Net Area and Green's Theorem

Given the coordinates of the vertices of a triangle in the plane (with respect to an orthogonal xy coordinate system), what is the enclosed area of the triangle? Consider an arbitrary triangle such as the one shown below

We are given the coordinates of the vertices (x₁,y₁), (x₂,y₂), and (x₃,y₃), and we wish to determine the enclosed area? It's helpful to inscribe the triangle inside a square as shown below

Clearly the area inside the triangle is just the area of the enclosing rectangle minus the areas of the three surrounding right triangles. Thus we have

which answers our original question. Now, we could expand the products in the above expression and try to simplify it, but its easier to observe that the three exterior right triangles are each half of a rectangle (cut diagonally), and that those three rectangles fill up the overall rectangle, with two of them overlapping in one rectangular corner. Therefore, the area of the three exterior triangles is simply half of the difference between the overall rectangle and the overlapped rectangle. Subtracting this from the area of the overall rectangle immediately gives the area of the original triangle

In this same way it's easy to deduce that the area enclosed by a general quadrilateral can be expressed in terms of the coordinates of its vertices as

It's worth noting that, assuming all the vertices are in the (++) quadrant of the xy coordinate system (i.e., all the coordinates are positive), these formulas give the positive area only if the vertices are numbered clockwise around the perimeter. If they are counter-clockwise, the computed area is negative. Of course, a quadrilateral can have crossing edges, such that the vertices are clockwise around one region and counter-clockwise around the other. Thus, the computed area of a non-degenerate quadrilateral can vanish, as in the case of the quadrilateral shown in Figure 3.

It's interesting to consider the correspondence between the "net area" (as defined by the above formula) associated with four points in the plane and the squared intervals in geometries with non-positive-definite metrics.

Continuing with this same approach, we find that the "net area" enclosed by a general n-sided polygon is given by

with the understanding that x_n+1 = x₁ and y_n+1 = y₁. Taking this to the limit as n goes to infinity, the signed net area enclosed by any closed loop in the plane is

where the integral is evaluated once around the loop. This formula expresses the (net) area within any closed curve on the plane, such as the curve shown below.

Given a parametric representation x = x(s), y = y(s) of the curve, where s is a path parameter, we can write the net area formula as

For example, consider an ellipse with major radius R and minor radius r. Centered at the origin and oriented appropriately, the boundary of this ellipse is given by the parametric equations

where the parameter α is the angle that varies from 0 to 2π as we proceed around the loop (counter clockwise, so our area will be negative). The derivatives of x and y are

so the enclosed area of the ellipse is

where we’ve made use of the fact that the integrand is identically 1. This illustrates how, because of the symmetry between x and y, the above area formula sometimes leads to very simple (even trivial) integrals when evaluating the areas enclosed within parametrically defined curves.

The preceding is essentially just an application of Green's Theorem, which states that if we have two differentiable scalar fields P(x,y) and Q(x,y) over a (regular) region R of the xy plane enclosed by the curve C, then we have the following equality between a double integral over the region and a line integral around the boundary curve

To show how this reduces to the simple area formula, notice that if we set Q = x and P = –y then ∂Q/∂x = 1 and ∂P/∂y = –1, and Green's Theorem reduces to

The left side is twice the area of the enclosed region R, and the right side is twice our previous formula for the area of R (with the opposite sign convention).

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