proof for the formular of volume of cone/pyramid

Carson 的答案很好。我這個答案要解決的是一個更general的情況，一個任意形狀底的，也可以是斜的錐體。
我的方法generalize Carson他解決長方錐體體積公式的方法，想像錐體放在一個三維空間裏，把底放在x-y plane上。設錐體的頂點在(x,y,h)上，h是錐體的高。設這個錐體的底面積為A。
我現在把這個錐體水平切開成n份，每份的高度就是h/n，最低一份的體積必然小於Ah/n，也必然大於上一層的底面積 乘　h/n。
那麼上一層的底面積是甚麼呢，它是A[(n-1)/n]^2，證明如下：
如(p, q,0)是在底面積上邊上的任何一點，(p',q',h/n)是在第一層上的對應一點，則
(p,q,0), (p',q',h/n) 和 (x,y,h) 應該是在一條直線上，不難求出：
p' = p + (x - p) /n = (n-1)p/n +x/n
q' = q + (y - q) /n = (n-1)q/n +y/n
用2D變換的想法，這是變換了一次縮少和平移，平移是不會影響面積的，縮少則影響跟據比例的平方，所以第一層的面積為A[(n-1)/n]^2。
如此類推每層的體積上限是 A[(n - i+1) / n]^2 * h/n，而下限則為A[(n - i) / n]^2 * h/n。i 是由底數起，總共有n 層。
把每層的上限加起來就是錐體體積的上限，而每層的下限加起來就是錐體體積的下限。
加起來的上限是Ah(n)(n+1)(2n+1)/6n^3，而下限則是Ah(n)(n-1)(2n-1)/6n^3。 加法很容易，跳步了。當n愈大，上下限的差距愈少。用Squeezing Principle，我們可以把錐體體積定在上下限的共同極限，即Ah/3。
證明結束，QED。

2007-10-14 17:47:45 補充：
當 n 接近無限大，上限和下限都會接近Ah/3，這是因為所有非n^3都會因為分母變大得太快而變無限小。所以Squeezing Principle可以用來證明體積就是Ah/3。

Imagine a step-pyramid made up of small square prisms with dimensions
a-by-a-by-b, arranged in square layers. If the layers are k-by-k for
k = 1, 2, 3, ..., n, then the total volume of the step-pyramid is:
V = a^2*b*[1^2 + 2^2 + 3^2 + ... + n^2],
= a^2*b*n*(n + 1)*(2*n + 1)/6 (which can be proved by induction),
= a^2*b*n^3*(1/3 + 1/[2*n] + 1/[6*n^2]).
Now, the height of the step-pyramid is h = b*n, and the area of the
base is:
B = a^2*n^2,
so the volume is:
V = h*B*(1/3 + 1/[2*n] + 1/n^2).
Now let n -> infinity, so the steps get smaller and smaller, and the
volume of the step-pyramid approaches the volume of the inscribed
pyramid. Then the right-hand side approaches h*B/3, as the volume V
approaches the volume of the pyramid.
This same proof works for a pyramid with any polygonal base. For a
cone, take a pyramid with a regular m-gon for a base, and let
m -> infinity.
The base approaches a circle, and the pyramid approaches a cone, and
the formula V = h*B/3 still holds in the limit.
There is another proof, which cuts a rectangular solid into six pieces
of equal volume, each of which is a triangular pyramid (or irregular
tetrahedron). This proves that the volume of each is 1/6th the volume
of the starting rectangular solid, and each has base of area 1/2 the
area of one of the faces of the solid, and height equal to the
dimension of the solid perpendicular to that face.
r is the radius of the base of the cone
h is the height of the cone
s is the radius of the cross-section of the cone (0<s<r)
x is the distance between the specific cross-section to the tip of the cone (0<x<h)

圖片參考：http://www.mph.net/coelsner/calcapps/cone_exprn_files/eq0001L.gif

I hope this can help your understanding. :)