Why do we discuss the area inside a circle again? Why we shouldn’t confound the notions of intuition and rigor? Do calculus books, even today, still resort to circular reasoning? This paper is an attempt to elucidate these questions 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, which is 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.
In the following we present a brief review of the proofs of area inside a circle. A typical rigorous proof requires knowledge of integral calculus, seeing example  . But even in these proofs presented by calculus books, seeing the example  , the authors resort to circular reasoning. To prove the area inside a circle, they set up the integral followed by trigonometric substitution which requires knowing that the derivative of is But this latter fact requires
proving that For this proof, they resort to a geometric
argument, bounding the area of a sector of a unit circle between the areas of two triangles and showing that . They then apply the Squeeze Theorem. But for computation of the sector’s area, they resort to a standard
formula, which is based on knowing the area of a circle. So they
prove the area by assuming the area. This is obviously circular argumentation! For an excellent critique of this method we can see  .
There are also a number of intuitive proofs intended to provide an insight to the derivation of the area with just the knowledge of geometry and limits. In this short piece we begin by proving a preliminary result showing that
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
aside from the aforementioned limit, the function itself plays an
important role not only in mathematics but also in other fields of science such as physics and engineering.
Consider a circle of radius 1 which is centered at the origin, as shown in Figure 1; see  .
Figure 1. A unit circle in standard position.
Theorem 1. Let be an angle measured in radians. Then,
Proof. Since the magnitude of equals the length of the arc it subtends and
since we have that or This establishes a
lower bound. To establish the upper bound, we can show that This can be proved by the standard method of estimating an arc length of a rectifiable curve by the linear approximation of the lengths of the chords it subtends through partitioning. The result follows by applying the triangle inequality in each partition in  . Noting that we get
But So it follows that,
Combining this result with the previous lower bound gives,
Letting in the last expression completes the upper bound, resulting in
Finally, applying the Squeeze Theorem we get,
An interesting question related to the foregoing bounding of the angle is that if we define the derivatives of trigonometric functions of analytically (i.e., by infinite series of complex numbers or solutions of differential equations), can we arrive at the bounding of the angle? The next theorem follows.
Theorem 2. If and then
Proof. Let than and giving for
Therefore, This gives a lower bound for .
To find an upper bound for let
and giving for
Therefore, This gives an upper bound for . Combining the above results we get
Theorem 3. Area inside a circle of radius is
Proof. Consider a circle of radius centered at the origin in Figure 2.
Partition the circle into equal slices and consider a slice with central angle
radians. We know that the area of a triangle is one-half times 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 becomes
Because there are inscribed triangles in the circle, the total area of all these
triangles would be As we increase the number of slices
by increasing the 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 ,
as So, using (1), and since as the area of the circle
Figure 2. A circle of radius R with n equal slices.
1Area stretching is a result from geometry stating that if we stretch a region in the coordinate plane vertically by a factor of and horizontally by a factor of then its area will stretch by the factor
An intuitive and interesting method of proving the area inside a circle which requires area stretching1 and mapping from an annulus to a trapezoid is discussed in  . The subtle point in this method, as expressed in  , is that it assumes as evident the area preservation from a circular to a simply connected region. For a discussion of transformation of different regions by using complex variable method, we can see  .
There are many other intuitive approaches also, some of which involve slicing or opening up a circle. Below we offer a simple intuitive proof which is not based on area stretching, but assumes area preservation under mappings. Consider two concentric circles with radii and and corresponding areas and . Cut the annulus open in the shape of a right angle trapezoid as in Figure 3.
We can see that the area of the annulus equals the area of the trapezoid. So, , or
We can choose as small as we please and so, in particular, if we let approach 0, the area of the inner circle approaches 0 and we get,
Thus Note that shrinking to 0, shrinks the trapezoid to the right triangle whose area is
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 is
Proof. Consider a circle of radius centered at the origin and partition it
into equal sectors, each having central angle and the corresponding arc length Assume the area of a sector is If we stretch the radius
by a factor of we create a circle with radius So, the corre-
sponding streched sector will have an arc length equal to and the its area
Figure 3. An annulus of a circle cut open in the shape of a trapezoid.
If we connect the center to the point which is the midpoint of the triangles and become right angle congruent triangles with right angles at the point As a result, central angles and
will each equal Obviously triangles and are also congruent
having right angles at the point To calculate the area of the trapezoid, we
note that its larger base has length and its smaller base has length The height of the trapezoid is:
Therefore the area of the trapezoid becomes:
Figure 4. A sector of a circle stretched by a factor k > 1 to an enhanced sector.
Setting , gives,
or This approximation can be improved by increasing
Now, multiplying both sides of the above by gives:
Since there are exactly identical sectors in the circle of radius its area
becomes Therefore, Now, taking the limit of
both sides as and applying our earlier result (1) and the fact that as , we get:
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.