## Finding the Area of an Irregular Polygon

Finding the area of an irregular polygon via geometry can be a bit of a chore, as the process depends heavily on the shape of the polygon.  It turns out, however, that there’s a formula that can give you the area of any polygon as long as you know the polygon’s vertices.  That formula is based on Green’s Theorem.

Green’s Theorem is a powerful tool in vector analysis that shows how to convert from a line integral around a closed curve to a double integral over the region enclosed by the curve and vice versa.  More specifically, the circulation-curl form of Green’s Theorem states that, if D is a closed region with boundary curve C and C is oriented counterclockwise, then $\displaystyle \oint_C (M dx + N dy) = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA$.

The special case of Green’s Theorem that can generate our formula for the area of a polygon has $N = x$ and $M = 0$.  This yields $\displaystyle \oint_C x dy = \iint_D dA$.

Since the double integral here gives the area of region D, this equation says that we can find the area of D by evaluating the line integral of $x dy$ counterclockwise around the curve C.  The boundary curve for a polygon consists of a finite set of line segments, though, and so to obtain our formula we just need to find out what $\oint_L x dy$ is for a line segment L that runs from a generic point $(x_1, y_1)$ to another generic point $(x_2, y_2)$.  Let’s do that now.

We need a parameterization of L.  That requires a point on the line segment L and a vector in the direction of L.  The starting point $(x_1, y_1)$ and the vector $(x_2 - x_1){\bf i} + (y_2 - y_1){\bf j}$ from the starting point to the ending point work well, giving us the parameterization $\displaystyle {\bf r}(t) = (x_1 + t(x_2-x_1)){\bf i} + (y_1 + t(y_2-y_1)){\bf j}, \: 0 \leq t \leq 1.$

With $x = x_1 + t(x_2-x_1)$ and $y = y_1 + t(y_2-y_1)$, we have $dy = (y_2 - y_1)dt$.  This gives us the integral $\displaystyle \int_0^1 (x_1 + t(x_2-x_1))(y_2-y_1)dt \\ = (y_2-y_1)\int_0^1 (x_1 + t(x_2-x_1))dt \\ = (y_2-y_1)\left[tx_1 + \frac{t^2}{2}(x_2-x_1)\right]_0^1 \\ = (y_2-y_1)\frac{x_1+x_2}{2}.$

In other words, the line integral of $x dy$ from $(x_1, y_1)$ to $(x_2, y_2)$ is $\displaystyle (y_2-y_1)\frac{x_1+x_2}{2},$ the difference in the y coordinates times the average of the x coordinates.

Let’s put all this together to get our formula.  Suppose we have a polygon D with n vertices. Start with any vertex.  Label it $(x_1, y_1)$.  Then move counterclockwise around the polygon D, labeling successive vertices $(x_2, y_2), (x_3, y_3)$, and so forth, until you label the last vertex $(x_n, y_n)$.  Finally, give the starting vertex a second label of $(x_{n+1}, y_{n+1})$.  Then the area of D is given by $\displaystyle \text{Area of } D = \sum_{i=1}^n (y_{i+1} - y_i) \frac{x_i + x_{i+1}}{2}.$

In words, this means that to find the area of polygon D you can just take the difference of the y coordinates times the average of the x coordinates of the endpoints of each line segment making up the polygon and then add up.

This formula isn’t the only such formula for finding the area of a polygon.  In fact, a more common formula that can also be proved with Green’s Theorem is $\displaystyle \text{Area of } D= \frac{1}{2} \sum_{i=1}^n (x_i y_{i+1} - x_{i+1}y_i)$.

This second formula looks a little nicer to the eye, but I prefer the first formula for calculations by hand.

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### 1 Response to Finding the Area of an Irregular Polygon

1. velikod says:

WoW! That is beautiful!