# Summing n-number of Terms to Find the Area of a Polygon

1. Nov 23, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
Let C be the line segment connecting the points (x1,y1) and (x2,y2). More over let the line integral over C of (x dy - y dx) = x1y2 - x2y1.

Suppose the vertices of a polygon, listed in counter-clockwise order, are (x1y1), (x2y2), ... , (xnyn). Show that the area of the polygon is

(1/2) * ((x1y2 - x2y1) + (x2y3 - x3y2) + ... + (xny1 - x1yn))

2. Relevant equations

I don't know a relevant equation, but I suspect there probably is one.

3. The attempt at a solution

So, basically, I just want to say something like, let C* be the set of all line segments that connect, with positive orientation, (x1y1), (x2y2), ... , (xnyn). Then using the fact that the line integral over C of (x dy - y dx) = x1y2 - x2y1, and by repeatedly applying this fact, I would have something like:

$\sum$ * = 1 n of ($\int$C* (x dy - y dx)). I think this gives the desired result except for the 1/2, which still eludes me.

Also, how do you format these sigmas? It's supposed to read "Sigma from *=1 to n"

2. Nov 23, 2011

### Staff: Mentor

$$\sum_{i = 1}^n \int_{C^*} (x~dy - y~dx)$$

The LaTeX looks like this (with the spaces in the tex tags removed):
[ tex]\sum_{i = 1}^n \int_C^* (x~dy - y~dx)[ /tex]

3. Nov 23, 2011

### TranscendArcu

Excellent! That's absolutely helpful. Thank you.

Suppose I had integrated line integral over C of (x dy - y dx) to prove that x1y2 - x2y1. Then the parameterization of the curve is r(t) = <x1 + (x2 - x1)*t, y1 + (y2 - y1)*t>. I can calculate a line integral:

C x dy - y dx = ∫(0≤t≤1) (x1 + (x2 - x1)*t)*(y2 - y1) - (y1 + (y2 - y1)*t)*(x2 - x1) dt.

Calculating gives,

(x1 + (1/2)(x2 - x1))*(y2 - y1) - (y1 + (1/2)(y2 - y1))*(x2 - x1).

This introduces at least some kind of (1/2) into my calculation. Not sure if that helps though, since the 1/2 just cancels.

4. Nov 23, 2011

### Dick

The area is the integral of (1/2)(xdy-ydx). I think you left the (1/2) out from the very beginning.

5. Nov 23, 2011

### TranscendArcu

6. Nov 23, 2011

### Dick

No, no typo really. It just says the integral along a line segment is x1*y2-y1*x2 and asks you to prove it. It doesn't say that that is the area. I think you were supposed to figure out that you should put the (1/2) in. Did they tell you someplace else in the book that dA=(1/2)(xdy-ydx)?

7. Nov 23, 2011

### TranscendArcu

Indeed they did! Guess I should probably start reading the book, huh?

8. Nov 23, 2011

### Dick

BTW the easiest proof I know is that the cross product of the three dimensional vectors (x,y,0) and (x+dx,y+dy,0) gives you the area of the parallelogram generated by (x,y,0) and (x+dx,y+dy,0) and the origin. That's xdy-ydx. The area of the triangle from the origin is (1/2) of that. That's dA. I'm sure there is a way to do it without introducing the third dimension, but that's the way I think about it.

9. Nov 23, 2011

### Dick

You took the words out my mouth. :)