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
Comments:
the "co-ordinates of the centroid"
problem is straightforward
(if we stoop to co-ordinates;
i haven't tried proving the result
using "classical" methods).
put the triangle in the first quadrant
with the right angle at the origin.
the other two vertices are now
at (0,h) and (r,0). draw the medians
(connect these points, respectively,
with (r/2, 0) and (0,h/2)).
the equations of the median lines
(via "slope-intercept form")
are now
y =(-2h/r)X + h
and
y = -h/(2r)*X + h/2.
equate the right-hand sides;
solve for X; X = r/3.
"plug in" on either equation:
Y = h/3 (or invoke symmetry;
this calculation can be skipped).
vlorbik
Much appreciated, sir.
A more geometric way to find the location of the centroid as being 2/3 of the way from the vertex to the midpoint (but that clearly implies that it's at 1/3 of the height in your example):
The medians land at the midpoints of the opposite sides. Draw the line connecting those midpoints. It's well-known that the resulting segment is parallel to the third side and half its length. Then consider the two triangles created in the trapezoid between those parallel lines. Those triangles are clearly similar (because of the parallel lines) and stand in a ratio of 1:2 (since one base of the trapezoid is half the length of the other). But then the other sides of the triangle also stand in this ratio, making the intersection of the medians at a point 2/3 of the way from the vertex to the opposite midpoint.
I’m not sure whether anybody read my comment to the problem of the volume of the sphere, if my demonstration was understandable.
A cone can be subdivided into many smaller pyramids with triangular base. Each pyramid has a volume equal to the base multiplied by the height, and divided by three. All we have to demonstrate is that a pyramid with triangular base is 1/3 the volume of the corresponding prism.
Let’s call the vertices of the prism’s base ABC, and the vertices of the opposite base (the roof) A’B’C’. The prism can be sliced into two pyramids: one pyramid with base ABC and vertex A’; the second pyramid with base A’B’C’ and vertex B. Each pyramid has identical base and identical height, therefore they have identical volume. What is left is a wedge with one edge BC and the opposite edge A’C’. This edge has volume identical to the other two pyramids. Why?
Well, the wedge can be thought as a pyramid with base CA’C’, and vertex B. Conversely pyramid ABCA’ can be thought as a pyramid with base ACA’ and vertex B. ACA’ and A’C’C are identical bases and the height is the same since the vertex B is in common.
Therefore we have demonstrated that the three pyramids have identical volume and their sum makes the prism: hence the volume of the pyramid is 1/3 the volume of the prism. This finding can be extended to pyramids with square base or any shape of the base, by subdividing the base in ever tinier bases. Therefore a cone can also be subdivided in many smaller pyramids with triangular base.
Giovanni Ciriani
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