# A note on Green’s theorem in the plane and finding areas enclosed by parametric curves

Green’s theorem in the plane says that if $P$$Q$$\partial P/\partial y$ and $\partial Q/\partial x$ are single-valued and continuous in a simple connected region $\mathfrak{R}$ bounded by a simple closed curve $C$, then

$\oint_C P \mathrm{d}x + Q \mathrm{d}y = \iint_{\mathfrak{R}}\big(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\big) \mathrm{d}x \mathrm{d}y$

where the line integral along $C$ is in the anti-clockwise direction, as shown in the sketch. The theorem allows one to replace a double-integral over the region $\mathfrak{R}$ by a line integral around the boundary curve $C$, 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 $P$ and $Q$. For example, in the context of integration of vector functions, if we write a two-dimensional vector $V$ in unit-vector form as $V = i V_x + j V_y$, then putting $Q = V_x$ and $P = - V_y$ in Green’s theorem gives the divergence theorem in two dimensions, whereas putting $Q = V_y$ and $P = V_x$ gives Stokes’ theorem in two dimensions. Another result like this is obtained by putting $Q = x$ and $P = -y$ in Green’s theorem. This yields a formula for calculating the area enclosed by the simple closed curve $C$ in the sketch above:

$\oint_C x\mathrm{d}y - y \mathrm{d}x = \iint_{\mathfrak{R}} \big(\frac{\partial x}{\partial x} - \frac{\partial (-y)}{\partial y}\big) \mathrm{d}x \mathrm{d}y = 2 \iint_{\mathfrak{R}} \mathrm{d}x \mathrm{d}y$

$\implies$

$\iint_{\mathfrak{R}} \mathrm{d}x \mathrm{d}y = \frac{1}{2} \oint_C x\mathrm{d}y - y \mathrm{d}x$

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 $C$ 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 $C$ is defined parametrically by $x(t)$ and $y(t))$, then simply changing variables from $x$ and $y$ to $t$ in the formula above gives

$\iint_{\mathfrak{R}} \mathrm{d}x \mathrm{d}y = \frac{1}{2} \oint_{C} x \mathrm{d}y - y \mathrm{d}x = \frac{1}{2} \oint_C (x \dot{y} - y \dot{x}) \mathrm{d}t$

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 $r$ is $\pi r^2$. The circle is described by the Cartesian equation $x^2 + y^2 = r^2$ but has a parametric representation $x = r\mathrm{cos} (t)$, $y = r \mathrm{sin} (t)$, with $t$ ranging from $0$ to $2 \pi$. Therefore $\dot{x} = -r\mathrm{sin} (t)$ and $\dot{y} = r\mathrm{cos} (t)$. Putting these into the formula we get

$\frac{1}{2}\oint_C (x \dot{y} - y \dot{x}) \mathrm{d}t = \frac{1}{2}\int_{t=0}^{2 \pi} (r^2\mathrm{cos}^2(t) + r^2\mathrm{sin}^2(t)) \mathrm{d}t = \frac{r^2}{2} \int_{t=0}^{2 \pi} \mathrm{d} t = \pi r^2$

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 $x = \frac{1}{2} \mathrm{cos}(t) - \frac{1}{4} \mathrm{cos}(2t)$, $y = \frac{1}{2} \mathrm{sin}(t) - \frac{1}{4} \mathrm{sin}(2t)$ with $t$ ranging from $0$ to $2 \pi$ (see, e.g., Weisstein, Eric W., Mandelbrot set, From MathWorld – A Wolfram Web Resource). We find that $\dot{x} = -\frac{1}{2} \mathrm{sin}(t) + \frac{1}{2} \mathrm{sin}(2t)$ and $\dot{y} = \frac{1}{2} \mathrm{cos}(t) - \frac{1}{2} \mathrm{cos}(2t)$, and therefore

$x\dot{y} = \frac{1}{4} \mathrm{cos}^2(t) - \frac{3}{8} \mathrm{cos}(t) \mathrm{cos}(2t) + \frac{1}{8}\mathrm{cos}^2(2t)$

$y\dot{x} = -\frac{1}{4} \mathrm{sin}^2(t) + \frac{3}{8} \mathrm{sin}(t)\mathrm{sin}(2t) - \frac{1}{8}\mathrm{sin}^2(2t)$

$x\dot{y} - y\dot{x} = \frac{3}{8} - \frac{3}{8}\mathrm{cos}(t)$

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

$\frac{1}{2}\oint_C (x \dot{y} - y \dot{x}) \mathrm{d}t = \frac{1}{2}\int_{t=0}^{2 \pi} \big(\frac{3}{8} - \frac{3}{8}\mathrm{cos}(t)\big) \mathrm{d}t = \frac{3\pi}{8}$

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 $C$ to be closed. As an illustration of this situation, consider the problem of finding the area $A$ in the diagram. The arc joining the two points $(x_1, y_1)$ and $(x_2, y_2)$ is assumed to have parametric representation $x = f(t)$, $y = g(t)$, such that $(x_1, y_1) = (f(t_1), g(t_1))$ and $(x_2, y_2) = (f(t_2), g(t_2))$, with $t_2 > t_1$. The claim is then that the area $A$ in the diagram is given by the same formula as before, but applied only along the arc joining $(x_1, y_1)$ and $(x_2, y_2)$ rather than all the way around the enclosed region. Thus, we are claiming that

$A = \frac{1}{2} \int_{t_1}^{t_2} (x \dot{y} - y \dot{x}) \mathrm{d}t$

To prove this we first note that since $x = f(t)$, we can write $t = f^{-1}(x)$, so $y$ can be written as a function of $x$ as $y = g(f^{-1}(x))$. The area under the arc joining $(x_1, y_1)$ and $(x_2, y_2)$ is then given by

$\int_{x_1}^{x_2} g(f^{-1}(x)) \mathrm{d}x$

Changing variables in this integral from $x$ to $t$ we find that $t_1 = f^{-1}(x_1)$$t_2 = f^{-1}(x_2)$, $g(f^{-1}(x)) = g(t) = y$ and $\mathrm{d}x = f^{\prime}(t) \mathrm{d}t = \dot{x} \mathrm{d}t$. Thus, we find that the area under the arc is given by

$\int_{x_1}^{x_2} g(f^{-1}(x)) \mathrm{d}x = \int_{t_1}^{t_2} y \dot{x} \mathrm{d}t$

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

$\frac{1}{2}x_2y_2 - \frac{1}{2}x_1y_1 - \int_{t_1}^{t_2} y \dot{x} \mathrm{d}t$

where $\frac{1}{2}x_2y_2$ is the area under the line joining $0$ and $(x_2, y_2)$, and $\frac{1}{2}x_1y_1$ is the area under the line joining $0$ and $(x_1, y_1)$.

Next, we imagine flipping the graph over so that $y$ is now along the horizontal axis and $x$ 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

$\int_{y_1}^{y_2} f(g^{-1}(y)) \mathrm{d}y$

and upon changing variables from $y$ to $t$ this becomes $\int_{t_1}^{t_2} x \dot{y} \mathrm{d}t$. But, returning to the original graph, we  see that the sum of the two areas $\int_{t_1}^{t_2} y \dot{x} \mathrm{d}t$ and $\int_{t_1}^{t_2} x \dot{y} \mathrm{d}t$ is the same as the difference $x_2y_2 - x_1y_1$, where $x_2y_2$ is the area of the rectangle with vertices at $(0, 0)$, $(0, y_2)$, $(x_2, y_2)$ and $(x_2, 0)$, and $x_1y_1$ is the area of the smaller rectangle with vertices at $(0, 0)$, $(0, y_1)$, $(x_1, y_1)$ and $(x_1, 0)$. Thus, we can write

$x_2y_2 - x_1y_1 = \int_{t_1}^{t_2} (x \dot{y} + y \dot{x}) \mathrm{d}t$

$\iff$

$\frac{1}{2}x_2y_2 - \frac{1}{2}x_1y_1 - \int_{t_1}^{t_2} y \dot{x} \mathrm{d}t= \frac{1}{2}\int_{t_1}^{t_2} (x \dot{y} - y \dot{x}) \mathrm{d}t$

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

As a simple application of this result, suppose the arc in the above scenario is part of a parabola with Cartesian equation $y = x^2$ and we want to find the area $A$ when $(x_1, y_1) = (1, 1)$ and $(x_2, y_2) = (2, 4)$. The quadratic equation has a parametric representation $x = t$, $y = t^2$ and $t$ ranges from $1$ to $2$ along the arc. We have $\dot{x} = 1$ and $\dot{y} = 2t$, so putting these into the formula we find that

$A = \frac{1}{2} \int_{t_1}^{t_2} (x \dot{y} - y \dot{x}) \mathrm{d}t = \frac{1}{2}\int_1^2 t^2 \mathrm{d}t = \frac{7}{6}$