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How do I evaluate double integral as the limit of a sum

  1. Aug 2, 2009 #1
    How do I evaluate double integral as the limit of a sum: [tex]\int\int[/tex] 1 dA with a snowflake region constructed as follows:
    Step 1: Start with a square of area 1 unit2.
    Step 2: Divide each edge into 3 and construct a smaller square on the middle third, thus creating new edges.
    Step 3: Repeat step 2.
    I know to use a simple geometric series but not sure how to use it or work put this integral.
    Please help.
  2. jcsd
  3. Aug 3, 2009 #2


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    First try to find the answer to this: in the nth step, how many new triangles do you add. What is the length of the sides of these new triangles? Then what is their surface? What is the total surface area added in this step?
  4. Aug 3, 2009 #3
    Ignore the integral crap. What you're doing here is simply finding the area inside a fractal.

    The three steps you have been given produces the iterations of the fractal. You can think of this as a sequence {S_i} of shapes (where S_0 is a square). From this sequence, you can produce another sequence {A_i}, where A_i is the area of S_i (so A_0 = 1, since the area of a unit square is 1).

    The question is to find the limit of {A_i}.

    To do this, there are two steps:

    Step 1) Find the formula for A_i.
    Step 2) Find the limit of A_i.

    Step 1 is going to be finding a recursive formula for A_i. That means something like...

    A_0 = 1
    A_(i+1) = A_i + (something something), for i > 0

    Once you have this formula, choose a really big number n and figure out what A_n is on your calculator. This will give you an idea of what lim {A_i} is. Once you think you know the limit, try and find a proof.
  5. Aug 6, 2009 #4
    Did you get there in the end? There's a nice walkthough here of the process of calculating the area of the classic triangular version.


    Tac-Tics's step 1 is the same, in principle, for a square with square growths, it's just the numbers that are different. It's simpler for a square because the area is 1 when the side-length is 1. Step 2, in either case, is a matter of finding the sum of a geometric series [ http://en.wikipedia.org/wiki/Geometric_series#Sum ]. I used Wolfram Alpha to check my answer.


    (One quirk I've found though is that sometimes, with a complicated formula, it only gives me the infinite sum when I don't type ", for n=0 to infinity" after it!)

    Another page about the Koch snowflake (the classic triangular variant of the Koch island fractal) that I came across purports to calculate the area for the case where the area of the initial triangle is 1. I think the principles are right here, but I reckon they've made a numerical mistake (it should be 48 triangles added on the third iteration) which affects their result. I get 1.6 times the area of the initial triangle.

    Last edited by a moderator: Apr 24, 2017
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