IT has beenoftenremarked that,though themethodof exhaustion exemplified in Euclid xn. 2 reallybrought theGreek geometersface to face with the infinitely great and the infinitely small, they never allowed themselves to use such conceptions. It is true that Antiphon, a sophist who is said to have often had disputes with Socrates, had stated* that, if one inscribed any regular polygon, say asquare, ina circle, then inscribed an octagon byconstructing isosceles triangles in the four segments, then inscribed isosceles triangles in the remaining eight segments, and so on, "until the whole area of the circle was by this means exhausted, a polygon would thusbe inscribedwhose sides,in consequence of their small- ness,would coincidewith the circumference of the circle." But as against this Simplicius remarks, and quotes Eudemus to the same
effect, that the inscribed polygon will never coincide with the circumference of the circle, even though it be possible to carry the division of the area to infinity, and to suppose that it would
istosetasidea geometrical principlewhich lays down that magni- tudes are divisibleadinfinitum^. The time had, in fact,riot come forthe acceptance of Antiphon's idea, and, perhaps as theresult of the dialecticdisputes towhich the notion of the infinite gave rise, the Greek geometers shrank from the use of such expressions as infinitelygreatandinfinitelysmalland substitutedtheidea of things greater orlessthananyassigned magnitude. Thus,as HankelsaysJ, they neversaid that acircle is a polygonwithan ir finitenumberof
* Bretschneider,p. 101.
t Bretschneider,p.102.
Hankel, Zur Geschichte der Mathematik im Alterthum und Mittelalter, p. 123.
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS, cxliii infinitely smallsides
; they alwaysstood stillbefore theabyssof the infinite and never ventured to overstep the bounds of clear con- ceptions. Theynever spoke of an infinitely closeapproximation or a limiting value of the sum of a series extending to an infinite
number of terms. Yet they must have arrived
practically atsuch aconception, e.g., inthe case of the proposition that circles are to one anotherasthe squareson their diameters, theymust have been inthe firstinstance led to inferthetruth of the proposition bythe ideathat the circle could be regarded as the limit of an inscribed regular polygon with an indefinitely increased number of corre- spondingly small sides. They did not, however, rest satisfied with suchan inference; theystrove afteran irrefragable proof, and this, from the nature of the case, could only be an indirect one. Ac- cordingly we always find, in proofs by the method of exhaustion, a demonstration that an impossibility is involved by any other assumption than that which the proposition maintains. Moreover
this stringent verification, by means of a double reductio ad ab- surdum, is repeated inevery individual instance of the use of the method of exhaustion; there is no attempt to establish, in lieu of this part of the proof, any general propositions which could be simply quoted in any particular case.
The above general characteristics of the Greek method of exhaustion are equally present in the extensions of the method found in Archimedes. To illustrate this, it will be convenient, before passing to thecases where he performs genuine integrations, tomentionhis geometrical proof of the property that the area of a parabolicsegment is four-thirds of the trianglewith the same base and vertex. Here Archimedes exhauststhe parabola by continually drawing, in each segment left over, a triangle with the same base and vertex as the segment. If A be the area of the triangle so inscribed intheoriginalsegment, theprocess gives a seriesof areas
A, {A, (D'A,...
and theareaof thesegmentis reallythe sumoftheinfinite series
ButArchimedesdoes not expressit inthisway. He first proves
that, ifA19 A2,...Anbe anynumberofterms of such a series, sothat
Al= 442, AZ-4tASJ ... , then
Al+ At +A.,+ ...+ An+ $An= 4lf or A {i + i+ a)2+ ... + ay-' + jar1}=$A.
Cxliv INTRODUCTION.
Having obtained this result, we should nowadays suppose n to increase indefinitely and should infer at once that (j-)*"1 becomes indefinitely small,andthatthe limit ofthesum on theleft-hand side is the area of the parabolicsegment, whichmust thereforebeequal to $A. Archimedes does not avow that he inferred the result in this way; he merely states that the area of the segment is equal to A9 and thenverifies it in the orthodox manner byproving that
it cannot beeither greater orlessthan A.
I pass now to the extensions by Archimedes of the method
of exhaustion whichare theimmediate subject of this chapter. It will be noticed, as an essential feature of all of them, that Archimedes takes both.an inscribed figure and a circumscribed figure in relation to thecurve or surface ofwhichheis
investigating the areaor the solid content, and then, as it were, compresses the two figures into one so that they coincide with one another and with the curvilinear figure to be measured; but again it must be understood that he does not describehismethod in thiswayor
sayat anytime that the given curve or surfaceisthelimitingform of the circumscribed or inscribed figure. I will take the cases in the order inwhich theycome in the text of thisbook.
1. Surface ofasphere or spherical segment.
The first stepis toprove (On theSphereand CylinderI. 21, 22) that, if in a circle or a segment of a circle there be inscribed polygons, whose sides AB, BC, CD, ... are all equal, as shown
in the respective figures, then (a) forthe circle
(BB
1
+CC'+ .
..) : AA' = A'B: BA,
(b) forthe segment
(BB' +CC'+...+KK'+LM) :AM=A'B:BA.
Next it is proved that, if the polygons revolve about the diameter AA', the surface described by the equal sides of the polygon in a complete revolution is [i. 24, 35]
(a) equaltoacircle withradius<JAB (BBf+ CC1+ ... + YY') or (b) equal toacirclewith radius J~ATT(B]fr
+C~C7+... +LAf~).
Therefore, by means of the above proportions, the surfaces described by the equal sides are seen to be equal to
ARCHIMEDES'ANTICIPATIONS OF THE INTEGRAL CALCULUS, cxlv (a) acircle with radiusJAA'.A'JJ,
and (b) acirclewithradiusJAM.All
theyare therefore respectively [i. 25, 37] less tJian (a) acirclewith radiusAA\
(b) acirclewithradiusAL.
Archimedesnowproceeds totake polygons circumscribedto the circleorsegment ofacircle(supposed in thiscase tobe less than a semicircle) sothat their sidesareparallel to those of the inscribed polygons before mentioned (cf. the
figures on pp. 38, 51); and he proves bylike steps[i. 30,40] that,if thepolygons revolveabout the diameteras before, thesurfaces described bythe equal sides during a complete revolutionare greaterthan the samecircles respectively.
Lastly, having proved these results for the inscribed and circumscribed figuresrespectively,Archimedes concludesand proves
[i. 33, 42, 43]that the surface of the sphere orthe segment of the sphere isequal tothefirstorthesecondofthe circles respectively.
In order to see theeffect of the successive steps, let us express the several resultsbymeansof trigonometry. If, in thefigureson pp. 33, 47 respectively, we suppose 4n to be thenumberof sidesin the polygon inscribed in thecircle and 2>ithe numberof the equal sides in the polygon inscribed in the segment, while in the latter case the angle AOL is denoted by a, the proportions given above are respectively equivalent to the formulae*
IT . 2?T _ 7T 7T
sin +sin -- 4- ...+sin(2n. 1) =cot ,
2n '2n ^ '"2n 4?i
- f . a . 2a .
, ,.a)
2 {sin +sin + ... +sin (n 1)-
[ +sina
. I n n n] a
and ,
=cot-^ .
1 cosa 2?i
Thusthetwoproportionsgive in fact a summation oftheseries sin6+sin20+...+sin (n 1
)
bothgenerally where nO isequal to anyangle a less than TT, andin theparticular case wheren iseven and =
TT/H.
Again, theareas of the circles which are equal to the surfaces described by the revolution of the equal sides of the inscribed
* These formulaeare taken,withaslightmodification,fromLoria,IIperiodo aureodellageometriagreca, p.108.
H. A. k
Cxlvi INTRODUCTION.
polygons are respectively (if a be the radius of the great circle of the sphere)
o
and
If ( T 2?T . -iv 71") A 2 V
sin - xsm _-+sm +...+sin(2n-1
)h k or bira cos--
4n [ 2n 2n v ' 2n) 4?i,
o ~ . a I", ( . a . 2a . . nx a) "1
a2. 2sm-^- 2 ^sin- +sm + ... 4-sin(n-l)- x+sma ,
2n\_ [ n n ^ 'n) J
a
or Tra2.2 cosT-(1 cosa).
2nv '
Theareas of thecircleswhichare equal to thesurfaces described bythe equal sidesof the circumscribed polygons are obtained from the areas ofthe circlesjustgivenby dividingthem by cos27r/4?iand cos8a/2n respectively.
Thusthe resultsobtainedby Archimedes are the same aswould be obtained bytaking the limiting valueof theabove trigonometri- cal expressions when nis indefinitely increased, andwhen therefore cos7r/4?iand cosa/2nare both unity.
Butthefirst expressions forthe areas ofthe circles are (when n
is indefinitely increased) exactly what we represent by the integrals
Cir
ira2.i I sin dO, or 47ra
2 - ,
Jo
and no,2. / 2 sin dO, or 2ira2(I-cos
a).
7o
Thus Archimedes' procedure is the equivalent of a genuine integration in each case.
2. Volumeofa sphere orasectorofasphere.
The methoddoes not needto beseparatelysetoutin detail here, because it depends directly011 the preceding case. The investiga-
tionproceeds concurrently withthat ofthe surface of a sphereora segment ofasphere. Thesame inscribedand circumscribedfigures are used, thesector of a sphere being of course compared with the
solidfiguremadeup of thefigure inscribed or circumscribed tothe segment and ofthe conewhichhasthe same baseasthatfigure and has its vertex at the centre of the sphere. It is then proved, (1) forthefigure inscribed or circumscribed to the sphere, that its volume isequal tothat of acone with base equal to the surface of the figure and height equal to theperpendicularfrom thecentre of the sphere on anyone of the equal sides of the revolving polygon, (2) forthefigure inscribed or circumscribed to the sector, that the
ARCHIMEDES'ANTICIPATIONS OF THE INTEGRAL CALCULUS, cxlvii volume isequal tothatof a cone with base equal to the surface of the portion of the figurewhich isinscribed or circumscribed to the segmentofthe sphere included in thesectorand whoseheightis the perpendicular from the centre on one of the equal sides of the polygon.
Thus, when the inscribed and circumscribed
figures are, so to speak, compressed into one, the takingof the limit is practically the same thing in this case as in the case of the surfaces, the resulting volumes being simply the before-mentioned surfaces multiplied in each case by \a.
3. Area ofan ellipse.
This case again is not strictly in point here, because it does not exhibit any of the peculiarities of Archimedes' extensions of the method of exhaustion. That method is, in fact, applied in the same manner, mutatis mutandis, as in Eucl. XH. 2. There
is no simultaneous use of inscribed and circumscribed figures,but only the simple exhaustion of the ellipse and auxiliary circle by increasing to any desired extent the number of sides in polygons inscribed to each(On ConoidsandSpheroids, Prop. 4).
4. Volume ofa segment ofa paraboloid ofrevolution.
Archimedesfirststates,as aLemma,a resultproved incidentally inaproposition ofanother treatise(On Spirals, Prop. 11), viz.that,
ifthere be ntermsofan arithmetical progressionh, '2h, 37*, ...,then h+2k + 3h+... -fnh>
\tfh\
and h +2h+'3h+ ...+ (w-1)A <\ri-h}
(a''
Next he inscribes and circumscribes to the segment of the paraboloid figuresmade upof small cylinders(asshown inthefigure of On Conoids and Spheroids, Props. 21, 22) whose axes lie along the axis of the segment and divide it into any number of equal parts. If c is the length of the axis AD of the segment, and if
there are ncylinders inthe circumscribed figure and their axes are eachoflengthh, so thatc=nh,Archimedesprovesthat
cylinderCE _ u-h
(1) inscribedfig. h-4-2h+ 3h+... 4-(/t 1)7*
>2, by the Lemma,
cylinder (7#_ _ n'2h
^' circumscribedfig.
~
h+2h +37*+...+nh
2
Cxlviii INTRODUCTION.
Meantimeithasbeen proved[Props. 19, 20] that, by increasing
n sufficiently, the inscribed and circumscribed figure can be made
to differ by less than any assignable volume. It is accordingly concludedand provedbythe usual rigorousmethod that
(cylinder CE) ='2(segment), so that (segment AJ2C)= (coneAJ3C).
The proof is therefore equivalent to the assertion, that if h is indefinitelydiminished andnindefinitely increased,whilenhremains equalto c,
limit ofh \h+ 2A+37*+ ... +(n-1)h] -ic2
;
thatis, inour notation,
i'C
I xdx -ic2. Jo
Thus the method is essentially the same as ours when we
express the volume ofthesegment ofthe paraboloidinthe form
K
.'0
where Kisa constant, which does not appear inArchimedes' result for the reason that he does not give the actual content of the segmentof the paraboloid but only the ratio which it bears to the circumscribedcylinder.
5. Volume ofa segment of a hyperbola id of revolution.
Thefirststep in this case istoprove[On ConoidsandSpheroids, Prop. 2] that, ifthere be a seriesofnterms,
ah+h\ a.2h+ (2/<)2, a.37*+ (37*)2, ... a.nh +(w7*)2, and if (ah+tf)+ \a.2h+(27t)2J + ... + !. nh +(nlif] =SM,
then n{a.nh +(nh)*}/Mn<(a+nh)i
( +
/V2 3;)
and n[a.nh +(n/t)*J/.?_,>( +nh) ?+ -
Next [Props. 25, 26] Archimedes draws inscribed and circum- scribed figuresmade up of cylinders as before(figureon p. 137),and
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS, cxlix proves that, ifADisdividedinto nequalparts of length h, so that
nh -AD, and ifAA'=a, then
cylinderEB' _ n [a.nh+(nhf\
inscribed tigure
~~
_!
fa nh\
. cylinder KB' n{a.nh +(nhf\
and .-J-----7 nr- - -- c.
circumscribed
hg. on
a nh
Theconclusion, arrived at in thesame manneras before, is that cylinderEB' . ..//a nh\
A !>/=
(a +n'0/(^+ "^~)'
segmentABB ^ ' \2 3/
This isthe sameas sayingthat, if nk=b, andifh beindefinitely diminished while )iis
indefinitely increased, limit ofn(ab+b*)/SH=
(a+ b)/(^
-t-
|), r -4. f& v ifa 'A
or limit of- ^ =6-
(^-
+ -
) .
Now >S'H-= re(//-f '2/i+ ... +nh)+ J/r-f
(2/*)
2+...+(nh)-\,
so that A,^-a/*,
(A+2A-f...+nh
)+h\h2+(2h)2+...+(nhf\. The limit ofthe lastexpressionis what we should writeas
f h
I (a.v+
ar)rf.r, JQ
which isequal to 62(-+
j;
and Archimedes has given the equivalentof this integration.
6. Volume of a segment ofa spheroid.
Archimedesdoes not heregive the equivalent ofthe integration
fb
(*-rf),
.'o
presumably because, with his method, it would have required yet another lemmacorresponding tothatinwhichthe results
(/?)above are established.
cl INTRODUCTION.
Supposethat, in the case ofasegmentlessthanhalfthespheroid (figureon p.
142), AA' =a, 07)=ic, AD =6; and let ADbe divided intonequalparts of lengthh.
The gnomonsmentioned in Props.29, 30arethen the differences betweenthe rectangle cb+lrandthe successive rectangles
andinthis casewe have the conclusions that (if Sn be the sum of
ntermsof theseries representing thelatter rectangles) cylinder EB' n(cb+b2) inscribed figure n(cb+b2)-*V
/t
. I/c -2b
cyUnderEB' n(cb+b*
circumscribed fig. n(cb+bs
)
i- .1 T ^ cylinderEB' . ,x//c 2/A and inthe limit J .inn/=-"(c+^)/ (-, +-,-).
segment J/?J5 x / \2 3/
Accordingly \ve havethe limittaken of theexpression
and theintegrationperformed isthe sameas that in thecase of the hyperboloid above, with csubstituted fora.
Archimedes discusses, as a separate case, the volume of half a spheroid [Props. 27, 28]. It differs from that just given in that c
vanishes andb=
<&, sothat it isnecessaryto find the limit of
(3/4)^f_.~..+(nilf and this is done by means of a
corollary to the lemma given on pp. 107 9 [OnSpirals, Prop. 10] whichproves that
A9+(2h)*+ ... + (nhf > \n(nh)\
and h2+(2A)2+...+ \(n~1)7/j2<\n(nh)\
Thelimit of course correspondstothe integral
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS, cli 7. Area ofa spiral.
(1) Archimedes finds the area bounded by the first complete turnof aspiralandtheinitialline by meansoftheproposition just quoted, viz.
A2+(2h)2+... +(nltf> \n(nhf,
hr+(2A)2+... + {(w-1)A}
8<Jn(nh)*.
He proves [Props. 21, 22, 23] that a figure consisting of similar sectorsof circles can be circumscribed about anyarc ofaspiralsuch that the areaofthe circumscribedfigure exceeds that of the spiral by less than anyassigned area,and alsothat a figure of the same kind can beinscribed such that the area of the spiral exceeds that oftheinscribed figureby less than anyassigned area. Then, lastly, he circumscribes andinscribesfigures of this kind[Prop. 24]; thus
e.g. in the circumscribed figure, if there are n similar sectors, the radii will be n lines forming an arithmetical progression, as A, 27i, 3//, ... nh, and nh will be equal to a, where a is the length inter- cepted 011the initial line bythespiral at the end of the first turn.
Since, then, similar sectors are toone anotheras the squareof their radii, andn times thesector of radiusnhorais equal to the circle with the same radius, thefirstof theaboveformulae proves that
(circumscribed fig.)>
J-?ra 2
.
A similar procedurefor the inscribed figure leads, by the useof the second formula, tothe result that
(inscribedfig.)<
J^rrr.
Theconclusion, arrived at in theusual manner, is that (area ofspiral) $7ra~\
and the proofisequivalent totaking the limit of
or of -*
[A
3+(-IKf+...+
[(
-
1)A}
2
a ],
which lastlimit weshould expressas
Clii INTRODUCTION.
[It is clear that this method of proof equally gives the area bounded bythespiraland any radius vector of length b not being greater than a
; for we have onlyto substitute trb/a for TT, and to
rememberthat inthis casenh-b. We thus obtainforthe area
fit
^ Px*dx, or
(2) To find the area bounded by an arc on any turn of the
spiral (not being greater than a complete turn) and the radii vectores to its extremities, of lengths b and c say, where c>/;, Archimedes uses the proposition that, if there be an arithmetic progression consisting of the terms
6, b+h, b+'2h, ... 6+(M-!)/, and if /? -62+
(6+h)~+(b+2h)'2+... + [b+(n-
1)7*J a
,
and
[0/iSpirals, Prop. 1 1 and note.]
TheninProp. 26 he circumscribesandinscribes figures consisting of similar sectors of circles, as before. There are n\ sectors in
each figureand thereforenradii
altogether, including both band c,
sothatwecan takethem tobe the terms ofthe arithmetic progres- sion given above, where
[b+(n 1)k\=c. It is thus proved, by means of the above inequalities, that
sector OB'C \b+(n-
1) A|
2
^ sectorOJfC circumscribed fig. {b+(n-l)h\b + \(n-l)/i^ inscr. fig.
and it isconcludedafter the usual mannerthat {b+ (n-})h}'2 _
spiral OUC
Remembering that n -1=(c-
&)//*, we see that the result is the
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS, cliii samethingasprovingthat, inthelimit,when nbecomesindefinitely greatand hindefinitely small,whileb+(n-
1)h=c,
limitofh[b*+(b+h)3+ ...4-
[b+(n-
2)h\*]
thatis, with ournotation,
=i
(c 3-
A f
(3) Archimedes works out separately [Prop. 25], by exactly thesamemethod, theparticular casewherethe areaisthatdescribed in any one complete turn of the spiral beginning from the initial line. This is equivalentto substituting (n I)aforband naforc,
whereais the radiusvector to theendofthe first complete turn of the spiral.
It will be observed that Archimedes does not use the result correspondingto
f x2dx- {' x2dx^ Cx"d.c.
Jo Jb JO
8. Area of a parabolic segment.
Ofthe twosolutions which Archimedesgives of the problem of squaring a parabolic segment, it is the mechanical solution which gives the equivalent of a genuine integration. In Props. 14, 15 of the Quadrature of the Parabola it is proved that, of two figures
inscribed andcircumscribed to the segment and consisting in each case of trapeziawhose parallel sides are diameters of the parabola, the inscribed figure is less, and the circumscribed figure greater, than one-third of a certain triangle (EqQ in the figureonp. 242).
Then inProp. 1Cwe have the usual process whicli is equivalent to taking the limit whenthe trapezia become infinite in number and theirbreadth infinitely small, and it isproved that
(area ofsegment)=
\AEqQ.
Theresult isthe equivalentofusing theequationofthe parabola referred to Qq as axis of xand the diameter through Q as axis of
2/, viz.
p\j= x(2a-
o?)f
which can, as shownon p.236, be obtainedfromProp.4,andfinding
/o ydx,