Green’s Theorem as a planimeter

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A planimeter computes the area of a region by tracing the boundary.

Green’s Theorem may seem rather abstract, but as we will see, it is a fantastic tool for computing the areas of arbitrary bounded regions. In particular, Green’s Theorem is a theoretical planimeter. A planimeter is a “device” used for measuring the area of a region. Ideally, one would “trace” the border of a region, and the planimeter would tell you the area of the region.

How is Green’s Theorem a planimeter?

Recall Green’s Theorem:

Green’s Theorem If the components of $\vec {F}:\R ^2\to \R ^2$ have continuous partial derivatives and $C$ is a boundary of a closed region $R$ and $\vec {p}(t)$ parameterizes $C$ in a counterclockwise direction with the interior on the left, then $\iint _R \curl \vec {F}\d A = \oint _C \vec {F}\dotp \d \vec {p}$

Given a vector field $\vec {F}:\R ^2\to \R ^2$ , if $\curl \vec {F} = 1$ , then the left-hand side of the conclusion of Green’s Theorem gives the area of the region $R$ : $\iint _R \curl \vec {F}\d A = \iint _R \d A$

So now the question becomes, which vector fields have $\curl \vec {F} = 1$ ? Here are three basic candidates:

$\vec {F}(x,y) = \vector {0,x}$
$\vec {F}(x,y) = \vector {-y,0}$
$\vec {F}(x,y) = \vector {-y/2,x/2}$

The key idea that connects the three vector fields above is:

Their curl is $1$ . They are conservative fields. They are gradient fields. When used in combination with Green’s Theorem, they help compute area.

Once we have a vector field whose curl is $1$ , we may then apply Green’s Theorem to use a line integral to compute the area.

You must parameterize $C$ with $\vec {p}(t)$ for $a\le t\le b$ such that:

$C$ is drawn in a counterclockwise direction.
$C$ is drawn exactly once.
The interior of $R$ is to the left of the direction of $\vec {p}'(t)$ .

Computing areas with Green’s Theorem

Now let’s do some examples.

Compute the area of the trapezoid below using Green’s Theorem.

In this case, set $\vec {F}(x,y) = \vector {0,x}$ . Since $\curl \vec {F} = 1$ , Green’s Theorem says: $\iint _R \d A = \oint _C \vector {0,x}\dotp \d \vec {p}$ We need to parameterize our paths in a counterclockwise direction. We’ll break it into four line segments each parameterized as $t$ runs from $0$ to $1$ :

where: $\begin{align*} \vecl_1(t) &= \vector{4t,\answer[given]{0}}\\ \vecl_2(t) &= \vector{\answer[given]{4-t},2t}\\ \vecl_3(t) &= \vector{3-2t,\answer[given]{2}}\\ \vecl_4(t) &= \vector{\answer[given]{1-t},2-2t} \end{align*}$

and each draws the line as $t$ runs from $0$ to $1$ . Write:

$\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} = \int_0^1 &\vec{F}(\vecl_1(t))\dotp \vecl_1'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_2(t))\dotp \vecl_2'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_3(t))\dotp \vecl_3'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_4(t))\dotp \vecl_4'(t) \d t \end{align*}$

For each of the integrands above, say $\vecl (t) = \vector {x(t),y(t)}$ , we will write $\vec {F}(\vecl (t))\dotp \vecl '(t) = \vector {0,x(t)} \dotp \vector {x'(t),y'(t)}$ and combine them into a single integral. Write with me

$\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} &= \int_0^1 \left(\answer[given]{2(4-t) -2(1-t)} \right)\d t\\ &= \eval{\answer[given]{6t}}_0^1\\ &=\answer[given]{6}. \end{align*}$

So this is the area of the trapezoid.

Compute the area of the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ using Green’s Theorem.

To start, we’ll set $\vec {F}(x,y) = \vector {-y/2,x/2}$ . Since $\curl \vec {F} = 1$ , Green’s Theorem says: $\iint _R \d A = \oint _C \vector {-y/2,x/2}\dotp \d \vec {p}$ We can parameterize the boundary of the ellipse with $\begin{align*} x(t) &= a \cos(t)\\ y(t) &= b \sin(t) \end{align*}$

for $0\le t<2\pi$ . Write with me

$\begin{align*} \iint_R \d A &= \oint_C \vector{-y/2,x/2}\dotp\vector{x'(t),y'(t)}\d t\\ &= \int_0^{2\pi} \vector{\answer[given]{-b\sin(t)/2},\answer[given]{a\cos(t)/2}}\dotp\vector{\answer[given]{-a\sin(t)},\answer[given]{b\cos(t)}}\d t\\ &= \int_0^{2\pi} (1/2)\left(ab\sin^2(t) + ab\cos^2(t)\right) \d t\\ &= \int_0^{2\pi} (1/2)ab \d t\\ &= \answer[given]{\pi ab}. \end{align*}$

Done. Green’s Theorem for the win!

Finally, what do you do if you have a very strangely shaped curve? You approximate it with a polygonal curve. Check out next example.

Compute the area of the polygonal region below using Green’s Theorem.

In this case, set $\vec {F}(x,y) = \vector {0,x}$ . Now we’ll break this curve into six line segments that draw this curve in a counterclockwise fashion as a parameter $t$ runs from $0$ to $1$ .

where: $\begin{align*} \vecl_1(t) &= \vector{1+2t,\answer[given]{-1+2t}}\\ \vecl_2(t) &= \vector{\answer[given]{3-3t},1}\\ \vecl_3(t) &= \vector{-t,\answer[given]{1+2t}}\\ \vecl_4(t) &= \vector{-1-t,\answer[given]{3-4t}}\\ \vecl_5(t) &= \vector{\answer[given]{-2+3t},-1-2t}\\ \vecl_6(t) &= \vector{1,\answer[given]{-3+2t}} \end{align*}$

and each draws the line as $t$ runs from $0$ to $1$ . Write:

$\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} = &\int_0^1 \vec{F}(\vecl_1(t))\dotp \vecl_1'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_2(t))\dotp \vecl_2'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_3(t))\dotp \vecl_3'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_4(t))\dotp \vecl_4'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_5(t))\dotp \vecl_5'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_6(t))\dotp \vecl_6'(t) \d t \end{align*}$

$\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} &= \int_0^1 \left(\answer[given]{12} \right)\d t\\ &= \eval{\answer[given]{12t}}_0^1\\ &=\answer[given]{12}. \end{align*}$

So this is the area of the region.

Green’s Theorem gives a fairly easy method for computing any the area of any polygonal region. Any region with a “smooth” border can be approximated by a polygonal region. The upshot? Green’s Theorem is a powerful tool for computing area.

The shoelace algorithm

Green’s Theorem can also be used to derive a simple (yet powerful!) algorithm (often called the “shoelace” algorithm) for computing areas. Here’s the idea: Suppose you have a two-dimensional polygon, where the vertices are identified by their $(x,y)$ -coordinates:

Here we see a polygon with $n$ vertices, and the “dashed-line” means that there could be more to this polygon than “meets the eye.” So to compute the area, here is a neat trick. Write all of the coordinates in a column, writing the starting coordinate twice, both at the beginning and at the end:

Now multiply entries of the columns diagonally down from the left to the right and add them together

to obtain: $x_1y_2 + x_2y_3 + x_3y_4 + \dots + x_ny_1$ Now, multiply entries of the columns diagonally down from the left to the right and subtract them from our previous sum

to obtain: $\begin{align*} S=x_1y_2 &+ x_2y_3 + x_3y_4 + \dots + x_ny_1 \\ &-x_2y_1 - x_3y_2 - x_4y_3 - \dots - x_1y_n \end{align*}$

The area of the polygon in question will be: $A = \frac {|S|}{2}$ The algorithm is called the “shoelace” algorithm because of the crisscrossing pattern you see above.

Compute the area of the following polygon:

The area is $\answer {47}$ square units.

Why does the shoelace algorithm work?

Now we are going explain why the shoelace algorithm works via Green’s Theorem. The restrained young mathematician may protest that we are using a “crane to crush a fly,” but whatever. We like Green’s Theorem.

Shoelace Given a polygon $P$ with vertices at $(x_1,y_1),(x_2,y_2),\dots ,(x_n,y_n)$ we may compute the area of the polygonal region $R$ by setting $(x_{n+1},y_{n+1}) = (x_1,y_1)$ and computing: $\text {area} = \frac {1}{2}\left |\sum _{i=1}^n \left (x_iy_{i+1} - x_{i+1}y_i\right )\right |$

To start, recall that if $\vec {F} = \vector {0,x}$ , then $\curl \vec {F}(x,y) = 1$ . Hence Green’s Theorem states: $\iint _R \d A = \oint _P x\d y$ This means we can compute the area of the region $R$ , by evaluating the line integral on the right along the polygonal boundary $P$ . Since we’re supposing that the vertices of the polygon are $(x_1,y_1),(x_2,y_2),\dots ,(x_n,y_n)$ $P$ can be broken into $n$ edges, (if a polygon has $n$ vertices, it has $n$ edges). We can parameterize each edge as $\begin{align*} E_1 &: \vector{x_1,y_1} + t \vector{x_2-x_1,y_2-y_1}\\ E_2 &: \vector{x_2,y_2} + t \vector{x_3-x_2,y_3-y_2}\\ E_3 &: \vector{x_3,y_3} + t \vector{x_4-x_3,y_4-y_3}\\ &\vdots\\ E_n &: \vector{x_n,y_n} + t \vector{x_n-x_1,y_n-y_1} \end{align*}$

with $0\le t<1$ . Moreover

$\begin{align*} \iint_R \d A &= \oint_P x\d y\\ &= \sum_{i=1}^n \int_{E_i} x\d y, \end{align*}$

this is just saying that the line integral along the perimeter of the polygon is the sum of the line integrals along the edges. Now write with me

$\begin{align*} \int_{E_i} x \d y &= \int_0^1 (x_i + t x_{i+1}-t x_i) (y_{i+1}-y_i) \d t\\ %&= \int_0^1 (x_i y_{i+1}+ t x_{i+1} y_{i+1}-t x_iy_{i+1} - x_iy_i - tx_{i+1}y_i+tx_iy_i) \d t\\ %&= x_i y_{i+1}+ \frac{x_{i+1} y_{i+1}}{2}- \frac{x_iy_{i+1}}{2} - x_iy_i - \frac{x_{i+1}y_i}{2}+\frac{x_iy_i}{2} \\ &=\frac{1}{2}(x_i y_{i+1}+ x_{i+1} y_{i+1} - x_iy_i - x_{i+1}y_i) \end{align*}$

So now we may write

$\begin{align*} \iint_R \d A &= \oint_P x\d y\\ &= \sum_{i=1}^n \int_{E_i} x\d y\\ &= \frac{1}{2}\sum_{i=1}^n (x_i y_{i+1}+ x_{i+1} y_{i+1} - x_iy_i - x_{i+1}y_i) \end{align*}$

Noting that the terms $x_{i+1} y_{i+1}$ and $x_iy_i$ will cancel with each other as we cycle through the sum, we find that

$\begin{align*} \iint_R \d A &= \oint_P x\d y\\ &= \frac{1}{2} \sum_{i=1}^n \left(x_iy_{i+1} - x_{i+1}y_i\right) \end{align*}$

Since this value relies on $P$ being parameterized in a counterclockwise fashion, we take the absolute value to ensure a correct answer (just in case the young geometer accidentally parameterized in a clockwise fashion). Thus we have completed the explanation of the shoelace algorithm.

A student is working with a pentagon:

and using the shoelace algorithm:

computes the area of the pentagon as: $\begin{align*} S&=2+10+9+8+24-8-6-20-18-2\\ &=-1 \end{align*}$

So the student concludes that the area is $\text {Area} = 1/2~\text {square units}$ Is this correct?

Yes No

What is the correct answer? $\text {Area} = \answer {6.5}~\text {square units}$

For some interesting extra reading check out:

The Surveyor’s Area Formula, B. Braden, College Math Journal, September 1986.