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Math Help - Integration as summation

  1. #1
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    Integration as summation

    Hi,

    i'm a third year engineering student and still I haven't got rid of the maths, so a word of advice: start loving maths as you'll need it for the rest of your career.

    anyway, here's my problem:

    http://img189.imageshack.us/img189/3096/imgelm.jpg

    this is strictly speaking a fluid dynamics question, and usually we are given the relationship of the velocity and integrate accordingly. this time though we are given a table and are expect to use summation as a means of integration. I think i have got the hang of it but I still have some doubts, especially for the inner circle. I wrote down my difficult on the paper too.


    thanks for the help gurus, much appreciated
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  2. #2
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    Your paper is huge and takes up too much space. That is probably why no one has responded. Just saying.
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  3. #3
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    Quote Originally Posted by cincin View Post
    no one??
    In addition,why don't type out the question.
    I for one, do not open attachements in general.
    I think that you should be willing to type it out in LaTeX.
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  4. #4
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    how can i type in laTeX please? can I include diagrams also? because thats why i scanned it too so you can see what my problem is as it isn't strictly speaking mathematics problem.

    or maybe some other imagehosting websites you prefer?

    thanks for the replies
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  5. #5
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    Quote Originally Posted by cincin View Post
    how can i type in laTeX please? can I include diagrams also? because thats why i scanned it too so you can see what my problem is as it isn't strictly speaking mathematics problem.

    or maybe some other imagehosting websites you prefer?
    LaTeX

    Yes, you can use imagehosting websites.
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  6. #6
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    cheers, here it goes:

    I made the image much smaller showing only the table and diagram.
    http://img13.imageshack.us/img13/8528/imgcopyu.jpg

    Since the pipe is circle, A = \pi r^2, therefore dA = 2 \pi r \, dr

    Using the above table of results therfore:

    \int u \, dA = 2 \pi \int u dr = \sigma 2 \pi u dr

    Now

    \sigma 2 \pi u dr = 2 pi R_1 (R_1-0) + 2 \pi R_1 (R_2-R_1) + 2 \pi R_2 (R_3-R_2) + 2 \pi R_3 (R_4-R_3)

    I am understanding more or less what is happening, but what I am not understanding is why the first part wasnt taken as \pi r^2 since the strip is a solid circle not "hollow tubes" if you understand what i mean.

    thanks
    Last edited by mr fantastic; January 15th 2010 at 03:39 PM. Reason: Fixed the latex as best as I can ( I can't absolutely guarantee the equations are all exactly as intended).
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