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

[TranscendArcu]

Homework Statement

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), ... , (x_ny_n). Show that the area of the polygon is

(1/2) * ((x₁y₂ - x₂y₁) + (x₂y₃ - x₃y₂) + ... + (x_ny₁ - x₁y_n))

Homework Equations

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

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), ... , (x_ny_n). 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"

 

[Mark44]

\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]

 

[TranscendArcu]

Excellent! That's absolutely helpful. Thank you.

I'm thinking about this integral here:

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.

 

[Dick]

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

 

[TranscendArcu]

Could you please show the calculations that lead you to that conclusion? Certainly there is no 1/2 preceding the "xdy-ydx" in the book, so if there is a typo, it'd be good to know about.

If you'd like to see the problem, it can be found on google books: http://books.google.com/books?id=Vo...he line segment connecting the point"&f=false

 

[Dick]

Could you please show the calculations that lead you to that conclusion? Certainly there is no 1/2 preceding the "xdy-ydx" in the book, so if there is a typo, it'd be good to know about.

If you'd like to see the problem, it can be found on google books: http://books.google.com/books?id=Vo...he line segment connecting the point"&f=false

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)?

 

[TranscendArcu]

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

 

[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.

 

[Dick]

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

You took the words out my mouth. :)