we substitute the value of C (C=2*PI*r) then we get:

(1/2) * r * 2 * pi * r

which can be reduced to: Area = Pi * r^2

And there you have it.

Why didn't you just type that?

[Flumper Barflusterbutt]

Why didn't you just type that?

Because, then I would not have had the need to explain how you can derive that formula using the miracles of calculus.

[leeman]

Because, then I would not have had the need to explain how you can derive that formula using the miracles of calculus.

Nuthon miraculus with calculus ..

Just facts.

[Flumper Barflusterbutt]

Nuthon miraculus with calculus ..

Just facts.

We could also use calculus to derive the formula for the volume of a cone. Say a cone has a base radius of "r" and a height of "H". If we come up with a formula that is a function of h to describe a horizontal section of that cone in terms of r and h (the height from the base), then we can integrate that area from 0 to H and get the volume.

The slope of the edge of the cone is: m = H/r

So, the radius of any slice that is a distance h above the base is: r-(h*r/H)

So, the area of any slice that is h above the base is: (r - hr/H)^2*PI

That expands to: (r^2 - 2hr^2/H + h^2r^2/H^2) * PI

If we integrate that from 0 to H we get: PI x (r^2h - 2h^2r^2/2H + h^3r^2/3H^2)

Substituting H for h = PI x (r^2H - H^2r^2/H + H^3r^2/3H^2)

Which we can reduce to: PI x (r^2H - r^2H + r^2H/3)

Which we can reduce to: PIr^2H/3

And that is the forumula for the volume of the cone.

[Dagger]

Long way around the block for the first formula, but correct in the end.
The cone volume was forgotten many years ago, thanks for the refresher.