Area of a Parabolic Segment: Archimedes' Theorem
A parabolic segment is the region enclosed by a parabola and a chord connecting two points on the curve. More than two thousand years ago, Archimedes discovered a remarkably elegant way to calculate its area without using calculus.
According to Archimedes' theorem, the area \( S \) of a parabolic segment is equal to two-thirds of the area \( A \) of the rectangle formed between the chord and the tangent to the parabola that is parallel to the chord. $$ S = \frac{2}{3} \times A $$
The theorem can also be expressed in another, more famous form. The area of a parabolic segment is exactly four-thirds of the area of a particular triangle inscribed within it.
In this triangle, the base is the chord AB, while the third vertex C is the point where the tangent to the parabola is parallel to the chord.
If the inscribed triangle has area \( A \), then the area of the parabolic segment is:
$$ S = \frac{4}{3} \times A $$
This result is one of the masterpieces of ancient Greek mathematics and highlights the extraordinary geometric insight of Archimedes.
Note: Today, the exact area of a parabolic segment is usually computed using definite integrals. Archimedes arrived at the same result nearly two millennia before the development of calculus.
A Worked Example
Let's apply Archimedes' theorem to a specific parabola.
Consider the function:
$$ y = x^2 $$
Draw the chord AB joining the points A(-1, 1) and B(2, 4).
Our goal is to find the area enclosed by the parabola and the chord AB.
Following Archimedes' method, we first determine the tangent line that is parallel to the chord.
The chord AB lies on the line:
$$ y = x + 2 $$
Any line parallel to it can be written as:
$$ y = mx + q $$
The intersection points between the parabola and a line from this family satisfy:
$$ x^2 = mx + q $$
To find the tangent line, we require the equation to have exactly one solution. This occurs when the discriminant is zero.
Rewriting the equation:
$$ x^2 - mx - q = 0 $$
Its discriminant is:
$$ \Delta = b^2 - 4ac $$
Substituting \( a = 1 \), \( b = -m \), and \( c = -q \):
$$ \Delta = (-m)^2 - 4(1)(-q) = 0 $$
$$ m^2 + 4q = 0 $$
Solving for \( q \):
$$ q = -\frac{m^2}{4} $$
Therefore, the tangent line parallel to the chord has equation:
$$ y = mx - \frac{m^2}{4} $$
Since the chord has slope \( m = 1 \), we obtain:
$$ y = x - \frac{1}{4} $$
This is the tangent to the parabola \( y = x^2 \) that is parallel to AB.
Next, draw perpendicular segments from A and B to the tangent line. Together with the chord and the tangent, they form rectangle ABCD.
According to Archimedes' theorem, the area of the parabolic segment is two-thirds of the area of this rectangle.
Step 1: Find the Length of AB
The distance between A and B is:
$$ \overline{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} $$
$$ \overline{AB} = \sqrt{(2 - (-1))^2 + (4 - 1)^2} $$
$$ \overline{AB} = \sqrt{3^2 + 3^2} $$
$$ \overline{AB} = \sqrt{18} $$
Step 2: Find the Length of AC
AC is the perpendicular distance from point A to the tangent line.
$$ \overline{AC} = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} $$
The tangent line can be written as:
$$ -x + y + \frac{1}{4} = 0 $$
Therefore:
$$ a = -1,\quad b = 1,\quad c = \frac{1}{4} $$
Substituting these values:
$$ \overline{AC} = \frac{|-x + y + \frac{1}{4}|}{\sqrt{2}} $$
Using the coordinates of A(-1,1):
$$ \overline{AC} = \frac{|-(-1) + 1 + \frac{1}{4}|}{\sqrt{2}} $$
$$ \overline{AC} = \frac{\frac{9}{4}}{\sqrt{2}} $$
$$ \overline{AC} = \frac{9}{4\sqrt{2}} $$
Step 3: Calculate the Area of the Rectangle
$$ A = \overline{AB} \cdot \overline{AC} $$
$$ A = \sqrt{18} \cdot \frac{9}{4\sqrt{2}} $$
$$ A = 3\sqrt{2} \cdot \frac{9}{4\sqrt{2}} $$
$$ A = \frac{27}{4} $$
Step 4: Apply Archimedes' Theorem
$$ S = \frac{2}{3}A $$
$$ S = \frac{2}{3} \cdot \frac{27}{4} $$
$$ S = \frac{27}{6} $$
$$ S = 4.5 $$
The area of the parabolic segment is therefore:
$$ S = 4.5 $$
square units.
Verification Using Integration
We can verify the result by calculating the same area with integral calculus.
The region AEFB is made up of a rectangle and a triangle:
$$ Area(AEFB) = Area(AHF) + Area(AEFH) $$
$$ Area(AEFB) = \frac{3 \cdot 3}{2} + 3 \cdot 1 $$
$$ Area(AEFB) = 7.5 $$
The area under the parabola is:
$$ \int_{-1}^{2} x^2 \, dx $$
$$ \int_{-1}^{2} x^2 \, dx = \left[\frac{x^3}{3}\right]_{-1}^{2} $$
$$ = \frac{8}{3} - \left(-\frac{1}{3}\right) $$
$$ = 3 $$
Therefore, the area under the parabola is 3 square units.
The area of the parabolic segment is the difference between these two regions:
$$ S = Area(AEFB) - \int_{-1}^{2} x^2 \, dx $$
$$ S = 7.5 - 3 $$
$$ S = 4.5 $$
The result agrees perfectly with the value obtained using Archimedes' theorem.
Proof
Archimedes' original proof is one of the most celebrated achievements in the history of mathematics.
Consider a parabolic segment and the triangle inscribed within it, whose base is the chord AB and whose third vertex lies at the point where the tangent is parallel to the chord.
To establish the theorem, Archimedes used what is now called the method of exhaustion, a technique that anticipated many of the ideas later formalized in integral calculus.
He began with the largest inscribed triangle and then repeatedly constructed new triangles inside the remaining parabolic segments. Each new generation of triangles contributed a fixed fraction of the area contributed by the previous one.
As a result, the areas form an infinite geometric series. Summing this series shows that the total area of the parabolic segment is exactly:
$$ \frac{4}{3} $$
times the area of the original inscribed triangle.
Therefore:
$$ S = \frac{4}{3}A $$
This classical result is known as the quadrature of the parabola, one of the greatest achievements of ancient geometry.
