Volume of a Cone, Finale
Okay, so here we go. To generate a cone, we can rotate a right triangle with a width of r (radius) and a height of h about its height.
As you can see, we have the same problem we did with the rotating rectangle in coming up with a volume formula for the figure that is generated. We can't simply take the area of the right triangle (1/2rh) and multiply that by the circumference of the circular base (2&pir), because not every point in the area of the triangle moves a distance of 2&pir.
However, as was the case with the rotating rectangle, we can treat the rotating right triangle as a collection of data points. In order to find the total distance all of the points would move at different distances from the center of rotation, we can find the average distance every point would move—the distance all of the points would move if they were all the same distance from the center and covered the same total distance.
Consider again the right-triangle-shaped "cat" data we looked at in the previous post:
Just as we did with the rotating rectangle, we bisected first all the horizontal lines (all the points of the right triangle) and then all the vertical lines (again, all the points of the right triangle), to come up with a single point, the centroid, that could "describe" all the points in the right triangle.
As it turns out, there's a name for these "average lines" in a triangle--medians (good name). And, wouldn't you know it, but all the medians of a triangle (there are actually three, but we only need to show two) intersect at a point called the centroid of the triangle.
And you want to know something else? The centroid of a triangle with a width of r (radius) and a height of h is always located at (1/3r, 1/3h)!
So, for our right triangle, the total distance every point would move at different distances from the center of rotation is the same as the total distance all the points would move if they were all 1/3r from the center.
Thus, we take the area of our triangle (1/2rh)—all the points—and multiply this area by the distance all the points would move (2&pi[1/3r]). The result, of course, is 1/3&pir2h.
Also, a big hat tip to that Greek I mentioned in the first post, Pappus of Alexandria. In putting together all four parts of this explanation, I have really just been working backward from his second centroid theorem:
As you can see, we have the same problem we did with the rotating rectangle in coming up with a volume formula for the figure that is generated. We can't simply take the area of the right triangle (1/2rh) and multiply that by the circumference of the circular base (2&pir), because not every point in the area of the triangle moves a distance of 2&pir.
However, as was the case with the rotating rectangle, we can treat the rotating right triangle as a collection of data points. In order to find the total distance all of the points would move at different distances from the center of rotation, we can find the average distance every point would move—the distance all of the points would move if they were all the same distance from the center and covered the same total distance.
Consider again the right-triangle-shaped "cat" data we looked at in the previous post:
Just as we did with the rotating rectangle, we bisected first all the horizontal lines (all the points of the right triangle) and then all the vertical lines (again, all the points of the right triangle), to come up with a single point, the centroid, that could "describe" all the points in the right triangle.
This Is It!
It's important to see that no matter what kind of right triangle we have, these "average lines," (actually, they're a lot like regression lines) will look the exact same. That is, they will always extend from the midpoint of a side to the opposite vertex. If you study the right-triangle-shaped cat data above, you will see why this is so. Each "average line" must extend from 0 (a vertex) and bisect the opposite side (the midpoint).As it turns out, there's a name for these "average lines" in a triangle--medians (good name). And, wouldn't you know it, but all the medians of a triangle (there are actually three, but we only need to show two) intersect at a point called the centroid of the triangle.
And you want to know something else? The centroid of a triangle with a width of r (radius) and a height of h is always located at (1/3r, 1/3h)!
So, for our right triangle, the total distance every point would move at different distances from the center of rotation is the same as the total distance all the points would move if they were all 1/3r from the center.
Thus, we take the area of our triangle (1/2rh)—all the points—and multiply this area by the distance all the points would move (2&pi[1/3r]). The result, of course, is 1/3&pir2h.
Postscript
I apologize for leaving this demonstration well short of perfect by omitting an elementary explanation for why the centroid is always located where it is in a triangle, but, on the bright side, I'm fairly certain that the fact can be demonstrated without the use of calculus, so I'm still considering the original challenge to have been met. (Let me issue a counter-challenge to anyone willing, however, to provide an explanation for this very handy fact.)Also, a big hat tip to that Greek I mentioned in the first post, Pappus of Alexandria. In putting together all four parts of this explanation, I have really just been working backward from his second centroid theorem:
The volume . . . of a solid of revolution generated by the revolution of a lamina about an external axis is equal to the product of the area . . . of the lamina and the distance . . . traveled by the lamina's geometric centroid.
Labels: mathematics