I've had a quick search on-line and found two that attempt this. But both use the underlying ideas of calculus whilst avoiding the formal integral approach. The nrich pages are full of interesting maths so it's well worth a look.

http://mathforum.org/library/drmath/view/55041.html

Bob

]]>The two blue shapes are similar. So there is a scale factor that transforms lengths in the large shape into the corresponding lengths in the small shape.

The area scale factor in such a situation is the square of the length scale factor.

http://www.mathsisfun.com/geometry/tria … orems.html

section 3.

Bob

]]>This is what I was eager for.

I still want to know, how did you get:

B

]]>That works for any 'pyramid', whatever the shape of the base.

Many mathematicians insist that to be a pyramid, the base should be a polygon, but for the following proof, let's make the base any shape at all. (see picture) Then to be a pyramid like that, all I require is that any length, d, at distance x from the vertex and the corresponding distance in the base, D, are related by this formula:

where H is the height of the pyramid.

Because the dark blue base region is mathematically similar to the light blue region the areas, small:large, will be in this proportion:

Now for a little integral calculus:

Imagine the solid divided into thin slices parallel to the base with thickness delta x.

So let the slices become infinitesimally thin with thickness 'dx', and sum all slices from the top to the base

Furthermore, the pyramid need not be 'right', ie. the axis at right angles to the base. If a right pyramid is sheared parallel to the base, so that it leans over, the slices still have the same area, so the volume formula continues to work, provided H is the perpendicular height. (picture in next post)

Bob

]]>Which is Area of the base * Height * 1/3

]]>I am not getting the question. Do you have a sketch?

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