Sandra Philps
16. April 2015
26. EUCLID’S THEORY OF VOLUME
INTRODUCTION
3D Geometry begins with Volume of solid figures.
There is no Definition of Volume ( as there isn’t for Area)
It isn’t said that equality is no longer = to congruence
Here EQUALITY OF SOLID FIGURES = EQUALITY OF VOLUME
To begin we take Volume as a notion similar to CONTENT for plane figures
TO SIMPLIFY
We suppose that:
we are working over a Field F
we see volume as a function v: P  v(P)  F which associates to congruent figures the same volume
v is additive for Unions with nonoverlapping interiors
FOCUS
How Euclid proves that certain figures have equal volume
TERMINOLOGY DEFINITIONS
Pyramid: is a solid figure formed by taking a plane figure (ABCD) and joining it’s
vertices to a point outside this plane
(note: if the Base is a Triangle it is a triangular pyramid, of the base is a square it is a
square pyramid
Prism: Is a solid figure formed by taking two congruent figures in parallel planes,
with parallel edges, and joining their corresponding vertices
(note: if the base is triangular than it is a triangular prism)
Parallelepiped: is a prism whose base is a parallelogram.
(note: 3 pairs of parallelograms in parallel planes)
EUCLID’S FIRST RESULTS CONCERNING VOLUME
Proposition XI.28 A Parallelepiped is bisected by the plane (CDEF) (passing through the diagonals (CF,DE)
of a pair of opposite planes) through two opposite edges.
(Remember: Bisection is the division of something into two equal or congruent parts)
Prove:
By prop. I.34 ( In Parallelograms figures the opposite sides and angles are equal
to one another, and a diagonal cuts them in half) triangle CGF & CFB are equal
and triangle ADE & DEH are equal.
By prop.XI.24(If a solid figure is contained by (six) parallel planes its opposite
Sandra Philps
16. April 2015
planes are both equal and parallelogramic) Parallelogram CA is equal to EB and parallelogram GE is equal to
CH.
It follows that Prism ADECFG (contained by the two triangle DAE&CGF and the 3 parallelograms GE,AC&CE)
is equal to Prism DEHBCF (contained by the two triangles CFB&DEH and the 3 parallelograms CH,BE&CE)
Since the two prism are contained by Planes which are equal in number and in magnitude (def. XI.10 equal
and similar solid figures are contained by similar planes equal in number and in magnitude)
the SOLID AB IS CUT IN HALF BY THE PLANE CDEF.
NOTES ON THE PROOF
Euclid proves that the two halves have congruent faces and angles, BUT they can’t be superimposed on
each other in 3D spaces because they are mirror images. (as our hands are)
Anyway LEGENDRE (disciple of Euclid, in 1794 it comes up a geometry book in which he simplifies and
modernise Euclid’s book “EUCLID’S ELEMENTS OF GEOMETRY” , his name is engraved on the “Tour Eiffel”)
gives a proof that they are equivalent in sense of dissection.
Gerling later proves that any solid figure is equivalent by dissection to its mirror image.
EUCLID’S FIRST RESULTS CONCERNING VOLUME
Proposition XI.29 : Parallelepiped solids which are on the same base, and have the same height, and in
which the standing(ends of the straight-lines) up are on the same straight-lines, are equal to one another
Proposition XI.30 : Parallelepiped solids which are on the same base, and have the same height, and in
which the standing up are not on the same straight lines, are equal to one another
It follows that Parallelepipeds having the same bases and of the same high have the same volume
More in general:
Proposition XI.31: Parallelepipeds having equal bases (in the sense of same content) and of the same
height have equal volume.
Corollary: Any parallelepiped has the same volume as a rectangular parallelepiped with sides 1,1 and,a for
some a in F.
(In fact if we assume to be in an Archimedean field we can “show” this by dissection)
PROPOSITION XI.39
Given two triangular prism, one lying on its side, one standing up. The
parallelogram (AF) is twice the triangular base (GHK) (content) and they have
the same height. Then they have equal volume.
Prove:
ABCDEF & GHKLMN are equal height prism. (follows from assumption)
Sandra Philps
16. April 2015
We double both to get parallelepipeds.
Parallelogram AF is twice triangle GHK (by assumption). Parallelogram HK
is double triangle GHK(because we double)
So by I.34 ( In Parallelograms figures the opposite sides and angles are
equal to one another, and a diagonal cuts them in half) Parallelogram AF is equal to HK. (doubles of equals
are equal)
And so by XI.31 (Parallelepipeds having equal bases (in the sense of same content) and of the same height
have equal volume) parallelepiped AO is equal to GP.
Since prism ABCDEF is half of solid AO and prism GHKLMN is half of solid GP (because we doubled them),
the two prism are equal, since halves of equals are equal.
NOTES ON PROOF OF XI.39
This proof works by using the principle that: «halves of equals are equal»
That’s Ok if we consider Volume as a function with values in a field. But it needs justification if we want a
purely geometric theory of volume by dissection or by content.
THEORY OF VOLUME OF PARALLELEPIPEDS AND PRISM
The theory of volume of parallelepipeds and prism can be handled by methods familiar from the theory of
area, which means we have to translate notions from the plane into 3D space.
VOLUE OF PIRAMIDS
We have a different situation for the study of the volume of pyramids. For this the key result is:
Proposition XII.5: Triangular pyramids having equal bases (in sense of content) and equal height
have equal volume.
Proof:
Euclid uses the “method of exhaustion” (is a proof by examining all individual cases)
Write both pyramids as infinite unions of subfigures having equal volume.
Each subfigure equal to more than ½ of what was left over after removing the previous subfigures.
In this way the subfigures “exhaust” the entire figure.
And since the subfigures have the same volume it follows that also the two pyramids have equal volume
For this last step we use Archimedes Axiom: “Given two magnitudes having a relation, one can find a
multiple of either which will exceed the other”. This principle was the basis for the method of exhaustion,
which Archimedes invented to solve problems of area and volume.
If we apply this axiom to our prove we see that if the two pyramids were different, we can go on with this
exhaustion process until the remainder left over is less than the difference of the two pyramids. And this is
only possible because each remainder is less than one half of the previous remainder.
So we can conclude (in modern language) that the volume of the entire figure is the sum of all the
subfigures.
APPLICATION OF THE PROOF
P= ABCD is one of the given pyramids.
E,F,G,H,J,K are the midpoints of the sides.
We can decompose P in 4 pieces: 2 Pyramids P1=AEFG and P2=FBHK, which are congruent to each other
and their edges are ½ the edges of P, and 2 triangular prism T1=FHKGJD and T2=EFGCHJ. By prop. XI.39 (T1
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16. April 2015
is lying on its side, T2 is standing up/the base of T1 is a parallelogram which is twice the triangle base of T2
[remember: if you take the midpoint of any triangle and conjunct them the formed parallelogram will have
the same content as the sum of the other two triangles]) T1 and T2 have the same volume.
Prism T2 & pyramid P1have the same base and the same height it follows that: T2 > P1 (volume) (and
thatT1 > P2).
So T1+T2 > ½ P.
Now we do the same thing for the second Pyramid P’.
We divide P’ into 4 pieces P1’,P2’,T1’,T2’.
Since the base triangles of P and P’ are equal in content also the parallelograms on the base of T1 and T1’
have the same content. And so the parallelepipeds T1 and T1’ which obviously have the same height have
also the same volume.
So T1+T2 and T1’+T2’ have the same volume.
And both are more than ½ of P and P’.
It remains P1+P2 and P1’+P2’ which both are unions of two pyramids having same base and same high.
So know we can inductively go on with this process dividing each P1,P1’,P2,P2’ as unions of 4 paces.
Since the prism constructed at each step have the same volume and these exhaust the pyramids, in the
limit we find that P and P’ have the same volume.
COROLLARY XII.7
A triangular pyramid has volume equal to one-third of a triangular prism on the same
base of the same height.
Proof:
Let’s take the triangular prism T=ABCDEF.
We can rewrite T as the union of three pyramids: P1=ABCD, P2=BCDE, P3=CDEF.
We note that P1 and P2 have the same vertex C (so they have the same height) and
their bases are both halves of the parallelogram BADE (So by proposition I.34 triangle base
BDE is equal in content to the triangle base BAD).
Therefore by proposition XII.5 (Triangular pyramids having equal bases (in sense of
content) and equal height have equal volume) P1 and P2 have the same volume.
Now we can do exactly the same thing for P2 and P3. In this case both pyramids have
vertex D (so they have the same height) and their bases are halves of the parallelogram BCEF (So by
proposition I.34 triangle base BCE is equal in content to the triangle base CEF).
Therefore as before by proposition XII.5 we can say that P2 and P3 have the same volume.
And so we can conclude that P1=P2=P3 . And this means that if you want the volume of only one you can
take 1/3 of the volume of the starting prism.
NOTES ON THE PROOF OF XII.5
As in the other proves we notice that this proves is good if we make the supposition we did in the beginning
(that we are working over an Archimedean Field and that we see volume as a function). But if we want a
purely geometrical theory of volume we need to consider his method of exhaustion together with the
earlier notions of dissection and complementation.
Sandra Philps
16. April 2015
TO CONCLUDE
There were found letters in which Gauss and Gerling in 1844 discussed whether or not this method of
exhaustion is really necessary in Euclid’s proof of XII.5 .
This discussion goes so on that through Hilbert this question was included in the list of the 23 most
important problems facing mathematics in the XX century which Hilbert presented at the international
congress of mathematicians in 1900. In fact this problem is known as “Hilbert’s Third Problem”, which will
be discussed in the next section.