When dealing with areas, we have seen that there exists a sharp division between the computation of areas of figures, bounded by lines or by curves. For the latter, the elementary methods of decomposition and augmentation were not enough and we encountered infinite processes, that is, limit methods. Therefore we should expect that also in the case of volumes limit methods only arise when curved surfaces are involved.
It is noteworthy that one should already encounter problems in an attempt of determining the volume of a pyramid in the elementary manner, which can be applied to cubes or prisms. This has been known a long time ago; in fact, the Greek mathematicians knew already that for pyramids limiting processes were required and developed an appropriate method. The so called method of exhaustion was proposed by Eudoxos of Knidos (408 - 355 b.C.). Euclid included it in his Elements and made it thus accessible to all mathematicians. Archimedes improved the method later on and applied it to the circle, sphere and bodies with curved surfaces.
The real reason, for which the computation of the volume of the pyramid led to limit considerations, has only been recognized in recent times. While you can compute the area of a triangle by elementary methods, because every triangle can be completed into a parallelogram, it is impossible to reduce every pyramid by decomposition and augmentation to a prism.
If you subdivide the height of a pyramid
into n layers and place through the points of
subdivision planes parallel to the base, you obtain an internal
and external staircase
structure, which comprises
parallelepiped shaped plates. If you denote the volumes of the
external and internal staircases by Ve and Vi,
respectively, then
Ve > V > Vi .
With n = 2, 4, 8, . . . , the bounds come closer, so that
Ve > V'e > V"e > . . . > V > . . . > V"i > V'i > Vi,
and the doubling can be continued indefinitely, whence follows the Rule:
The volume of the pyramid is the limit, which is simultaneously approached by the external and internal staircases during continued refinement of the subdivision.
An exact execution
of the limit process demands mathematics, which are 
beyond the range of the present
treatment, whence
you will have to be satisfied with a proof, based on visual
insight, proposed by
Bonaventura
Cavalieri in his Principle:
If two bodies can be placed between two parallel planes in such a manner, that all corresponding cuts (that is, at equal height parallel to the base) have the same area, they have the same volume.
For two pyramids, this principle is obvious. The corresponding external and internal staircases of both bodies have the same volume as prisms with equal base (assumption) and height. Hence also the complete external and internal staircases are equal. The volumes of both pyramids lie between the same bounds; since one can draw the limits arbitrarily, the volumes must be the same.
In a similar manner, you can also decompose two other arbitrary bodies between parallel planes, the corresponding cross-sections of which are equal, into external and internal staircases and arrive at the same result.
By means of Cavalieri's Principle, we will prove first of all the Rule:
Pyramids with equal base and height have the same volume.
It will be sufficient to consider an especially simple pyramid, in order to arrive at a formula for all of them.
Let two pyramids
with equal base and height lie between the same parallel planes.
The ratio of any parallel cross-sections of a
pyramid to its base is: G'1:G1=h'1²:h1²,
and that of the corresponding cross-section to the base of
another pyramid: G'2:G2=h'2²:h2².
As G1= G2, h1=
h2 and h '1=h '2,
, then also G '1 : G'2.
This is true for any corresponding cross-sections. Therefore,
according to Cavalieri's
Principle, the two
pyramids have the same volume.
It has already been pointed out, that the reason why it is not possible to compute the volume of the pyramids without a limiting process must be sought in the fact, that not every pyramid can be augmented into a prism; however, this is possible for the especially simple three-faced pyramid. In fact, it can be shown that
a
straight, three-faced pyramid can be decomposed
into three pyramids with the same volume.
If ABC is the top face, DEF the base of a straight, three-faced prism, then you can decompose it by a cut from D to the opposite edge BC into the three-faced pyramid D(ABC) and a four-faced pyramid D(BEFC).
If you now cut from D to the diagonal CE of the base BEFC, the four-faced pyramid becomes two three-faced pyramids D(BEC) and D(EFC), which have the same volume, because their bases are half the parallelogram BEFC and their heights are the same.
Similarly, the first and third pyramids above have the same volume, because they can also be viewed to be pyramids, the bases of which are the base and cover of the prism and the heights of which are equal to that of the prism. Hence the three pyramids have the same volume, that is, a third of the volume of the prism V = Gh/3.
Since pyramids with equal base and height have the same volume, the volume of every pyramid is
V = Gh/3.