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Area in the plane

We consider areas of plane sets bounded by closed curves. In the more general cases, the concept of area becomes theoretically very difficult.

The area of a planar set is defined by reducing to the areas of simpler sets. The area cannot be "calculated", unless we first have a definition of "area" (although this is common practice in school mathematics).

Starting point

The area of a rectangle
The area of a rectangle is base $$\times$$ height: $A=ab.$

Definition: Area of a Parallelogram

The area of a parallelogram is base $$\times$$ height: $A=ah.$

Definition: Area of a triangle

The area of a triangle is (by definition) $A=\frac{1}{2}ah.$

Polygon

A (simple) polygon is a plane set bounded by a closed curve that consists of a finite number of line segments without self-intersections.

Definition: Area of a polygon

The area of a polygon is defined by dividing it into a finite number of triangles (called a triangulation of the polygon) and adding the areas of these triangles.

Theorem.

The sum of the areas of  triangles in a triangulation of a polygon is the same for all triangulations.

General case

For a plane set $$\color{red} D$$ bounded by a closed curve we can construct inner polygons $$\color{blue}P_i$$ and outer polygons $$P_o$$: $$\color{blue}P_i\color{black} \subset \color{red}D\color{black}\subset P_o$$.

A bounded set $$D$$ has an area if for every $$\varepsilon >0$$ there is an inner polygon $$P_i$$ and an outer polygon $$P_o$$, whose areas differ by less than $$\varepsilon$$: $A(P_o)-A(P_i)<\varepsilon.$ This implies that between all areas $$A(P_i)$$ and $$A(P_o)$$ there is a unique real number $$A(D)$$, which is (by definition) the area of $$D$$.

A surprise: The condition that $$D$$ is bounded by a closed curve (without self-intersections) does not guarantee that it has an area! Reason: The boundary curve can be so "wiggly", that it has positive "area". The first such example was constucted by [W.F. Osgood, 1903]:

Wikipedia: Osgood curve

Example

Derive the formula $$A=\pi R^2$$ for a circle with radius $$R$$ by choosing regular inscrided and circumscribed $$n$$-gons as inner and outer polygons, and let $$n\to\infty$$.

The solution is a voluntary exercise, where you need the limit $\lim_{x\to 0}\frac{\sin x}{x} = 1.$ Hint: Show that the inscribed and circumscribed areas are $\pi R^2\frac{\sin (2\pi/n)}{2\pi/n} \ \text{ and }\ \pi R^2\frac{\tan \pi/n}{\pi/n}.$