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Invariance of Area

  1. Jun 10, 2010 #1

    Quite some time ago I'd asked for help with a proof that proves that area of a closed curve is invariant i.e : its independent of the way it is spliced into.

    Say we splice a closed curve into one set of rectangles with parallel sides and we then splice an identical curve with rectangles with some different orientation, I basically sought to prove that area calculated by summing up areas in both cases would be equal.

    Here i present a proof

    I'd be grateful to members who could comment on the proof and check it for validity.
    Thanks :D

    Attached Files:

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  2. jcsd
  3. Jun 11, 2010 #2
    Please Help! 127 Views and not a single reply! :S

    Kindly Help!
  4. Jun 12, 2010 #3


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    I have no idea what you mean by "spliced into". "Sliced into"? Do you mean "divided into non-overlapping subregions"? Also, how are you defining the "area" of a plane region?
  5. Jun 12, 2010 #4
    Thanks a Lot for the reply! by splice i mean ''slice'' into non overlapping regions.

    Firstly defining the area of a rectangle as its length*breadth, and then for any general closed curve defining its area as the sum of areas of 'n',non overlapping rectangles that it can be divided into.Where n--> infinity
  6. Jun 12, 2010 #5
    Your proof fails because you can never get all the rectangles to be true rectangles so the formula lnwn + ln+1wn+1 + ... = A is not true. This is why you need a limit as w -> 0, and then you simply have integration which needs no proof.
  7. Jun 12, 2010 #6
    @ mu naught : yes that is why i said 'n' rectangles where n-->infinity.In integral calculus nowhere do we prove that the area is invariant (i mean irrespective of orientation of coordinate axes in the case of integration)
  8. Jun 12, 2010 #7
    It's tricky, but you need to be more precise than saying your definition of the area of a region A is the sum of the areas of n non-overlapping rectangles in A as n tends to infinity. What if I give you 10 rectangles that roughly provide the shape of the outer boundary of A and and then keep dividing the inner rectangles? The number of rectangles approaches infinity, but the area never gets closer to the intuitive area.
  9. Jun 12, 2010 #8
    Even if you do get this to work, all you will have proven is that the sum of the areas of some rectangles tends to the same limit as that of some differently oriented rectangles, as their number goes to infinity. If you're interested in defining area correctly and proving its properties, you need to learn about measure.
  10. Jun 13, 2010 #9
    @werg22 : Yes! This is exactly what i intended to prove,for which i have never seen a proof.
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