Area plays an important role in modern mathematics. In addition to its obvious importance in geometry and calculus, area is related to the definition of determinants in linear algebra, and is a basic property of surfaces in differential geometry. In analysis, the area of a subset of the plane is defined using Lebesgue measure, though not every subset is measurable. In general, area in higher mathematics is seen as a special case of volume for two-dimensional regions.
Area can be defined through the use of axioms, defining it as a function of a collection of certain plane figures to the set of real numbers. It can be proved that such a function exists.
An approach to defining what is meant by "area" is through axioms. "Area" can be defined as a function from a collection M of special kind of plane figures (termed measurable sets) to the set of real numbers which satisfies the following properties:
For all S in M, a(S) ≥ 0.
If S and T are in M then so are S ∪ T and S ∩ T, and also a(S∪T) = a(S) + a(T) − a(S∩T).
If S and T are in M with S ⊆ T then T − S is in M and a(T−S) = a(T) − a(S).
If a set S is in M and S is congruent to T then T is also in M and a(S) = a(T).
Every rectangle R is in M. If the rectangle has length h and breadth k then a(R) = hk.
Let Q be a set enclosed between two step regions S and T. A step region is formed from a finite union of adjacent rectangles resting on a common base, i.e. S ⊆ Q ⊆ T. If there is a unique number c such that a(S) ≤ c ≤ a(T) for all such step regions S and T, then a(Q) = c.
It can be proved that such an area function actually exists.
Subsequently, Book I of Euclid's Elements dealt with equality of areas between two-dimensional figures. The mathematician Archimedes used the tools of Euclidean geometry to show that the area inside a circle is equal to that of a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius, in his book Measurement of a Circle. (The circumference is 2πr, and the area of a triangle is half the base times the height, yielding the area πr2 for the disk.) Archimedes approximated the value of π (and hence the area of a unit-radius circle) with his doubling method, in which he inscribed a regular triangle in a circle and noted its area, then doubled the number of sides to give a regular hexagon, then repeatedly doubled the number of sides as the polygon's area got closer and closer to that of the circle (and did the same with circumscribed polygons).
Heron (or Hero) of Alexandria found what is known as Heron's formula for the area of a triangle in terms of its sides, and a proof can be found in his book, Metrica, written around 60 CE. It has been suggested that Archimedes knew the formula over two centuries earlier, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.
The development of integral calculus in the late 17th century provided tools that could subsequently be used for computing more complicated areas, such as the area of an ellipse and the surface areas of various curved three-dimensional objects.
where when i=n-1, then i+1 is expressed as modulusn and so refers to 0.
The area of this rectangle is lw.
The most basic area formula is the formula for the area of a rectangle. Given a rectangle with length l and width w, the formula for the area is:
A = lw (rectangle).
That is, the area of the rectangle is the length multiplied by the width. As a special case, as l = w in the case of a square, the area of a square with side length s is given by the formula:
Most other simple formulas for area follow from the method of dissection.
This involves cutting a shape into pieces, whose areas must sum to the area of the original shape.
For an example, any parallelogram can be subdivided into a trapezoid and a right triangle, as shown in figure to the left. If the triangle is moved to the other side of the trapezoid, then the resulting figure is a rectangle. It follows that the area of the parallelogram is the same as the area of the rectangle:
A = bh (parallelogram).
Two equal triangles.
However, the same parallelogram can also be cut along a diagonal into two congruent triangles, as shown in the figure to the right. It follows that the area of each triangle is half the area of the parallelogram:
The formula for the area of a circle (more properly called the area enclosed by a circle or the area of a disk) is based on a similar method. Given a circle of radius r, it is possible to partition the circle into sectors, as shown in the figure to the right. Each sector is approximately triangular in shape, and the sectors can be rearranged to form an approximate parallelogram. The height of this parallelogram is r, and the width is half the circumference of the circle, or πr. Thus, the total area of the circle is πr2:
A = πr2 (circle).
Though the dissection used in this formula is only approximate, the error becomes smaller and smaller as the circle is partitioned into more and more sectors. The limit of the areas of the approximate parallelograms is exactly πr2, which is the area of the circle.
This argument is actually a simple application of the ideas of calculus. In ancient times, the method of exhaustion was used in a similar way to find the area of the circle, and this method is now recognized as a precursor to integral calculus. Using modern methods, the area of a circle can be computed using a definite integral:
Archimedes showed that the surface area of a sphere is exactly four times the area of a flat disk of the same radius, and the volume enclosed by the sphere is exactly 2/3 of the volume of a cylinder of the same height and radius.
Most basic formulas for surface area can be obtained by cutting surfaces and flattening them out. For example, if the side surface of a cylinder (or any prism) is cut lengthwise, the surface can be flattened out into a rectangle. Similarly, if a cut is made along the side of a cone, the side surface can be flattened out into a sector of a circle, and the resulting area computed.
where r is the radius of the sphere. As with the formula for the area of a circle, any derivation of this formula inherently uses methods similar to calculus.
Areas of 2-dimensional figures
A triangle: (where B is any side, and h is the distance from the line on which B lies to the other vertex of the triangle). This formula can be used if the height h is known. If the lengths of the three sides are known then Heron's formula can be used: where a, b, c are the sides of the triangle, and is half of its perimeter. If an angle and its two included sides are given, the area is where C is the given angle and a and b are its included sides. If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of . This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points (x1,y1), (x2,y2), and (x3,y3). The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use calculus to find the area.
A simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points: , where i is the number of grid points inside the polygon and b is the number of boundary points. This result is known as Pick's theorem.
Area in calculus
Integration can be thought of as measuring the area under a curve, defined by f(x), between two points (here a and b).
The area between two graphs can be evaluated by calculating the difference between the integrals of the two functions
The area between a positive-valued curve and the horizontal axis, measured between two values a and b (b is defined as the larger of the two values) on the horizontal axis, is given by the integral from a to b of the function that represents the curve:
To find the bounded area between two quadratic functions, we subtract one from the other to write the difference as
where f(x) is the quadratic upper bound and g(x) is the quadratic lower bound. Define the discriminant of f(x)-g(x) as
By simplifying the integral formula between the graphs of two functions (as given in the section above) and using Vieta's formula, we can obtain
The above remains valid if one of the bounding functions is linear instead of quadratic.
Surface area of 3-dimensional figures
Cone:, where r is the radius of the circular base, and h is the height. That can also be rewritten as  or where r is the radius and l is the slant height of the cone. is the base area while is the lateral surface area of the cone.
and equality holds if and only if the curve is a circle. Thus a circle has the largest area of any closed figure with a given perimeter.
At the other extreme, a figure with given perimeter L could have an arbitrarily small area, as illustrated by a rhombus that is "tipped over" arbitrarily far so that two of its angles are arbitrarily close to 0° and the other two are arbitrarily close to 180°.
For a circle, the ratio of the area to the circumference (the term for the perimeter of a circle) equals half the radiusr. This can be seen from the area formula πr2 and the circumference formula 2πr.
The area of a regular polygon is half its perimeter times the apothem (where the apothem is the distance from the center to the nearest point on any side).
Doubling the edge lengths of a polygon multiplies its area by four, which is two (the ratio of the new to the old side length) raised to the power of two (the dimension of the space the polygon resides in). But if the one-dimensional lengths of a fractal drawn in two dimensions are all doubled, the spatial content of the fractal scales by a power of two that is not necessarily an integer. This power is called the fractal dimension of the fractal.
There are an infinitude of lines that bisect the area of a triangle. Three of them are the medians of the triangle (which connect the sides' midpoints with the opposite vertices), and these are concurrent at the triangle's centroid; indeed, they are the only area bisectors that go through the centroid. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.
Any line through the midpoint of a parallelogram bisects the area.
All area bisectors of a circle or other ellipse go through the center, and any chords through the center bisect the area. In the case of a circle they are the diameters of the circle.