Area Inside A Circle: Intuitive and Rigorous Proofs

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by  M. Vali Siadat

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AREA INSIDE THE CIRCLE:

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INTUITIVE AND RIGOROUS
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1 V. Siadat
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Richard J. Daley College
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Chicago, IL 60652, USA
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AREA INSIDE THE CIRCLE 1
Abstract:
Inthis article I conduct ashort review of theproofs of thearea
inside a circle. These include intuitive as well as rigorous analytic proofs.
This discussion is important not just from mathematical view point but also
because pedagogically the calculus books still today use circular reasoning
to prove the area inside a circle (also that of an ellipse) on this important
historicaltopic,firstillustratedbyArchimedes. Iofferaninnovativeapproach,
through introduction of a theorem, which will lead to proving thearea inside
a circle avoiding circular argumentation.
Keywords: Area, circle, ellipse, circular reasoning, intuitive proof, rig-
orous proof.
ACKNOWLEDGMENTS
I wish to thank Professor Cyrill Oseledets of Richard J. Daley College
for his review of the initial manuscript and making helpful suggestions
to improve it.
2 V. Siadat
1 INTRODUCTION
Why areainsideacircleagain? Why weshouldn’tconfoundthe notions
of intuition and rigor? Do calculus books, even today, still resort to
circularreasoning? Thispaperisanattempttoelucidatethesequestions
by walking the reader through the path of intuitive to solid analytical
reasoning, pointing out the gaps that often occur, on the proof of this
ancient and well known problem, first illustrated by Archimedes. The
motivation behind writing of this piece was to engage the reader in
further thinking about mathematical proofs and the level of rigor at
which they are presented.
Inthe followingwe presenta briefreview ofthe proofsofareainside
acircle. Atypicalrigorousproofrequiresknowledgeofintegralcalculus,
seeforexample[5]. Butevenintheseproofspresentedbycalculusbooks,
seeforexample[2],theauthorsresorttocircularreasoning. Toprovethe
area inside a circle, they set up the integral 1√1 x2dx followed by
0 −
trigonometricsubstitutionwhichrequiresknowingthatthederivativeof
R
sinθ
sinθ is cosθ. But this latter fact requiresprovingthat limθ→0 =1.
θ
Forthisproof,theyresorttoageometricargument,boundingtheareaof
a sector of a unit circle between the areas of two triangles and showing
that sinθ < θ < tanθ. They then apply the Squeeze Theorem. But
for computation of the sector’s area, they resort to a standard formula,
A= 1θ, which is based on knowing the area of a circle. So, they prove
2
theareabyassumingthearea. Thisisobviouslycircularargumentation!
Foranexcellentcritiqueofthismethodsee[4]. Therearealsoanumber
ofintuitiveproofsintendedtoprovideaninsighttothederivationofthe
areawith just the knowledgeof geometryandlimits. In this shortpiece
sinθ
webeginbyprovingapreliminaryresultshowingthatlimθ→0 =1,
θ
without, a priori, assuming the area of a sector. This limit is central
to the proof of the derivatives of trigonometric functions. We note that
sinθ
aside from the aforementioned limit, the function itself plays an
θ
important role not only in mathematics but in other fields of science
such as physics and engineering.
AREA INSIDE THE CIRCLE 3
2 PROOFS
Consider a circle of radius 1, centered at the origin, as shown in Fig. 1;
see [6].
Figure 1.
π
Theorem 1. Let 0<θ < be an angle measured in radians. Then,
2
sinθ
lim =1.
θ→0 θ
Proof. Since the magnitude ofθ equalsthe lengthofthe arcit subtends
θ
and since sinθ < AD, we have that sinθ < θ, or 1 < . This estab-
sinθ
θ
lishes a lower bound for . To show the upper bound, observe that
sinθ
bythetriangleinequality,θ <sinθ+BD.Thiscanbeestablishedbythe
4 V. Siadat
standardmethodofestimatinganarclengthofarectifiablecurvebythe
linear approximation of the lengths of the chords it subtends through
partitioning. The result follows by applying the triangle inequality in
each partition; see [4]. Noting that sinθ <tanθ we get θ <tanθ+BD
But BD =1 cosθ. So,
−
θ 1 1 cosθ
< + −
sinθ cosθ sinθ
and
1 cosθ (1 cosθ)2 1 cosθ
− = − = −
sinθ s 1 cos2θ 1+cosθ
− r
Combining this result with the previous lower bound gives,
θ 1 1 cosθ
1< < + −
sinθ cosθ 1+cosθ
r
Letting θ 0 in the last expression completes the upper bound, re-
→ θ
sulting in 1 1. Finally, applying the Squeeze Theorem we
≤ sinθ ≤
get,
sinθ
lim =1 (1)
θ→0 θ
Aninterestingquestionrelatedtotheforegoingboundingoftheangle
θ is that if we define the derivatives of trigonometric functions of θ
analytically (i.e., by infinite series or complex numbers or solutions of
differential equations),can we arriveat the bounding of the angle? The
next theorem follows.
Theorem 2. If (sinθ)′ =cosθ, and (tanθ)′ =sec2θ, then
sinθ <θ <tanθ.
′
Proof. Let f(θ) = sin(θ) θ Than f(0)= 0 and f (θ) = cosθ 1 0,
− − ≤
′
giving f (θ)<0 for θ >0.
Hence,
θ
′
f(θ)= f (t)dt<0.
Z0
Therefore, sinθ <θ. This gives a lower bound for θ.
AREA INSIDE THE CIRCLE 5
To find an upper bound for θ, let g(θ)=tanθ θ.
−
g(0)=0 and g′(θ)=sec2θ 1 0, giving g′(θ)>0 for θ >0.
− ≥
Hence,
θ
′
g(θ)= g (t)dt>0.
Z0
Therefore, θ < tanθ. This gives an upper bound for θ. Combining
the above results we get sinθ θ tanθ.
≤ ≤
Theorem 3. Area inside a circle of radius R is πR2.
Proof. Consider a circle of radius R centered at the origin in Fig. 2.
Figure 2.
Partitionthecircleintonequalslicesandconsideraslicewithcentral
6 V. Siadat
angle 2π radians. We know that the area of a triangle is one-half times
n
the product of two of its sides times the sine of the angle between the
two sides. So, the area of the triangle subtended by the central angle
2π becomes A = 1R2sin(2π). Because there are n inscribed triangles
n 2 n
in the circle, the total area of all these triangles would be A =
total
n1R2sin(2π). As we increase the number of slices by increasing n, the
2 n
sum of the areas of the inscribed triangles get closer to the area of the
circle. To get the area of the circle, we need to find the limit of A ,
total
as n . So, using (1), and since 2π 0, as n , the area of the
n
→ ∞ → → ∞
circle becomes:
sin2π
Acircle =limn→∞ n21R2sin(2nπ)=πR2(limn→∞ 2πn )=πR2.
n
Hence, A =πR2.
circle
An intuitive and interesting method of proving the area inside a
circle which requires area stretching 1 and mapping from an annulus
to a trapezoid is discussed in [1]. The subtle point in this method, as
expressedin[1],isthatitassumesasevidenttheareapreservationfroma
circulartoasimplyconnectedregion. Foradiscussionoftransformation
of different regions using complex variable method, see [3].
There are many other intuitive approaches also, some of which in-
volve slicing or opening up a circle. Below we offer a simple intuitive
proof which is not based on area stretching, but assumes area preserva-
tionundermappings. Considertwoconcentriccircleswithradiir andR
andcorrespondingareasAr andAR. Cuttheannulusopenintheshape
of a right angle trapezoid ABCD as in Fig. 3.
1Area stretching is a result from geometry stating that if we stretch a region in
the coordinate plane vertically by a factor of k > 0 and horizontally by a factor of
l>0,thenitsareawillstretchbythefactorkl.
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Figure 3.
We canseethatthe areaofthe annulusequalsthe areaofthe trape-
zoid. So, A =A , or
annulus trapezoid
1
AR Ar = (2πR+2πr)(R r)=πR2 πr2.
− 2 − −
We can choose r > 0 as small as we please and so, in particular, if we
let r approach 0, the area of the inner circle approaches 0 and we get,
AR 0=πR2 0. Thus AR =πR2. Note that shrinking r to 0, shrinks
− −
the trapezoid to the right triangle ABC, whose area is 1(2πR)(R
2 −
0) = πR2 = AR. In the following we present an analytic proof of the
area inside a circle using area stretching, which does not assume area
preserving mapping of regions.
Theorem 4. Area inside a circle of radius r is πr2.
Proof. Consideracircleofradiusrcenteredattheoriginandpartitionit
2π
intonequalsectors,eachhavingcentralangle ,andthecorresponding
n
2π
arclength r.Assumetheareaofasectoriscn.If westretchtheradius
n
r by a factor of k > 1, we create a circle with radius R = kr. So, the
2π
correspondingstrechedsectorwillhaveanarclengthequalto kr and
n
theitsareawillbeincreasedbyafactorofk2tok2cn;seeFig. 4 Nowthe
area between the two sectors is Abetweensectors =k2cn cn =cn[k2 1]
− −
8 V. Siadat
which is approximately equal to the area of the trapezoid ABCD, in
Fig. 4.
Figure 4.
If we connect the center O to the point F which is the midpoint
of BC, the triangles ∆OBF and ∆OCF become right angle congruent
triangles with right angles at the point F. As a result, central angles
π
∠AOE and ∠DOE will each equal . Obviously triangles ∆OAE and
n
∆ODE are also congruent having right angles at the point E. To cal-
culate the area of the trapezoid, we note that its larger base has length
π π
BC = 2Rsin , and its smaller base has length AD = 2rsin . The
n n
AREA INSIDE THE CIRCLE 9
height of the trapezoid is:
π π
EF =Rcos rcos
n − n
Therefore the area of the trapezoid becomes:
1 π π π π
A = 2Rsin +2rsin Rcos rcos
trapezoid
2 n n · n − n
hπ π i h i
= sin cos [R2 r2]
n n −
π π
= sin cos r2[k2 1].
n n · −
Setting A A , gives,
betweensectors trapezoid
≈
π π
cn[k2 1] sin cos r2[k2 1],
− ≈ n n · −
π π
orcn sin cos r2.Thisapproximationcanbeimprovedbyincreas-
≈ n n·
ing n. Now, multiplying both sides of the above by n gives:
π π
ncn nsin cos r2
≈ n n ·
Since there are exactly n identical sectors in the circle of radius r, its
π π
areabecomes c=ncn. Therefore,c nsin cos r2. Now,taking the
≈ n n·
limit of both sides as n , and applying our earlier result (1) and
→ ∞
the fact that cosθ 1, as θ 0,we get:
→ →
π π
c = lim nsin cos r2
n→∞ n n ·
π
sin π
= lim π n cos r2
n→∞ · π n ·
n
π
sin π
= πr2( lim n) ( lim cos )
n→∞ π · n→∞ n
n
= πr2
Hence, c=πr2.