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

  • #1

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

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

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), ... , (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:

[itex]\sum[/itex] * = 1 n of ([itex]\int[/itex]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"
 

Answers and Replies

  • #2
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[tex]\sum_{i = 1}^n \int_{C^*} (x~dy - y~dx)[/tex]

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
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.
 
  • #4
Dick
Science Advisor
Homework Helper
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The area is the integral of (1/2)(xdy-ydx). I think you left the (1/2) out from the very beginning.
 
  • #5
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=Vou3MZu_7tcC&pg=PA940&lpg=PA940&dq="is+the+line+segment+connecting+the+point"&source=bl&ots=y720QtuErR&sig=tnQaVx7A5RbyzuylbM-gaws2Lp4&hl=en&ei=3sbNTu-ENsj10gHY7NhN&sa=X&oi=book_result&ct=result&resnum=3&ved=0CC0Q6AEwAg#v=onepage&q="is the line segment connecting the point"&f=false
 
  • #6
Dick
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Homework Helper
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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=Vou3MZu_7tcC&pg=PA940&lpg=PA940&dq="is+the+line+segment+connecting+the+point"&source=bl&ots=y720QtuErR&sig=tnQaVx7A5RbyzuylbM-gaws2Lp4&hl=en&ei=3sbNTu-ENsj10gHY7NhN&sa=X&oi=book_result&ct=result&resnum=3&ved=0CC0Q6AEwAg#v=onepage&q="is the 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)?
 
  • #7
Indeed they did! Guess I should probably start reading the book, huh?
 
  • #8
Dick
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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
Dick
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Indeed they did! Guess I should probably start reading the book, huh?
You took the words out my mouth. :)
 

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