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Calculating the perimeter of a region using integrals

  1. Feb 5, 2007 #1
    I was wondering,how do we calculate the perimeter of a region using integral calculus?I know that to calculate the area we have to draw the region and if we want the reunion of those regions we have to sum them and use some values to define the boundaries of the integral.
    Any help on this is really appreciated!Thanks in advance!
  2. jcsd
  3. Feb 5, 2007 #2


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    In the most general terms, the length of a perimeter (from the way you stated the question, I am assuming you are talking about a region of the plane, where the perimeter is a curve) is given as the integral of ds,where:


    In practice, you would use:

  4. Feb 5, 2007 #3


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    In other words, you are talking about arclength. Use mathman's formula or, if x and y are given in terms of a parameter, t, use
    [tex]ds= \sqrt{\left(\frac{dx}{dt}\right)+ \left(\frac{dy}{dt}\right)^2}ds[/tex]

    If the perimeter is given by two or more different formula (as a region defined to be between two curves) you will need to integrate each separately and then add.
  5. Feb 5, 2007 #4
    Sorry for my "ignorence" but, what does de ds means?Is it the integral?
  6. Feb 6, 2007 #5


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    No, "ds" is the length of the tiny tangent vector at one point of the curve.
    We add all those tiny lengths up to get the entire curve's length.
  7. Feb 6, 2007 #6

    Gib Z

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    I have a feeling esmeco wants a appraoch similar to the way he posted the integral for areas.

    For the area, we cut up the region into tiny rectangles yes? well the integral basically makes it perfectly accurate by making lots and lots of rectanles. For that to happen, the base has to be smaller. When its really small, we call it dx. a small change in x. a small change in height would then be dy.

    if we took those small changes in x and y, and applied pythagoras to them to get the straight line that connects them. That imitates the perimeter abit. Then we integrate it to sum all of the small ones. :)
  8. Feb 6, 2007 #7
    So, if I have a function like 0<=y<=sqrt(x+2) to find the perimeter of that region I have to make a tangent to the curve and use the pithagoras formula?Like in the case,the lenght of the catects would be two, so it should be something like:

    ds=sqrt(2^2 + 2^2)=sqrt8

    Is this correct?But what do I do now with the hipothenus value?
  9. Feb 7, 2007 #8

    Gib Z

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    Im going to ignore the < signs and the 0 because its not longer a function then.

    if [tex]y=\sqrt{x+2}[/tex]. Try drawing the graph. Then zoom in. It gets abit straighter yea? Then in even more, it'll look even more straight. Since a tangent straight line then becomes a good approximation of that part, we make a straight line that hugs the curve. To do that, we take the lines vertical height, and horizontal height. From those 2, and Pythagoras Theorem, we get the length of the line that hugs the curve. The tangent will vary in different places, and this method of doing this is very long, taxing and sort of in accurate. With Integral calculus, we can do this process perfectly accurately and very quickly.

    But you haven't done Differential Calculus yet, so your at least a good few months off this.
  10. Feb 7, 2007 #9

    Gib Z

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    Um actually i've seen some of your other posts, and it seems you might be further ahead in mathematics than i am lol. If your doing differential equations and asking for the primitive of sin^3 x, perhaps you'll understand this formula.

    [tex]\int^b_a \sqrt{1+ (y')^2} dx[/tex]. Thats the perimeter of the function y, from b to a. I tried use prime notation rather than Leibniz, but thats hard to do with Integrals lol.

    From your other forums, it seems you haven't encountered integrals, but rather only primitives.
  11. Feb 7, 2007 #10


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    That's the length of the graph of y=y(x), from x= a to x= b. The word "perimeter" is reserved for closed geometric figures.
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