Green’s theorem in the plane says that if , , and are single-valued and continuous in a simple connected region bounded by a simple closed curve , then

where the line integral along is in the anti-clockwise direction, as shown in the sketch. The theorem allows one to replace a double-integral over the region by a line integral around the boundary curve , or vice versa, whichever is the easier one to solve. It acts as a template for generating a multitude of useful formulas of this kind that can be tailored to suit particular situations by carefully choosing the forms of the functions and . For example, in the context of integration of vector functions, if we write a two-dimensional vector in unit-vector form as , then putting and in Green’s theorem gives the divergence theorem in two dimensions, whereas putting and gives Stokes’ theorem in two dimensions. Another result like this is obtained by putting and in Green’s theorem. This yields a formula for calculating the area enclosed by the simple closed curve in the sketch above:

In the present note I want to quickly record a couple of observations about how this last result extends to cases in which the relevant curve is defined *parametrically* rather than in terms of Cartesian coordinates.

First, the result can easily be adapted to obtain a very useful formula for finding the areas of closed parametric curves. If the curve is defined parametrically by and , then simply changing variables from and to in the formula above gives

The expression on the right-hand side can immediately be applied to a huge range of problems involving finding areas of closed parametric curves. As a simple initial illustration to check that it works, we can use it to confirm that the area of a circle of radius is . The circle is described by the Cartesian equation but has a parametric representation , , with ranging from to . Therefore and . Putting these into the formula we get

as expected.

As a slightly more interesting example, we can find the area of the main cardioid in the Mandelbrot set. This has parametric representation , with ranging from to (see, e.g., Weisstein, Eric W., Mandelbrot set, From MathWorld – A Wolfram Web Resource). We find that and , and therefore

Putting this into the formula we find that the area of the main cardioid in the Mandelbrot set is

The second observation I want to make here is that we can sometimes use the same formula to find the area of a region by integrating along a parametric curve that is *not* closed. This seems surprising at first because we obtained the formula using Green’s theorem which explicitly requires the curve to be closed. As an illustration of this situation, consider the problem of finding the area in the diagram. The arc joining the two points and is assumed to have parametric representation , , such that and , with . The claim is then that the area in the diagram is given by the same formula as before, but applied only along the arc joining and rather than all the way around the enclosed region. Thus, we are claiming that

To prove this we first note that since , we can write , so can be written as a function of as . The area under the arc joining and is then given by

Changing variables in this integral from to we find that , , and . Thus, we find that the area under the arc is given by

By simple geometry we can then see that the area is given by

where is the area under the line joining and , and is the area under the line joining and .

Next, we imagine flipping the graph over so that is now along the horizontal axis and is along the vertical axis. We can proceed in the same way as before to find the area under the curve from this point of view. In Cartesian form the area under the curve is given by

and upon changing variables from to this becomes . But, returning to the original graph, we see that the sum of the two areas and is the same as the difference , where is the area of the rectangle with vertices at , , and , and is the area of the smaller rectangle with vertices at , , and . Thus, we can write

This proves the result since we saw above that the expression on the left-hand side gives the area .

As a simple application of this result, suppose the arc in the above scenario is part of a parabola with Cartesian equation and we want to find the area when and . The quadratic equation has a parametric representation , and ranges from to along the arc. We have and , so putting these into the formula we find that