Chapter 5.1 The Elementary Notion of Area

Chapter 5 is all about the principle of finite approximation.

Finite Approximation

The principle of finite approximation is to understand infinite things and things with infinitely many parts, study their approximation by things that are finite and have finitely many parts.

Our introdution to this principle will be with understanding area and motion.

Motion and Area

Since ancient times, mathematicians have studied the properties of area and motion. These ideas are deeply rooted in the human experience of the world. However, natural language is not precise enough to explore these ideas.

In fact, only at the very end of the 19th century did mathematics finally become sufficiently developed to provide a framework for reasoning to support the investigation of the ideas that Euclid studied about 2,500 years earlier.

This part of the course focuses on developing a precise language to support reasoning that uses the principle of finite approximation. The language that this reasoning requires is the language of sequences and sequential convergence.

The principle of finite approximation is critical to the study of motion, change, and area. It is a powerful way to use and to extend human intuition about finite things and things that have finitely many parts to infinite things and to things with infinitely many parts.

This is a subset, \(R\), of the plane that is bounded by a curve \(\partial R\):

The curve \(\partial R\) is the boundary of \(R\) and it decomposes the plane into three parts, a part that is inside the curve, a part that is outside the curve, and the curve itself.

If we think of the curve to be the base of a tank with vertical walls that are one unit of length tall and you filled the tank with water to the top, you could use the measurement of the volume of water to determine the area of the base

This gives the intuition that the region \(R\) has area that is bounded by \(\partial R\).

However, what do the above statements mean? What is a curve? What is area? What does it mean for a curve to bound a region?

Although natural language is not precise enough for studying mathematics, the intuition it supports is essential. These words in our natural language facilitate the communication of ideas about human interaction with the environment.

Development of mathematical language is an incremental process that requires patience, care, and great attention to detail.

Path, Curve, Simple Curve, Simple Closed Curve, Parameterization

A path is a function from an interval to \(\mathbb R^d\).

A curve in \(\mathbb R^d\) is the range (we will often prefer to say image) of a path. For now, take \(d\) to be 2.

A simple curve, \(\Gamma\), is the image of a path, \(c\), where

• \(c\) is one-to-one, that is, \(c(x) = c(y)\) implies that \(x = y\).

A simple closed curve, \(\Gamma\), is the image of a path, \(c\), where:

• \(c\) is defined on an interval \([a,b]\), where \([a,b)\) is not empty;

• \(c\) is one-to-one on \([a,b)\);

• \(c(a) = c(b)\).

In both cases, the path \(c\) is a parameterization of \(\Gamma\), it parameterizes \(\Gamma\).

Use the images below to understand the definitions:

In Example 1, we will see our first example of a parameterization of a curve. In this case, the curve is the graph of the function \(f(x)=x^2\) restricted to a closed interval. Pay attention to how we can create more than one parameterization to a curve by changing the time domain using the principle of transformation.

Example 1

Take \(f\) to be the function from \([-1,2]\) to \(\mathbb R\) that is given by \[f(x)=3x^2.\] Find parameterizations for \(f\) (viewed as a subset of \(\mathbb R^2\)) that have domains equal to the following intervals:

1. \([-1,2]\);

2. \([0,1]\).

Is \(f\) a simple curve? Is \(f\) a simple closed curve?

The function \(f\) looks like this.

1. One parameterization for \(f\) is \(c \colon [-1,2]\to \mathbb R^2\), where \(c(t)=(t,3t^2)\).

2. Use transformation to find a parameterization. First, transform the interval \([0,1]\) to the interval \([-1,2]\). For example, \[ [0,1]\to\text{(scale by 3)}\to [0,3]\to\text{(translate by }-1)\to[-1,2].\] Mathematically, \(t\) transforms by \(T_{-1}\circ S_3(t)=3t-1\). So \(c(t)=(t,3t^2)\) becomes \(c(t)=(3t-1,3(3t-1)^2)\). A parameterization \(C \colon [0,1]\to \mathbb R^2\) is \[C(t)=(3t-1,3(3t-1)^2).\]

Our ultimate goal is to use curves and parameterizations to identify a way to describe the idea of inside and outside. This is very difficult to do in general.

However, the idea that a point is inside or outside a disk can be precisely understood as a statement about the point being in a certain feasible set.

Inside or Outside Disk

For any disk \(D\) of radius \(r\) with center \(p\), a point \(q\) is, respectively, inside \(D\) or outside \(D\) if \[ ||p-q|| < r \quad \text{or if}\quad ||p-q|| > r.\]

The disk \(D\) of radius \(r\) with center \(p\) is the set of all \(q\) in \(\mathbb R^2\) such that \[||p-q||\leq r.\]

The set \(\partial D\) (read: “boundary of \(D\)”) is the circle of radius \(r\) with center \(p\), and is a simple closed curve.

Use the pictures below to visually understand the idea of inside, outside and on the boundary of a disk.

In the next example, construct a parameterization of a circle. Use the principle of transformation to simplify the construction.

Example 2

Take \(D\) to be the disk of radius \(2\) with center at \((1, 3)\). Find a path \(c\) that parameterizes the circle \(\partial D\). Be sure that \(c\) moves clockwise around \(\partial D\).

Recall that \(c\colon [0,2\pi]\to \mathbb{R}^2\) given by \(c(t)=(\cos(t),\sin(t))\) parameterizes the unit circle in a counter-clockwise pattern. Transform it to get a parameterization for \(\partial D\): \[\ (\cos(t),\sin(t))\to\text{ scale by 2 for radius}\to (2\cos(t),2\sin(t))\to\text{ translate by }V=\langle 1,3\rangle\text{ to get center}\to (2\cos(t)+1,2\sin(t)+3) \]

This can be expressed as \(\langle 1,3\rangle +2c(t).\)

The new path parameterizes \(\partial D\), but does so counter-clockwise. Reverse the direction by changing the angle to negative angles. Therefore the final answer is \(C\colon [0,2\pi]\to\mathbb{R}^2\) \[C(t)=(2\cos(-t)+1,2\sin(-t)+3).\]

Because an ellipse is an asymmetrically scaled circle, we can use what we know about circles to define inside and outside for an ellipse.

Inside and Outside of Ellipse

A solid ellipse is an asymmetrically scaled disk and an ellipse is an asymmetrically scaled circle and the boundary of a solid ellipse.

If \(E\) is a solid ellipse and \(D\) is a disk so that \[S(D) = E,\] for some asymmetric scaling \(S\), then a point \(p\) is:

• inside \(E\) if and only if \(p\) is equal to \(S(q)\) for some \(q\) that is inside \(D\);

• outside \(E\) if and only if \(p\) is equal to \(S(q)\) for some \(q\) that is outside \(D\);

• in \(\partial E\) if and only if \(p\) is equal to \(S(q)\) for some \(q\) that is in \(\partial D\).

Use the picture below to understand the definition.

In the next problem, recognize the scaling hidden in the ellipse to simplify the problem of finding a point inside an ellipse to a problem of scaling the appropriate circle and point in the disk. That will also help with finding a path that parameterizes the ellipse.

Example 3

Take \(E\) to be the solid ellipse that is given by the equation \[\frac{(x-2)^2}{4}+(y-1)^2 \leq 4.\]

1. Find a point that is inside \(E\).

2. Find a path \(c\) that parameterizes the ellipse \(\partial E\).

An ellipse is a transformed version of a circle. Take \(D\) to be the disk centered at \((0,0)\) with radius \(r=2\), which is the set of all points \((x,y)\) for which \(x^2+y^2\leq 4\).

Take \(S=X_2\), the asymmetric scaling in the \(x\)-axis by \(2\).

Then \(S(D)=\{(2x,y)\colon (x,y)\in D\}=\{(2x,y)\colon x^2+y^2\leq 4\}.\) Set \(X=2x\). Solve for \(x\) to get \(\frac{X}{2}=x\). Rewrite so \(S(D)=\{(2x,y)\colon x^2+y^2\leq 4\}\) becomes \(S(D)=\left\{(X,y)\colon \left(\frac{x}{2}\right)^2+y^2\leq 4\right\}\) or \(S(D)=\left\{(X,y)\colon \frac{x^2}{4}+y^2\leq 4\right\}.\)

Thus \(\frac{x^2}{4}+y^2\leq 4\) is a solid ellipse centered at \((0,0)\).

To get the desired center, translate by \(V=\langle 2,1\rangle\) to get

\[ \begin{align*} V+S(D)&=\left\{\langle 2,1\rangle +(x,y)\colon \frac{x^2}{4}+y^2\leq 4\right\}\\ &=\left\{(x+2,y+1)\colon \frac{x^2}{4}+y^2\leq 4\right\}\\ &=\left\{(X,Y)\colon \frac{(X-2)^2}{4}+(Y-1)^2\leq 4\right\}&&\text{set }X=x+2\text{ and }Y=y+1 \end{align*} \]

the solid ellipse desired.

In summary, compute \(\langle 2,1\rangle+X_2\) to get the desired solid ellipse. Use this to find points inside \(E\).

1. Pick any point in \(D\) the disk centered at \((0,0)\) with radius \(2\). For example \((1,1)\). Now transform it: \(\langle 2,1\rangle +X_2(1,1)=(4,2)\) will be a point in \(E\). Confirm it works by doing the following calculation: \[ \begin{align*} \frac{(4-2)^2}{4}+(2-1)^2&=\frac{2^2}{4}+1^2\\ &=\frac{4}{4}+1\\ &=2\\ &\leq 4 \end{align*} \]

2. To find a path \(c\) that parameterizes the ellipse \(\partial E\), take the parameterization of the circle centered at \((0,0)\) with radius \(2\) and transform it:

\[\begin{align*} (2\cos(t),2\sin(t))&\to\text{ assymetric scale $x$ by 2}\\&\to (4\cos(t),2\sin(t))\\&\to\text{ translate by }V=\langle 2,1\rangle\text{ to get center}\\&\to (4\cos(t)+2,2\sin(t)+1) \end{align*}\]

Therefore the final answer is \(C\colon [0,2\pi]\to\mathbb{R}^2\) \[C(t)=(4\cos(t)+2,2\sin(t)+1).\]

General paths and their images are too poorly behaved to conform to physical intuition about what paths should be. To say anything meaningful requires that we initially restrict to a very basic collection of paths, which we will expand as we develop our mathematical tools.

First, we define an epoch.

Epoch

An epoch in an interval \([a,b]\), where \(a<b\), is an \(n\)-tuple \((t_1, \dots, t_n)\) so that:

• For each \(k\) in \(\{1,\dots,n-1\}\), \(t_k<t_{k+1}\)

• \(t_1=a\)

• \(t_n=b\)

Use the picture to understand what an epoch is.

The paths that we specialize to are ones that created by connecting linear pieces.

Continuous Piecewise Linear Path

A continuous, piecewise linear path \(\ell\) is a path with domain \([a,b]\), where \(a<b\), with the property that there is an epoch \((t_1, \dots, t_n)\) and linear functions \(\ell_1,\ell_2,\dots,\ell_{n-1}\) so that

• for each \(k\) in \(\{1, \dots, n-1\}\), \(\ell=\ell_k\) each interval \([t_k, t_{k+1}]\)

• for each \(k\) in \(\{1, \dots, n-2\}\), \(\ell_k(t_{k+1})=\ell_{k+1}(t_{k+1})\)

The ordered vertex set of \(\ell\) is the \(n\)-tuple \((\ell(t_1), \dots, \ell(t_n))\).

The vertex set of \(\ell\) is the set of distinct entries of the ordered vertex set of \(\ell\)

Use the picture below to visualize the definition.

In this example, we will construct a piecewise linear path with a given vertex set and domain. However, we do not specify the epoch in the problem. This means that there can be more than one piecewise linear path.

Example 4

Construct a continuous, piecewise linear path with domain equal to \([1, 4]\) and ordered vertex set equal \(\big((2, 0), (1, 1), (2, 4), (5, 2)\big)\).

How would your answer change if you wanted the domain to be \([0,5]\)?

Construct linear paths that go from one vertex to the another vertex by finding vectors that translate from one vertex to the other.

• Path from \((2,0)\) to \((1,1)\): \(t\langle -1,1\rangle+(2,0)\)

• Path from \((1,1)\) to \((2,4)\): \(t\langle 1,3\rangle+(1,1)\)

• Path from \((2,4)\) to \((5,2)\): \(t\langle 3,-2\rangle+(2,4)\).

Now stitch them together so that the domain of our continuous, piecewise linear path is \([1,4]\):

\[ c(t)=\begin{cases} (t-1)\langle -1,1\rangle+(2,0)&\text{ if }1\leq t<2\\ (t-2)\langle 1,3\rangle+(1,1)&\text{ if }2\leq t<3\\ (t-3)\langle 3,-2\rangle+(2,4)&\text{ if }3\leq t\leq 4\\ \end{cases} \]

Change the time intervals so get an answer that is defined on \([0,5]\). Here is one example:

\[ c(t)=\begin{cases} t\langle -1,1\rangle+(2,0)&\text{ if }0\leq t<1\\ \frac{(t-1)}{2}\langle 1,3\rangle+(1,1)&\text{ if }1\leq t<3\\ \frac{(t-3)}{2}\langle 3,-2\rangle+(2,4)&\text{ if }3\leq t\leq 5\\ \end{cases} \]

We do not yet have a way of defining what it means for a point to be inside a simple closed curve \(\Gamma\).

Restrictions must be made on such curves for a definition of inside and outside to be meaningful.

However, even without a precise definition, our intuition can help us to identify what inside and outside should be given a graphical representation of a curve.

First, understand whether or not the following curves form a bounded region by graphing the curves.

Example 5

Take \(\Gamma\) to be the curve in the plane that is the image of the path \(c\). Determine graphically whether or not \(\Gamma\) encloses a bounded region in the plane for these choices of paths:

1. \(c\colon [0,3\pi]\to\mathbb R^2\) by \(c(t)=(2t,\cos(t))\)

2. \(c\colon [0,\pi]\to\mathbb R^2\) by \(c(t)=(3\cos(2t),\sin(2t))\)

1. Not a bounded region

2. A bounded region: Ellipse

A simple example of a continuous piecewise linear path that we can easily study is a rectangle. Here is how to construct a rectangle:

• Take \(p\) to be a point in \(\mathbb R^2\) and take \(V\) to be a unit vector.
• Take \(a\) and \(b\) to be positive real numbers.
• The set \(S\) that is given by \[S = \{p, aV+p, bV_\perp+aV+p, bV_\perp +p\}\] is the vertex set of a rectangle.

Use the picture below to visualize the construction:

Using the definition of a piecewise linear path, we can define what a rectangle is.

Rectangle and Vertex Set

Take \(c\) to be the piecewise linear path that is given by \[c(t) = \begin{cases}taV + p&\text{ } 0\leq t <1\\(t-1)bV_\perp + aV+p&\text{ } 1\leq t <2\\-(t-2)aV + bV_\perp+aV+p&\text{ } 2\leq t <3\\-(t-3)bV_\perp + bV_\perp+p&\text{ } 3\leq t \leq 4.\end{cases}\hspace{3in}\] where \(a\) and \(b\) are real numbers and \(V\) is a unit vector.

Define \(\partial R\), the boundary of \(R\), to be the set \(c([0, 4])\)—The path \(c\) parameterizes \(\partial R\).

The union of all line segments with endpoints in \(\partial R\) is rectangle \(R\) with vertex set \(S\).

Note: Every rectangle is the feasible set for a system of linear inequalities.

A point \(p\) is inside \(R\) if it is in \(R\) but not in \(\partial R\) and is outside \(R\) if it is not in \(R\).

Use the next example to understand the definition of a rectangle via its parameterization, its description as a feasible set, and by finding a point inside the rectangle. It will be helpful to remember that all the angles inside of a rectangle are 90 degrees. So use vectors to help complete the example.

Example 6

Take \(R\) to be the rectangle with vertices \((2,5)\), \((5,6)\), \((3,12)\), and \((0,11)\).

1. Find a piecewise linear function \(c\) that parameterizes \(\partial R\) and simulate the motion of a particle whose position at time \(t\) is \(c(t)\).

2. Describe \(R\) as the feasible set for a system of inequalities.

3. Find a point that is inside \(R\).

1. Calculate \(V\) by computing \((5,6)-(2,5)\) and creating a unit vector: \(V=\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle\), \(p=(2,5)\), and \(V_\perp=\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle\). Calculate \(a\) and \(b\), which are the side lengths of the rectangle: \(a=\|(5,6)-(2,5)\|=\sqrt{10}\) and \(b=\|(3,12)-(5,6)\|=\|(-2,6)\|=\sqrt{40}\). The piecewise linear function is

\[ c(t) = \begin{cases}t\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + (2,5)&\text{if } 0\leq t <1\\(t-1)\sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)&\text{if } 1\leq t <2\\-\sqrt{10}(t-2)\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)&\text{if } 2\leq t <3\\-\sqrt{40}(t-3)\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+(2,5)&\text{if } 3\leq t \leq 4.\end{cases} \]

2. The rectangle is all \((x,y)\) that satisfy a system of inequalities. Here are two ways to do it:

   1. Create linear equations for the line that passes through \((2,5)\) and \((5,6)\), the line that passes through \((5,6)\) and \((3,12)\), the line that passes through \((3,12)\) and \((0,11)\), and the line that passes through \((0,11)\) and \((2,5)\). Then write the inequalities

   • the lines that passes through \((2,5)\) and \((5,6)\) is \(y=\frac{1}{3}(x-2)+5\)

   • the lines that passes through \((5,6)\) and \((3,12)\) is \(y=-3(x-5)+6\)

   • the lines that passes through \((3,12)\) and \((0,11)\) is \(y=\frac{1}{3}(x-3)+12\)

   • the lines that passes through \((0,11)\) and \((2,5)\) is \(y=-3(x-0)+11\)

     So the system of inequalities: \[ \begin{cases} y\geq \frac{1}{3}(x-2)+5\\ y\leq-3(x-5)+6\\ y\leq\frac{1}{3}(x-3)+12\\ y\geq-3(x-0)+11 \end{cases} \]

   2. Use the vector paths, rewrite them as linear equations, and then write the inequalities.

   • \(t\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + (2,5)=t\langle 3,1 \rangle+(2,5)\) can be written as \(y=\frac{1}{3}(x-2)+5\)

   • \[\begin{align*}&(t-1)\sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)\\&=t\left\langle -\frac{\sqrt{40}}{\sqrt{10}},\frac{3\sqrt{40}}{\sqrt{10}}\right\rangle +\left(\frac{\sqrt{40}}{\sqrt{10}}+5,-\frac{3\sqrt{40}}{\sqrt{10}}+6\right)\end{align*}\] can be written as \(y=\frac{\frac{3\sqrt{40}}{\sqrt{10}}}{-\frac{\sqrt{40}}{\sqrt{10}}}\left(x-\frac{\sqrt{40}}{\sqrt{10}}-5\right)-3\frac{\sqrt{40}}{\sqrt{10}}+6\)

   • \[\begin{align*}&-\sqrt{10}(t-2)\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)\\&=\langle-3t,-t \rangle+\left(-\frac{\sqrt{40}}{\sqrt{10}}+11,3\frac{\sqrt{40}}{\sqrt{10}}+8\right)\end{align*}\] becomes \(y=\frac{1}{3}\left(x+\frac{\sqrt{40}}{\sqrt{10}}-11\right)+3\frac{\sqrt{40}}{\sqrt{10}}+8\)

   • \[\begin{align*}&-\sqrt{40}(t-3)\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+(2,5)\\&=t\left\langle\frac{\sqrt{40}}{\sqrt{10}},-\frac{3\sqrt{40}}{\sqrt{10}} \right\rangle+\left(-\frac{3\sqrt{40}}{\sqrt{10}}-\frac{\sqrt{40}}{\sqrt{10}}+2,\frac{9\sqrt{40}}{\sqrt{10}}+\frac{3\sqrt{40}}{\sqrt{10}}+5\right)\end{align*}\] can be written as \(y=\frac{\frac{-3\sqrt{40}}{\sqrt{10}}}{\frac{\sqrt{40}}{\sqrt{10}}}\left(x+\frac{3\sqrt{40}}{\sqrt{10}}+\frac{\sqrt{40}}{\sqrt{10}}-2\right)+\frac{9\sqrt{40}}{\sqrt{10}}+\frac{3\sqrt{40}}{\sqrt{10}}+5\)

   So the system of inequalities: \[ \begin{cases} y\geq \frac{1}{3}(x-2)+5\\ y\leq\frac{\frac{3\sqrt{40}}{\sqrt{10}}}{-\frac{\sqrt{40}}{\sqrt{10}}}\left(x-\frac{\sqrt{40}}{\sqrt{10}}-5\right)-3\frac{\sqrt{40}}{\sqrt{10}}+6\\ y\leq\frac{1}{3}\left(x+\frac{\sqrt{40}}{\sqrt{10}}-11\right)+3\frac{\sqrt{40}}{\sqrt{10}}+8\\ y\geq\frac{\frac{-3\sqrt{40}}{\sqrt{10}}}{\frac{\sqrt{40}}{\sqrt{10}}}\left(x+\frac{3\sqrt{40}}{\sqrt{10}}+\frac{\sqrt{40}}{\sqrt{10}}-2\right)+\frac{9\sqrt{40}}{\sqrt{10}}+\frac{3\sqrt{40}}{\sqrt{10}}+5 \end{cases} \]

3. A point inside the rectangle is any \((x,y)\) that satisfies the system of inequalities given in part b. For example \((4,7)\) would work.

Now that we understand the idea of inside and outside for a rectangle, we can understand the idea of area.

In general, it is too difficult to ask that area be defined for any subset of the plane.

Instead, we focus on identifying properties that we want area to have.

For example, whatever area definition we come up with should agree with what we know the area of a rectangle should be.

Area of Rectangle

For any rectangle \(R\) with side lengths equal to \(a\) and \(b\), define the area of \(R\), \(\mathcal A(R)\), to be the product \[\mathcal A(R) = ab.\]

This notion of area in the plane agrees with the way in which we measure area. It is not changed by rigid motions.

In the next example, continue to solidify your understanding of rectangles by calculating the area and also determining a point that is inside of the rectangle.

Example 7

Take \(R\) to be the rectangle with vertices \((2,1), (7,6), (-8,21), (-13,16)\) Calculate the area that is bounded by \(\partial R\).

3. To find \(\mathcal{A}(R)\), find the side lengths of \(R\). Calculate the distance between the appropriate vertices. One side length will be \[a=\|(7,6)-(2,1)\|=\|\langle 5,5\rangle\|=\sqrt{50}\] while the other will be \[b=\|(-8,21)-(7,6)\|=\|\langle-15,15\rangle\|=\sqrt{450}.\] So the area is \[\mathcal{A}(R)=\sqrt{50}\cdot\sqrt{450}=150.\]

Now the next thing we want is to ensure that area satisfies certain property. First, here is notation that we will use for subsets of the plane that do have area.

Area of a Subset of \(\mathbb{R}^2\)

For any subset \(X\) of \(\mathbb R^2\) for which area is defined, denote by \({\mathcal A}(X)\) the area of \(X\).

Now we will talk about axioms we require.

Area Axioms

Area Axioms:

For any subsets of the plane \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, require the following:

1. (non-negativity) \({\mathcal A}(X)\) is non-negative;
2. (invariance) For any rigid motion of the plane \(T\), \[{\mathcal A}(T(X)) = {\mathcal A}(X);\]
3. (closure) \({\mathcal A}(X\cup Y)\), \({\mathcal A}(X\cap Y)\), and \({\mathcal A}(X\setminus Y)\) are defined where \(X\setminus Y=\{x\in X\text{ and } x\not\in Y\}\);
4. (additivity) If \(X\) and \(Y\) have empty intersection, then \[{\mathcal A}(X\cup Y) ={\mathcal A}(X) + {\mathcal A}(Y).\]

Use the pictures below to understand the axioms.

What follows are some immediate consequences. For example, the empty set has zero area.

Example 8

Show that the area of the empty set is \(0\).

Take \(X\) to be a square with side length 1 vertices at \((0,0)\), \((1,0)\), \((1,1)\) and \((0,1).\)

Take \(Y\) to be a square with side length 1 vertices at \((2,0)\), \((3,0)\), \((3,1)\) and \((2,1).\)

Both squares have well-defined area with \(\mathcal{A}(X)=1\) and \(\mathcal{A}(Y)=1\). Their intersection is \(X\cap Y=\emptyset\). By closure axiom, \(\mathcal{A}(\emptyset)\) is defined.

Suppose that \(X\) is a square with side length 1. Notice that \(X\cup\emptyset=X\) and \(X\) and \(\emptyset\) have empty intersection. The additivity axiom implies \[ \begin{align*} \mathcal{A}(X\cup \emptyset)&=\mathcal{A}(X)+\mathcal{A}(\emptyset)\\ \mathcal{A}(X)&=\mathcal{A}(X)+\mathcal{A}(\emptyset)\\ 0&=\mathcal{A}(\emptyset). \end{align*} \]

The example shows that area respects containment

Example 9

Show that for any subsets \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, \[X\subseteq Y \quad \text{implies that}\quad {\mathcal A}(X) \leq {\mathcal A}(Y).\]

Take \(X\) and \(Y\) to be subsets for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined and \(X\subseteq Y\).

Take \(W=Y\setminus X\).

Notice that \(X\cap W=\emptyset\) and \(X\cup W=Y\). Thee additivity axiom implies

\[ \begin{align*} \mathcal{A}(X\cup W)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \mathcal{A}(Y)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \end{align*} \]

Because of the non-negativity axiom, all quantities \(\mathcal{A}(Y), \mathcal{A}(X), \mathcal{A}(W)\) are non-negative. Therefore \[\mathcal{A}(X)+\mathcal{A}(W)\geq A(X).\]

Conclude that \(\mathcal{A}(Y)\geq A(X)\) or \({\mathcal A}(X) \leq {\mathcal A}(Y).\)

The next example shows that area has the subadditvity property. It also provides a way to estimate area of a region that is the union of two sets. This is different from additivity because here we do not assume that the intersection of the two sets is empty.

Example 10

Show that for any subsets \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, \[{\mathcal A}(X\cup Y) \leq {\mathcal A}(X) + {\mathcal A}(Y).\] This is known as the subadditivity of area.

Take \(X\) and \(Y\) to be subsets for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined.

Define \(W=Y\setminus X\). Notice \(X\cup Y=X\cup W\) and \(X\cap W=\emptyset\). The additivity axiom, implies that

\[ \begin{align*} \mathcal{A}(X\cup W)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \mathcal{A}(X\cup Y)&=\mathcal{A}(X)+\mathcal{A}(W). \end{align*} \]

where \(W=Y\setminus X\), \(W\subseteq Y\). From the pervious example, \(\mathcal{A}(W)\leq \mathcal{A}(Y)\). Use that to conclude that

\[ \begin{align*} \mathcal{A}(X\cup Y)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \mathcal{A}(X\cup Y)&\leq \mathcal{A}(X)+\mathcal{A}(Y) \end{align*} \]

The next example extends the additivity property to a union of more than two disjoint sets.

Example 11

Show that if \(\{X_1, X_2, \dots, X_n\}\) are all pairwise disjoint and if the area of each of the sets \(X_i\) is defined, then \[{\mathcal A}\!\left(X_1\cup\cdots\cup X_n\right) ={\mathcal A}(X_1) +\cdots+{\mathcal A}(X_n).\]

Take \(\{X_1, X_2, \dots, X_n\}\) to be pairwise disjoint. This means \(X_i\cap X_j=\emptyset\) for \(i\not=j\).

Take \(C_1=X_2\cup X_3\cup\cdots\cup X_n\). Then \(\left(X_1\cup\cdots\cup X_n\right)=X_1\cup C_1\) and \(X_1\cap C_1=\emptyset.\) The additivity axioms implies

\[ \begin{align*} \mathcal{A}\!\left(X_1\cup\cdots\cup X_n\right)&=\mathcal{A}(X_1)+\mathcal{A}(C_1)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2\cup X_3\cup\cdots\cup X_n).\\ \end{align*} \]

Take \(C_2=X_3\cup X_4\cup\cdots\cup X_n\). Then \(\left(X_2\cup X_3\cup\cdots\cup X_n\right)=X_2\cup C_2\) and \(X_2\cap C_2=\emptyset.\) The additivity axioms implies

\[ \begin{align*} \mathcal{A}\!\left(X_1\cup\cdots\cup X_n\right)&=\mathcal{A}(X_1)+\mathcal{A}(X_2\cup X_3\cup\cdots\cup X_n)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2 \cup C_2)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2)+\mathcal{A}(C_2)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2)+\mathcal{A}(X_3 \cup X_4\cup\cdots\cup X_n)\\ \end{align*} \]

Continue this process by defining \(C_j=X_{j+1}\cup\cdots\cup X_n\) so that \(\left(X_j\cup X_{j+1}\cup\cdots\cup X_n\right)=X_j\cup C_j\) and \(X_j\cap C_j=\emptyset\). The additivity axioms implies

\[{\mathcal A}\!\left(X_1\cup\cdots\cup X_n\right) ={\mathcal A}(X_1) +\cdots+{\mathcal A}(X_n).\]

This next example demonstrates the general additivity of area.

Example 12

Show that for any subsets \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, \[{\mathcal A}(X\cup Y) = {\mathcal A}(X) + {\mathcal A}(Y)-\mathcal{A}(X\cap Y).\]

Notice that \[X=(X\setminus Y)\cup (X\cap Y)\] and \[Y=(Y\setminus X)\cup (X\cap Y),\] where \((X\setminus Y)\cap (X\cap Y)=\emptyset\) and \((Y\setminus X)\cap (X\cap Y)=\emptyset.\)

The additivity axioms implies

\[\mathcal{A}(X)=\mathcal{A}(X\setminus Y)+\mathcal{A}(X\cap Y)\]

and

\[\mathcal{A}(Y)=\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y).\]

Since \((X\setminus Y)\cup (Y\setminus X)\cup(X\cap Y)=(X\cup Y)\), \((X\setminus Y)\cap (Y\setminus X)=\emptyset\), \((X\setminus Y)\cap (X\cap Y)=\emptyset\), and \((Y\setminus X)\cap(X\cap Y)=\emptyset\), the additivity axioms implies

\[ \begin{align*} \mathcal{A}(X\cup Y)&=\mathcal{A}((X\setminus Y)\cup (Y\setminus X)\cup(X\cap Y))\\ &=\mathcal{A}(X\setminus Y)+\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y). \end{align*} \]

Therefore,

\[ \begin{align*} \mathcal{A}(X)+\mathcal{A}(Y)&=\mathcal{A}(X\setminus Y)+\mathcal{A}(X\cap Y)+\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y)\\ &=\mathcal{A}(X\setminus Y)+\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y)+\mathcal{A}(X\cap Y)\\ &=\mathcal{A}(X\cup Y)+\mathcal{A}(X\cap Y), \end{align*} \]

implies that

\[{\mathcal A}(X\cup Y) = {\mathcal A}(X) + {\mathcal A}(Y)-\mathcal{A}(X\cap Y).\]

Because a line is “one-dimensional” it should have an area of zero. That is what we show in this example.

Example 13

Show that the area of any line segment is \(0\).

Take \(L\) to be a line segment. Create two rectangles \(X\) and \(Y\) of identical side lengths so that \(X\cap Y=L\) and \(\mathcal{A}(X)=\mathcal{A}(Y)\) and \(\mathcal{A}(X\cup Y)=2\mathcal{A}(X).\)

From the previous example, \[\mathcal{A}(X\cup Y)=\mathcal{A}(X)+\mathcal{A}(Y)-\mathcal{A}(X\cap Y).\]

Use the above to conclude that

\[ \begin{align*} \mathcal{A}(X\cup Y)&=\mathcal{A}(X)+\mathcal{A}(Y)-\mathcal{A}(X\cap Y)\\ 2\mathcal{A}(X)&=2\mathcal{A}(X)-\mathcal{A}(L)\\ 0&=\mathcal{A}(L). \end{align*} \]

A point has no “width” or “height” so its area also zero.

Example 14

Show that the area of any point in the plane is \(0\).

Take \(P\) to be a point in the plane . Create two line segments \(L_1\) and \(L_2\) of identical lengths so that \(L_1\cap L_2=P\).

The general additivity property implies that \[\mathcal{A}(L_1\cup L_2)=\mathcal{A}(L_1)+\mathcal{A}(L_2)-\mathcal{A}(L_1\cap L_2).\]

Use that to conclude

\[ \begin{align*} \mathcal{A}(L_1\cup L_2)&=\mathcal{A}(L_1)+\mathcal{A}(L_2)-\mathcal{A}(L_1\cap L_2)\\ 0&=0-\mathcal{A}(P)\\ 0&=\mathcal{A}(P). \end{align*} \]

Polygons and Polygonal Paths

Now we will work towards defining area for a class of regions known as polygons.

Polygonal Path and Curve

A polygonal path is a simple, closed, continuous, piecewise linear function from a closed interval \([a,b]\) to the plane.

A polygonal curve is the image of a polygonal path.

Use this picture to understand the definition.

Now we present the Jordan Curve Theorem of Polygons.

Jordan Curve Theorem for Polygons

The Jordan Curve Theorem for Polygons guarantees that a polygonal curve \(\Gamma\) decomposes the plane into three regions, a region that is inside \(\Gamma\), a region that is outside \(\Gamma\), and the curve \(\Gamma\) itself.

A polygon is the union of a polygonal curve and the set of points that are inside the curve.

Use this picture to understand the theorem and definition of polygon.

Next we present the boundary of a polygon, which divides the outside and inside.

Boundary of a Polygon

For any polygon \(P\), denote by \(\partial P\) the polygonal curve that defines \(P\).

Denote by \(\partial_o P\) the set of oriented line segments that is determined by a polygonal path that parameterizes \(\partial P\).

Note that a triangle and a rectangle are both examples of polygons, and a triangular (path) curve and a rectangular (path) curve are both examples of polygonal (paths) curves.

Understand the boundary with the picture below.

In this example, we construct a polygonal path with a given epoch.

Example 15

Take \(\big((-4,2), (-2, 5), (2,4), (-3,7)\big)\) to be the ordered vertex set of a polygonal path with epoch \((0, 2, 4, 5, 9)\). Identify the given path with a piecewise linear function, \(c\). Simulate a particle whose position at time \(t\) is \(c(t)\).

Construct linear paths that go from one vertex to the another vertex by finding vectors that translate from one vertex to the other with the specified time:

• Path from \((-4,2)\) at \(t=0\) to \((-2,5)\) at \(t=2\): \(\frac{t}{2}\langle 2,3\rangle+(-4,2)\)

• Path from \((-2,5)\) at \(t=2\) to \((2,4)\) at \(t=4\): \(\frac{t-2}{2}\langle 4,-1\rangle+(-2,5)\)

• Path from \((2,4)\) at \(t=4\) to \((-3,7)\) at \(t=5\): \((t-4)\langle -5,3\rangle+(2,4)\)

• Path from \((-3,7)\) at \(t=5\) to \((-4,2)\) at \(t=9\): \(\frac{(t-5)}{4}\langle -1,-5\rangle+(-3,7)\)

A continuous, piecewise linear path on \([0,9]\) is

\[ c(t)=\begin{cases} \frac{t}{2}\langle 2,3\rangle+(-4,2)&\text{ if }0\leq t<2\\ \frac{t-2}{2}\langle 4,-1\rangle+(-2,5)&\text{ if }2\leq t<4\\ (t-4)\langle -5,3\rangle+(2,4)&\text{ if }4\leq t< 5\\ \frac{(t-5)}{4}\langle -1,-5\rangle+(-3,7)&\text{ if }5\leq t\leq 9\\ \end{cases} \]

Another way to specify orientation is by encoding the information in the boundary. This definition tells us how to do it.

Oriented Line Segement and Oriented Boundary

Denote by \(\overrightarrow{pq}\) the oriented line segment with endpoints \(p\) and \(q\).

Any path that parameterizes the line segment \(\overline{pq}\) that is determined by \(\overrightarrow{pq}\) must start at \(p\) and end at \(q\).

A path \(c\) that parameterizes the boundary of a polygon \(P\) specifies an order on each line segment of the boundary by the order in which it reaches each of the endpoints and it specifies an ordered vertex set \((p_1, p_2, p_3, p_4)\) for \(\partial P\).

Denote by \(\partial_oP\) the oriented boundary of \(P\), where \[\partial_oP = \left\{\overrightarrow{p_1p_2}, \overrightarrow{p_2p_3}, \overrightarrow{p_3p_4},\dots, \overrightarrow{p_np_1}\right\}.\]

Here is an example of a polygon with an oriented boundary.

Example 16

Take \(\partial_oR\) to be the oriented boundary of a polygon that is given by \[\partial_oR = \left\{\overrightarrow{(2,2)(3,3)}, \overrightarrow{(3,3)(5,2)}, \overrightarrow{(5,2)(6,5)}, \overrightarrow{(6,5)(3,7)}, \overrightarrow{(3,7)(1,4)}, \overrightarrow{(1,4)(2,2)}\right\}.\] Take \((1,8,9,10,14, 20, 25)\) to be the epoch for a polygonal path \(c\) that parameterizes \(\partial R\) and respects \(\partial_oR\). Simulate a particle whose position at time \(t\) is \(c(t)\).

Attempt this example A path is given below

\[ c(t)=\begin{cases} \frac{t-1}{7}\langle 1,1\rangle+(2,2)&\text{ if }1\leq t<8\\ (t-8)\langle 2,-1\rangle+(3,3)&\text{ if }8\leq t<9\\ (t-9)\langle 1,3\rangle+(5,2)&\text{ if }9\leq t< 10\\ \frac{(t-10)}{4}\langle -3,2\rangle+(6,5)&\text{ if }10\leq t< 14\\ \frac{(t-14)}{6}\langle -2,-3\rangle+(3,7)&\text{ if }14\leq t< 20\\ \frac{(t-20)}{5}\langle 1,-2\rangle+(1,4)&\text{ if }20\leq t\leq 25\\ \end{cases} \]

Now we define what it means for an oriented boundary to be positively or negatively oriented.

Positively and Negatively Oriented

An oriented boundary is positively (respectively negatively) oriented if for each point \(p\) in an oriented line sgement of \(\partial_oP\) that is not a vertex, if \(V\) is a vector that moves points along the oriented line segment in the positive direction, then for any small enough positive real number \(\varepsilon\), \(\varepsilon V_\perp +p\) is inside (respectively outside) \(P\).

It turns out that only one of these two properties can be valid for a given polygon, and that one must be valid. This can help with simplifying the process of determining whether an orientation is positively or negatively oriented.

Another thing that is help are the following properties: take \(V\) to be a vector, then

• for any rotation or translation \(T\), \[T(V_\perp)=T(V)_\perp\],

• for any reflection \(T\), across a line in the plane, \[T(V_\perp)=-T(V)_\perp.\]

What this means is reflections reverse the orientation of an oriented boundary.

In this example, we will

Understand the definition by completing the next example. One thing to note is that finding whether a point is inside or outside a rectangle can be done in at least two ways. One is a transformation approach and the other is using the feasible set definition. For this example, we use the transformation approach.

Example 17

The set \(\{(4,1), (7,3), (5,6), (2,4)\}\) is a rectangular vertex set. Construct the positively oriented boundary of this rectangle and the negatively oriented boundary of this rectangle. Determine whether or not the point \(p\) is inside or outside \(R\), where \(p\) is given by:

1. \(p=(6,4)\)

2. \(p=(3,5)\)

A positively oriented boundary is \[\left\{\overrightarrow{(4,1)(7,3)}, \overrightarrow{(7,3)(5,6)}, \overrightarrow{(5,6)(2,4)}, \overrightarrow{(2,4)(4,1)}\right\}.\]

A negatively oriented boundary is \[\left\{\overrightarrow{(4,1)(2,4)}, \overrightarrow{(2,4)(5,6)}, \overrightarrow{(5,6)(7,3)}, \overrightarrow{(7,3)(4,1)}\right\}.\]

To determine if a point \(p\) is inside the rectangle, check if it lies to the left any oriented line segment of the oriented boundary. In some cases, this may be easier to see by translating the rectangle so that the lower left vertex is \((0,0)\) and one of its sides is parallel to the \(x\)-axis.

1. The lower left vertex is \((4,1)\). Translate all vertices by \(\langle -4,-1\rangle\). The new vertices are \((0,0)\), \((3,2)\), \((1,5)\), and \((-2,3).\)

2. Now rotate \((3,2)\) to the \(x\)-axis. Do that by first finding the appropriate angle: \[p=\left(\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\right).\] The angle is going to be \(p^{-1}=\left(\frac{3}{\sqrt{13}},-\frac{2}{\sqrt{13}}\right).\) Thus \((3,2)\star p^{-1}\) will rotate \((3,2)\) to the \(x\)-axis. Rotate the rest of the vertices by \(p^{-1}\):

• \((3,2)\star p^{-1}=\left(3\cdot \frac{3}{\sqrt{13}}-2\cdot\frac{-2}{\sqrt{13}},3\cdot\frac{-2}{\sqrt{13}}+2\cdot\frac{3}{\sqrt{13}}\right)=\left(\frac{13}{\sqrt{13}},0\right)\).

• \((1,5)\star p^{-1}=\left(1\cdot \frac{3}{\sqrt{13}}-5\cdot\frac{-2}{\sqrt{13}},1\cdot \frac{-2}{\sqrt{13}}+5\cdot\frac{3}{\sqrt{13}}\right)=\left(\frac{13}{\sqrt{13}},\frac{13}{\sqrt{13}}\right)\)

• \((-2,3)\star p^{-1}=\left(-2\cdot\frac{3}{\sqrt{13}}-3\cdot\left(\frac{-2}{\sqrt{13}}\right),-2\cdot\left(\frac{-2}{\sqrt{13}}\right)+3\cdot \frac{3}{\sqrt{13}}\right)=\left(0,\frac{13}{\sqrt{13}}\right)\)

1. To determine if \((6,4)\) is in \(R\), perform the translation and rotation from before to get \((V+(6,4))\star p^{-1}=\left(\frac{12}{\sqrt{13}},\frac{5}{\sqrt{13}}\right)\). Compare the coordinates of this point with the vertices \((0,0)\), \(\left(\frac{13}{\sqrt{13}},0\right)\), \(\left(\frac{13}{\sqrt{13}},\frac{13}{\sqrt{13}}\right)\),\(\left(0,\frac{13}{\sqrt{13}}\right)\). This is going to be inside of the rectangle. Hence the original point \((6,4)\) is inside of the original rectangle.

2. To determine if \((3,5)\) is in \(R\), perform the translation and rotation from before to get \((V+(3,5))\star p^{-1}=\left(\frac{5}{\sqrt{13}},\frac{14}{\sqrt{13}}\right)\). Compare the coordinates of this point with the vertices \((0,0)\), \(\left(\frac{13}{\sqrt{13}},0\right)\), \(\left(\frac{13}{\sqrt{13}},\frac{13}{\sqrt{13}}\right)\),\(\left(0,\frac{13}{\sqrt{13}}\right)\). The \(y\)-coordinate is bigger than the \(y\)-coordinate of the upper vertices, so it will not be inside of \(R\). Hence the original point \((3,5)\) is outside of the original rectangle.

In the next example, continue to solidify your understanding of rectangles by calculating the area and also determining a point that is inside of the rectangle.

Example 18

Take \(R\) to be the rectangle with vertices \((2,1), (7,6), (-8,21), (-13,16)\)

1. Construct both a positively and negatively oriented rectangular boundary for \(R\).

2. Find a point \(p\) that is inside \(R\) and show that \(p\) is inside of \(R\).

3. Calculate the area that is bounded by \(\partial R\).

1. A positively oriented boundary is \[\left\{\overrightarrow{(2,1)(7,6)}, \overrightarrow{(7,6)(-8,21)}, \overrightarrow{(-8,21)(-13,16)}, \overrightarrow{(-13,16)(2,1)}\right\}.\] A negatively oriented boundary is \[\left\{\overrightarrow{(2,1)(-13,16)}, \overrightarrow{(-13,16)(-8,21)}, \overrightarrow{(-8,21)(7,6)}, \overrightarrow{(7,6)(2,1)}\right\}.\]

2. To determine if a point \(p\) is inside the rectangle, check if it lies to the left any oriented line segment of the oriented boundary. In some cases, this may be easier to see by translating the rectangle so that the lower left vertex is \((0,0)\) and one of its sides is parallel to the \(x\)-axis.

   1. The lower left vertex is \((2,1)\). Translate all vertices by \(\langle -2,-1\rangle\). The new vertices are \((0,0)\), \((5,5)\), \((-10,20)\), and \((-15,15).\)
   2. Now rotate \((5,5)\) to the \(x\)-axis. Find the appropriate angle: \[p=\left ( \frac{5}{\sqrt{50}},\frac{5}{\sqrt{50}}\right) \]. The angle is going to be \(p^{-1}=\left ( \frac{5}{\sqrt{50}},-\frac{5}{\sqrt{50}}\right).\) Rotate all vertices by \(p^{-1}\).
   • \((5,5)\star p^{-1}=\left(\frac{50}{\sqrt{50}},0\right)\)
   • \((-10,20)\star p^{-1}=\left(\frac{50}{\sqrt{50}},\frac{150}{\sqrt{50}}\right)\)
   • \((-15,15)\star p^{-1}=\left(0,\frac{150}{\sqrt{50}}\right)\)

Based on the transformed vertices, a point \(q=(q_x,q_y)\) will be inside the rotated rectangle so long as \(0< q_x<\frac{50}{\sqrt{50}}\) and \(0<q_y<\frac{150}{\sqrt{50}}\). For example \(q=\left(\frac{1}{\sqrt{50}},\frac{1}{\sqrt{50}}\right)\) will be in the rotated rectangle. To get the corresponding point in the original rectangle, “reverse” the transformation: \(\langle 2,1\rangle+ (q\star p)=\left(2,\frac{10}{\sqrt{50}}+1\right)\). Thus \(q=\left(2,\frac{10}{\sqrt{50}}+1\right)\) is inside of the rectangle with vertices \((2,1), (7,6), (-8,21), (-13,16)\).

3. To find \(\mathcal{A}(R)\), find the side lengths of \(R\). Calculate the distance between the appropriate vertices of either the original rectangle or rotated rectangle. One side length will be \[a=\|(7,6)-(2,1)\|=\|\langle 5,5\rangle\|=\sqrt{50}\] while the other will be \[b=\|(-8,21)-(7,6)\|=\|\langle-15,15\rangle\|=\sqrt{450}.\] So the area is \[\mathcal{A}(R)=\sqrt{50}\cdot\sqrt{450}=150.\]

The next geometric shape we study is a triangle. In some ways, a triangle is simpler than a rectangle. In other ways, it is more challenging. First, understand how we define a triangle as well as understanding how we define inside and outside.

Inside and Outside of a Triangle

Denote by \(\partial R\) the union of all line segments with endpoints in \(\{p_1, p_2, p_3\}\), where \(p_1\), \(p_2\), and \(p_3\) are distinct points in the plane.

The triangle with vertex set \(\{p_1, p_2, p_3\}\) is the set \(R\) that is given by \[R = \big\{t(q_2 -q_1) + q_1\colon t\in [0,1] \;{\rm and}\; q_1, q_2 \in \partial R\big\}.\]

The set \(\partial R\) is the boundary of \(R\).

A point is inside \(R\) if it is in \(R\) and not in \(\partial R\).

A point is outside \(R\) if it is not in \(R\).

Just as with rectangles, any triangle is the feasible set for a system of linear inequalities.

Use the picture below to understand the information given above.

Just like with a rectangle, we can define an orientation for the triangle.

Orientated Triangular Boundary

Given a triangle with vertex set \(\{p_1, p_2, p_3\}\), an oriented triangular boundary \(\partial_oR\) is the set \[\partial_oR = \left\{\overrightarrow{p_1p_2}, \overrightarrow{p_2p_3}, \overrightarrow{p_3p_1}\right\}.\]

First practice constructing a continuous piecewise linear parameterization for a given triangle.

Example 19

Take \(R\) to be the triangle with ordered vertex set \(\big((1,5), (3,-1), (7,2)\big)\). Find a continuous, piecewise linear parameterization \(\ell\) for the boundary of \(R\), where the domain of \(\ell\) is equal to \([0,5]\). Is \(\partial R\) a simple closed curve?

Create linear paths that go to each vertex:

• The linear path from \((1,5)\) to \((3,-1)\) is \(t\langle 2,-6 \rangle+(1,5)\).

• The linear path from \((3,-1)\) to \((7,2)\) is \(t\langle 4,3 \rangle+(3,-1)\).

• The linear path from \((7,2)\) to \((1,5)\) is \(t\langle -6,3 \rangle+(7,2)\).

Now create a piecewise linear parameterization by modifying the \(t\) variable. Here is one example

\[ \ell(t)=\begin{cases} \frac{t}{2}\langle 2,-6\rangle +(1,5) &\text{if }0\leq t<2\\ \frac{(t-2)}{2}\langle 4,3 \rangle+(3,-1) &\text{if } 2\leq t<4\\ (t-4)\langle -6,3 \rangle+(7,2) &\text{if }4\leq t <5 \end{cases} \]

Based on this, conclude that \(\partial R\) is a simple closed curve.

Finally, determine whether a point is inside or outside of a triangle.

Example 20

Take \(R\) to be the triangle with vertex set \(\big\{(0,0), (2,3), (5,0)\big\}\). Determine whether or not the following points are inside \(R\), outside \(R\), or in \(\partial R\):

1. \((1,1)\);

2. \((4,1)\).

First find a way to describe the points on \(\partial R\)

• \((2t,3t)\) where \(t\in [0,1]\) describes points on the line segment from \((0,0)\) to \((2,3)\)
• \((3s+2,-3s+3)\) where \(s\in[0,1]\) describes points on the line segment from \((2,3)\) to \((5,0)\)
• \((5u,0)\) where \(u\in[0,1]\) describes points on the line segment from \((0,0)\) to \((5,0)\).

1. The point \((1,1)\) is \(R\) provided there are points on the boundary for which the line segment that connects the two points crosses \((1,1).\) There are many ways to go about it, but here is one way.

• The point \((1,0)\) is on the line segment from \((0,0)\) to \((5,0)\): take \(u=\frac{1}{5}\), then \((5u,0)\) becomes \((1,0)\)
• The point \(\left(1,\frac{3}{2}\right)\) is on the line segment from \((0,0)\) to \((2,3)\): take \(t=\frac{1}{2}\), then \((2t,3t)\) becomes \(\left(1,\frac{3}{2}\right)\).
• Obtain the point \((1,1)\) by creating the line segment that connects \(q_1=(1,0)\) and \(q_2=\left(1,\frac{3}{2}\right)\) together \(t(q_2 -q_1) + q_1=t\left\langle 0,\frac{1}{2}\right\rangle+(1,0)=\left(1,\frac{3}{2}t\right)\). Find \(t\) so that \((1,1)=\left(1,\frac{3}{2}t\right)\). The time value is \(t=\frac{2}{3}\), which means that \((1,1)\) is inside \(R\).

2. The point \((4,1)\) is in \(\partial R\) because it is on \((3s+2,-3s+3)\), where \(s\in[0,1]\), which describes points on the line segment from \((2,3)\) to \((5,0)\). Find \(s\) so that \((3s+2,-3s+3)=(4,1)\). Solve to get \(s=\frac{2}{3}\).

Here is one connection between triangles and circles.

Circumcircle

Given any triangle with ordered vertex set \((p_1,p_2,p_3)\), take \(\Delta p_1p_2p_3\) to be the triangle with oriented boundary \(\partial_o \Delta\) that is given by \[\partial_o \Delta =\left\{\overrightarrow{p_1p_2},\overrightarrow{p_2p_3},\overrightarrow{p_3p_1}\right\}.\] There is a unique circle \(C\), called the circumcircle of \(\Delta p1p2p3\), so that the points \(p_1,p_2\) and \(p_3\) all lie in \(C\).

In this example, we are given a circumcircle and we find the vertex set of its triangle.

Example 21

Identify the vertex set of a triangle whose circumcircle is given by the equation \[(x-2)^2+(y-1)^2=9.\] Construct both a positively and negatively oriented boundary for this triangle.

Here is the circle.

We can construct a triangle by picking three “easy” points on the circle.

So our triangle is \(\Delta (2,4)(5,1)(2,-2)\). This is a positively oriented boundary for the triangle: \[\partial_o\Delta((2,4)(5,1)(2,-2))=\left\{\overrightarrow{(5,1)(2,4)},\overrightarrow{(2,4)(2,-2)},\overrightarrow{(2,-2)(5,1)}\right\}.\] This is a negatively oriented boundary for the triangle: \[\partial_o\Delta((2,4)(5,1)(2,-2))=\left\{\overrightarrow{(5,1)(2,-2)},\overrightarrow{(2,-2)(2,4)},\overrightarrow{(2,4)(5,1)}\right\}.\]

In the next example, we construct the circumcircle for a given triangle. One way to do it is to find the midpoint of the line segment that corresponds to an edge of the triangle. Do this again with a line segment that corresponds to a different edge of the triangle. Now create a line that is perpendicular to the line that passes through the vertices of the end points of the edges and that passes through the midpoint. There will be two lines, one for each midpoint you found, and they will intersect at one point. This point will in fact be the center of the circle! The radius can be found by finding the distance from the center to any vertex on the triangle.

Example 22

Determine the circumcircle for the triangle \(\Delta\) with vertex set \(\{(2,3),(1,5),(-2,6)\}.\)

Here is a picture of the triangle and the circle we will ultimately find.

The midpoint of \(\overrightarrow{(-2,6)(2,3)}\) is \((0,\frac{9}{2}).\) The \(\overrightarrow{(2,3)(1,5)}\) is \((\frac{3}{2},4)\).

The line that passes through \((0,\frac{9}{2})\) and is perpendicular to the line that passes through \((-2,6)\) and \((2,3)\) is \[y=\frac{4}{3}x+\frac{9}{2}.\]

The line that passes through \((\frac{3}{2},4)\) and is perpendicular to the line that passes through \((2,3)\) and \((1,5)\) is \[y=\frac{1}{2}\left(x-\frac{3}{2}\right)+4.\]

The two lines intersect at \((x,y)\) if and only if

\[\frac{4}{3}x+\frac{9}{2}=\frac{1}{2}\left(x-\frac{3}{2}\right)+4.\]

Solve for \(x\) to obtain that \(x=-\frac{3}{2}\) and consequently \(y=\frac{5}{2}.\)

The center of the circumcircle is \(\left(-\frac{3}{2},\frac{5}{2}\right).\) The radius is the distance from the center of the circumcircle to any vertex of \(\Delta\): \[\begin{align*} r&=\|\left(-\frac{3}{2},\frac{5}{2}\right)-(2,3)\|\\ &=\|\langle -\frac{7}{2},-\frac{1}{2}\rangle\|\\ &=\sqrt{\frac{49}{4}+\fraC{1}{4}}\\ &=\sqrt{\frac{50}{4}}\\ &=\sqrt{\frac{25}{2}}. The circumcircle is thus $$ \left(x+\frac{3}{2}\right)^2+\left(y-\frac{5}{2}\right)^2=\frac{25}{2}. $$ \end{align*}\]

Triangulation and the Area of Polygons

We will use the properties of area to derive a formula for the area of a triangle in terms of the coordinates of its vertices. We can do this by using transformation.

First show that if the oriented triangle \(\Delta p_1p_2p_3\) is a right triangle with leg lengths \(a\) and \(b\), then \({\mathcal A}(\Delta p_1p_2p_3)\) exists and \[{\mathcal A}(\Delta p_1p_2p_3) = \frac{ab}{2}.\]

Create rectangle \(X\) and \(Y\) with the following characteristics:

• Rectangle \(X\) has length \(a\) and width \(b\).
• Rectangle \(Y\) has length \(c=\sqrt{a^2+b^2}\) and width \(b\).

Arrange \(X\) and \(Y\) in the xy plane via translation and or rotation so that

• Rectangle \(X\) has its vertices at \((0,0)\), \((a,0)\), \((a,b)\) and \((0,b)\)
• Rectangle \(Y\) has its vertices at \((0,0)\), \((c,0)\), \((c,-b)\) and \((0,-b)\)

Rotate \(Y\) so that vertex \((c,0)\) is at \((a,b)\), which we can do by rotating about the origin by the angle \(\left(\frac{a}{c},\frac{b}{c}\right).\)

When we take \(X\cap Y\), we will get a right triangle with leg lengths \(a\) and \(b\), meaning \(X\cap Y=\Delta p_1p_2p_3\)

By the closure axiom, \({\mathcal A}(X\cap Y)={\mathcal A}(\Delta p_1p_2p_3)\) exists.

Finally, a rectangle \(R\) with side lengths \(a\) and \(b\) can be cut diagonally so that we can create two identical copies of an oriented right triangle \(\Delta p_1p_2p_3\) with leg lengths \(a\) and \(b\). Thus, \[2\mathcal{A}(\Delta p_1p_2p_3)=\mathcal{A}(R)\] or \[\mathcal{A}(\Delta p_1p_2p_3)=\frac{ab}{2}.\]

Now we talk about the area of a general triangle.

Shoelace Formula for Triangles

Take \(p_1=(x(p_1),y(p_1))\), \(p_2=(x(p_2),y(p_2))\), \(p_3=(x(p_3),y(p_3)).\) The Shoelace Formula for the Area of Triangles states that if \(\Delta p_1p_2p_3\) is positively oriented, then \[\begin{align*}{\mathcal A}(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(x(p_1)y(p_2) + x(p_2)y(p_3) + x(p_3)y(p_1)\right.\\&\hspace{1.52in} \left.-y(p_1)x(p_2)-y(p_2)x(p_3)- y(p_3)x(p_1)\right).\end{align*}\]

The formula can be derived by doing the following.

Translate the triangle so that \(p_1=(x(p_1),y(p_1))\) is at the origin. Which we can do by translating by \[\langle -x(p_1),-y(p_1)\rangle .\]

So our new vertices are at \((0,0), \left(x(p_2)-x(p_1),y(p_2)-y(p_1)\right)\) and \((x(p_3)-x(p_1),y(p_3)-y(p_1))\).

Rotate the triangle so that \((x(p_2)-x(p_1),y(p_2)-y(p_1))\) is on the \(x\)-axis, which we do by rotating by the angle \(\left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)\), where \(V=\langle x(p_2)-x(p_1),y(p_2)-y(p_1)\rangle\).

The new vertices will be

• \((0,0)\star \left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)=Q_1\)
• \((x(p_2)-x(p_1),y(p_2)-y(p_1))\star \left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)=Q_2\)
• \((x(p_3)-x(p_1),y(p_3)-y(p_1))\star \left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)=Q_3\)

where \((a,b)\star (c,d)=(ac-bd,ad+bc).\)

Now we can use find the area of the triangle, by identifying the base and the height. The base will equal the \(x\)-coordinate of \(Q_2\) while the height will equal the \(y\)-coordinate of \(Q_3\).

The \(x\)-coordinate of \(Q_2\) will be

\[ \begin{align*} &(x(p_2)-x(p_1))\cdot \left(\frac{x(p_2)-x(p_1)}{\|V\|}\right)-(y(p_2)-y(p_1))\cdot\left(-\frac{y(p_2)-y(p_1)}{\|V\|}\right)\\&=\frac{(x(p_2)-x(p_1))^2}{\|V\|}+\frac{(y(p_2)-y(p_1))^2}{\|V\|}\\ &=\frac{(x(p_2)-x(p_1))^2+(y(p_2)-y(p_1))^2}{\|V\|}\\ &=\frac{\|V\|^2}{\|V\|}\\ &=\|V\|. \end{align*} \]

The \(y\)-coordinate of \(Q_3\) will be

\[ \begin{align*} &(x(p_3)-x(p_1))\cdot \left(-\frac{y(p_2)-y(p_1)}{\|V\|}\right)+(y(p_3)-y(p_1))\cdot\left(\frac{x(p_2)-x(p_1)}{\|V\|}\right)\\&=\frac{(x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1))}{\|V\|}+\frac{(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1))}{\|V\|}\\ &=\frac{(x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1)+(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1))}{\|V\|}. \end{align*} \]

Therefore, the area of the triangle is

\[ \begin{align*} {\mathcal A}(\Delta p_1p_2p_3) &=\frac{1}{2}x(Q_2)\cdot y(Q_3)\\ &=\frac{1}{2}\|V\|\cdot\frac{(x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1)+(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1))}{\|V\|}\\ &=\frac{1}{2}\left((x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1))+(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1)\right)\\ &=\frac{1}{2}\left(-x(p_3)y(p_2)+x(p_3)y(p_1)+x(p_1)y(p_2)-x(p_1)y(p_1)+y(p_3)x(p_2)-y(p_3)x(p_1)-y(p_1)x(p_2)+y(p_1)x(p_1)\right)\\ &=\frac{1}{2}\left(x(p_1)y(p_2)+x(p_2)y(p_3)+x(p_3)y(p_1)-y(p_1)x(p_2)-y(p_2)x(p_3)-y(p_3)x(p_1)\right) \end{align*} \] as desired.

The formula also provides a way to measure orientation.

Oriented Triangle

Given an oriented triangle \(\Delta p_1p_2p_3\), take \(\alpha(\Delta p_1p_2p_3)\) to be the quantity \[\begin{align*}\alpha(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(x(p_1)y(p_2) + x(p_2)y(p_3) + x(p_3)y(p_1)\right.\\&\hspace{1.52in} \left.-y(p_1)x(p_2)-y(p_2)x(p_3)- y(p_3)x(p_1)\right).\end{align*}\]

This is a positive quantity if \(\Delta p_1p_2p_3\) if positively oriented and is negative if \(\Delta p_1p_2p_3\) is negatively oriented.

Practice finding the area with the next example. Pay attention to the sign of the quantity \(\alpha\).

Example 23

For these choices of \(p_1\), \(p_2\), and \(p_3\), use the shoelace formula to determine the area of the triangle \(\Delta p_1p_2p_3\) and to determine whether or not the triangle is positively or negatively oriented:

1. \(p_1=(1,2)\), \(p_2=(5,4)\), \(p_3=(3,10)\);

2. \(p_1=(2,-4)\), \(p_2=(-3,-1)\), \(p_3=(-1,4)\).

In this example, \(x(p_1)=1, x(p_2)=5,x(p_3)=3\) and \(y(p_1)=2, y(p_2)=4,y(p_3)=10\). Use the formula to compute:

\[\begin{align*}\alpha(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(1\cdot 4 + 5\cdot 10 + 3\cdot 2-2\cdot 5-4\cdot 3- 10\cdot 1\right)\\ &=\tfrac{1}{2}\left(4+50+6-10-12-10\right)\\ &=\tfrac{1}{2}\left(28\right)\\ &=14.\end{align*}\] Since the quantity is positive, the triangle is positively oriented and the area is 14.

In this example, \(x(p_1)=2, x(p_2)=-3,x(p_3)=-1\) and \(y(p_1)=-4, y(p_2)=-1,y(p_3)=4\). Use the formula to compute:

\[\begin{align*}\alpha(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(2\cdot (-1) + (-3)\cdot (4) + (-1)\cdot (-4)-(-4)\cdot (-3)-(-1)\cdot (-1)- 4\cdot 2\right)\\ &=\tfrac{1}{2}\left(-2-12+4-12-1-8\right)\\ &=\tfrac{1}{2}\left(-31\right)\\ &=-\frac{31}{2}.\end{align*}\] Since the quantity is negative, the triangle is negatively oriented and the area is \(\left|-\frac{31}{2}\right|=\frac{31}{2}.\)

Triangles are uniquely determined by the lengths of their sides. The area should be computable only from the side lengths.

Heron’s Formula

Heron’s Formula states that if \(a\), \(b\), and \(c\) are the side lengths of a triangle, then \[{\mathcal A}(P) = \sqrt{S(S-a)(S-b)(S-c)}\quad{\rm where}\quad S=\frac{a+b+c}{2}.\]

In the next example, find the area of a triangle with given side lengths. If the side lengths are not given, then you would first need to find the lengths.

Example 24

A triangle \(P\) has side lengths 5, 9, and 7. Calculate \({\mathcal A}(P)\).

Here \(a=5,b=9,c=7\), so \(S=\frac{5+9+7}{2}=\frac{21}{2}.\) Use Heron’s Formula to get \[ \begin{align*} {\mathcal A}(P)& = \sqrt{S(S-a)(S-b)(S-c)}\\ &=\sqrt{\frac{21}{2}\left(\frac{21}{2}-5\right)\left(\frac{21}{2}-9\right)\left(\frac{21}{2}-7\right)}\\ &=\sqrt{\frac{21}{2}\left(\frac{11}{2}\right)\left(\frac{3}{2}\right)\left(\frac{7}{2}\right)}\\ &=\sqrt{\frac{4851}{16}}. \end{align*} \]

One of the reasons we study polygons is because we can decompose it into finitely many triangles. This is known as a triangulation.

Triangulation

Take \(P\) to be a polygon. A triangulation \({\mathcal T}(P)\) is a set of oriented triangles with the following property:

• Every oriented triangle in \({\mathcal T}(P)\) has the same orientation;
• Any two triangles in \({\mathcal T}(P)\) are either disjoint, intersect at a point, or share an edge;
• The union of all of the underlying triangles in \({\mathcal T}(P)\) is equal to \(P\).

Understand triangulation with the picture below.

Practice creating triangulations with the next example.

Example 25

Take \(\big((1,1), (5,5), (0,4), (-3,7), (-1,-2)\big)\) to be the ordered vertex set of a polygon \(P\). Sketch a triangulation for \(P\).

Here are the vertices of \(P\).

Here is one triangulation of \(P\): \[{\mathcal T}(P)=\{\Delta((1,1),(5,5)(0,4)), \Delta((-1,-2),(1,1)(0,4)), \Delta((0,4)(-3,7)(-1,-2))\}\]

\(\Delta((1,1),(5,5)(0,4))\)

\(\Delta((-1,-2),(1,1)(0,4))\)

\(\Delta((0,4)(-3,7)(-1,-2))\)

Based on the last example, it seems like it should be possible to create a triangulation. Is that true? What more can we say?

Now we present the following facts.

Take \(P\) to be a polygon.

• Fact 1: Every polygon has a triangulation.

• Fact 2: Any triangulation \({\mathcal T}(P)\) for a polygon \(P\) determines an orientation for \(\partial P\) and may be chosen to agree with \(\partial_oP\).

• Fact 3: If two oriented triangles in a triangulation \(T(P)\) share an edge \(\overline{pq}\), then one of the triangles will have \(\overrightarrow{pq}\) in its oriented boundary and one will have \(\overrightarrow{qp}\) in its oriented boundary. Furthermore, only two triangles may share the same edge

With these facts in mind, practice constructing triangulations with a specific orientation with the next example.

Example 26

Take \(\big((0,-1), (-3,1), (-1,4), (3,3), (2,0)\big)\) to be the ordered vertex set of a polygon \(P\).

1. Identify a positively oriented triangulation for \(P\) and identify all of the triangles in this triangulation.
2. Identify a negatively oriented triangulation for \(P\) and identify all of the triangles in this triangulation.

Here are the vertices of \(P\).

1. Here is one triangulation that is positively oriented: \({\mathcal T}(P)=\{\Delta((0,-1),(2,0)(3,3)), \Delta((3,3),(-1,4)(0,-1)), \Delta((-1,4)(-3,1)(0,-1))\}\)

2. Here is one triangulation that is negatively oriented: \({\mathcal T}(P)=\{\Delta((0,-1)(3,3)(2,0)), \Delta((3,3)(0,-1)(-1,4), \Delta((-1,4)(0,-1)(-3,1))\}\)

We now begin the first steps for defining the area of a polygon. First, we define the \(\alpha\) quantity for an edge and then a triangle.

Oriented Edge and \(\alpha\)

For any oriented edge \(\overrightarrow{pq}\), define \(\alpha\!\left(\overrightarrow{pq}\right)\) by \[\alpha\!\left(\overrightarrow{pq}\right) = x(p)y(q) - y(p)x(q).\]

For any triangle \(\Delta pqr\), \[\alpha\!\left(\Delta pqr\right) = \alpha\!\left(\overrightarrow{pq}\right) + \alpha\!\left(\overrightarrow{qr}\right) + \alpha\!\left(\overrightarrow{rp}\right).\]

The quantity \(\alpha\!\left(\Delta pqr\right)\) is actually a function on the oriented boundary of \(\Delta pqr\)

Introduce the following useful notation:

Take \(S\) to be a finite set and for each \(x\) in \(S\), take \(f(s)\) to be a real number.

Denote by the symbol \[\sum_{s\in S}f(s)\] the sum of all values \(f(s)\), where \(s\) is in \(S\).

Example 27

Take \(S\) to be the set that is given by \[S = \{(1,2), (1,4), (3,5), (2,1), (9,3), (-4, 2), (1, -5)\}.\] For any ordered pair \((a,b)\), take \(f(a,b)\) to equal \(a\) and compute the sum \[\sum_{p\in S}f(p).\]

In this example, \[f(1,2)=1,f(1,4)=1,f(3,5)=3,f(2,1)=2,f(9,3)=9,f(-4,2)=-4,f(1,-5)=1.\]

Therefore,

\[ \begin{align*} \sum_{p\in S}f(p)&=f(1,2)+f(1,4)+f(3,5)+f(2,1)+f(9,3)+f(-4,2)+f(1,-5)\\ &=1+1+3+2+9-4+1\\ &=13. \end{align*} \]

Now take \(P\) to be a polygonal curve with an ordered vertex set \((p_1, \cdots, p_n)\) and take \({\mathcal T}(P)\) to be a triangulation of \(P\).

All triangles in \({\mathcal T}(P)\) have the same orientation, and so \[\left|\sum_{\Delta pqr \in {\mathcal T}(P)} \alpha\!\left(\Delta pqr\right)\right| = \sum_{\Delta pqr \in {\mathcal T}(P)} \left|\alpha\!\left(\Delta pqr\right)\right|.\]

The area of \(P\) is the sum of the areas of all of the triangles in \({\mathcal T}(P)\), therefore \[{\mathcal A}(P) = \frac{1}{2}\left|\sum_{\Delta pqr \in {\mathcal T}(P)} \alpha\!\left(\Delta pqr\right)\right|.\]

The contribution of any internal edge of a triangle in the triangulation to the above sum is 0 because each such edge appears in two different triangles and has opposite orientation in each, as seen in this picture:

Shoelace Formula

Take \(P\) to be a polygonal curve with an ordered vertex set \((p_1, \cdots, p_n)\) and take \({\mathcal T}(P)\) to be a triangulation of \(P\). The area of \(P\) is, therefore, the function of \(\partial_oP\) that is given by \[\begin{align*}{\mathcal A}(P) = \frac{1}{2}\left(\alpha\!\left(\overrightarrow{p_1p_2}\right) + \cdots + \alpha\!\left(\overrightarrow{p_{n}p_1}\right)\right).\end{align*}\]

This is the Shoelace Formula for the area of a polygon.

If \(\big((x_1, y_1), \dots, (x_n, y_n)\big)\) is an positively oriented ordered vertex set for \(P\), then \[{\mathcal A}(P) = \frac{1}{2}\left(x_1y_2 + x_2y_3 + \cdots + x_ny_1 - y_1x_2 - y_2x_3 - \cdots - y_nx_1\right)\]

The evocative name of the theorem comes from the following visualization of the formula.

Practice using the shoelace formula with the following example.

Example 28

Take \(\big((-4,1), (-1,3), (2,2), (0,5), (-3,4)\big)\) to be the ordered vertex set of a polygon \(P\). Take \(\partial_oP\) to be the oriented boundary that is determined by the given ordered vertex set. Use the shoelace formula to determine the area of \(P\) and the orientation of \(\partial_oP\).

Here \(x_1=-4,x_2=-1,x_3=2,x_4=0,x_5=-3\) and \(y_1=1,y_2=3,y_3=2,y_4=5,y_5=4\). Use the shoelace formula to get

\[ \begin{align*} \mathcal{A}(P)&=\tfrac{1}{2}\left((-4)\cdot3+(-1)\cdot 2+(2)\cdot 5+(0)\cdot 4+(-3)\cdot(1)-(1)\cdot(-1)-(3)\cdot(2)-(2)\cdot(0)-5\cdot(-3)-4\cdot(-4)\right)\\ &=\tfrac{1}{2}\left(-12-2+10+0-3+1-6-0+15+16\right)\\ &=\tfrac{19}{2} \end{align*} \]

Use the above answer to conclude the orientation is positive and the area is \(9.5.\)

If a given polygon \(P\) is the union of two polygons \(P_1\) and \(P_2\) that overlap only along a part of their boundary, then the union of the triangulations of \(P_1\) and \(P_2\) is a triangulation of \(P\) and no triangle lies both triangulations.

The area of \(P\) is determined by the sum of areas in any triangulation of \(P\) and so \[{\mathcal A}(P) = {\mathcal A}(P_1) + {\mathcal A}(P_2).\]

Since the area of any triangle is unchanged by rigid motions, a fact that can be directly verified, the same is also true of the area of any polygon.

Example 29

Take \(\big((-4,1), (-1,3), (2,2), (0,5), (-3,4)\big)\) to be the ordered vertex set of a polygon \(P\).

1. Use the shoelace formula to determine the area of \(P\).
2. Sketch \(P\) and find a triangulation of \(P\).
3. Determine the area of \(P\) by calculating the area of the triangles in the triangulation that you found in b.

1. The conclusion of the previous example was \(\mathcal{A}(P)=\tfrac{19}{2}\).

2. Here is \(P\).

Here is an example of triangulation: \[{\mathcal T}(P)=\{\Delta((-1,3),(2,2)(0,5)), \Delta((0,5),(-3,4)(-1,3)), \Delta((-3,4)(-4,1)(-1,3))\}\]

\(\Delta((-1,3),(2,2)(0,5)\)

\(\Delta((0,5),(-3,4)(-1,3))\)

\(\Delta((-3,4)(-4,1)(-1,3))\)

3. \[\begin{align*}\mathcal{A}(P)&=\mathcal{A}(\Delta((-1,3),(2,2)(0,5))+\mathcal{A}(\Delta((0,5),(-3,4)(-1,3)))+\mathcal{A}(\Delta((-3,4)(-4,1)(-1,3)))\\ &=\frac{7}{2}+\frac{5}{2}+\frac{7}{2}\\ &=\frac{19}{2}. \end{align*}\]

Same answer as previous example.

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