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Riemann Sum of semi-sircle

  1. Feb 6, 2005 #1
    Can anyone please direct me in the right way on working out the approximate area of a semi-circle with equation y = (r^2 - x^2)^0.5, by using a Riemann Sum
     
  2. jcsd
  3. Feb 6, 2005 #2

    dextercioby

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    What do you know about Riemann sums (meaning their general formula for the case of simple definite integrals)...?
    Chose a system of coordinates with the center at the left end of the semicircle,so that the Ox axis in its posotive part to comprise entire diameter.Therefore your equation for the curve will be slightly modified.

    Daniel.
     
  4. Feb 6, 2005 #3

    quasar987

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    Here's what I suggest..

    First find the domain of that function ([-3,3]). Then create an arbitrary partition of that domain. That is to say, select an arbitrary number of points in that domain and label them [itex]\{x_0, x_1,...x_n\}[/itex]. (x0 has to be -3 and xn has to be 3). In principle, the more points you chose, the better the approximation.

    Then construct and evaluate the Riemann sum

    [tex]\Sigma_{i=1}^{n} y(t_i)(x_{i}-x_{i-1})[/tex]

    Where the ti are arbitrarily chosen points in the interval [itex][x_{i-1},x_{i}][/tex]
     
  5. Feb 6, 2005 #4

    dextercioby

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    Yes,but i suggested him that all values of the partition be positive,the way you took'em half are and half are not...I think that should create some avoidable problems...

    Daniel.
     
  6. Feb 6, 2005 #5

    quasar987

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    Really? Like what? I don't see what the difference will be, since [itex]x_{i}-x_{i-1}[/itex] will be positive anyway.
     
  7. Feb 7, 2005 #6

    Aki

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    I got a cool program on my graphing calculator that does that for me. It's handy when it comes to test. If you want it, just pm me.
     
  8. Feb 7, 2005 #7

    dextercioby

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    I don't know who you were talking to,but if,by chance,u were talking to me,learn that i don't have a graphing computer and that's why the software would be totally useless... :yuck:

    Thanks for the offer,though... :wink: :tongue2:

    Daniel.
     
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