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Math Help - problem with line integral

  1. #1
    Junior Member
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    problem with line integral

    hi, this is my problem:

    i need to find the mass of the curve (which is shown in parametric form):

    x = e^tcos(t), y = e^tsin(t), z = e^t

    from t=0 to t=a

    and the mass density of the line is reversely proportional to

    x^2 + y^2 + z^2

    and also \rho (1,0,1) = 1


    i am having trouble with it - is reversely proportional means

    \frac{1}{x^2+y^2+z^2},

    and if so what am is supposed to do with \rho?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by vonflex1 View Post
    hi, this is my problem:

    i need to find the mass of the curve (which is shown in parametric form):

    x = e^tcos(t), y = e^tsin(t), z = e^t

    from t=0 to t=a

    and the mass density of the line is reversely proportional to

    x^2 + y^2 + z^2

    and also \rho (1,0,1) = 1


    i am having trouble with it - is reversely proportional means

    \frac{1}{x^2+y^2+z^2},

    and if so what am is supposed to do with \rho?
    Well, I've never seen "reversely" proportional. In fact I don't believe that "reversely" is a word in the English language! I suspect this is a mis-translation of "inversely proportional" which means just what you say: a multiple of \frac{1}{x^2+ y^2+ z^2}.

    " \rho(x,y,z)" is the density function. While it is not a standard notation, since it is the only function given, I am sure that must be the case- and I suspect that previous examples in your text have used " \rho" to mean the density function. You are told two things:
    1) \rho= \frac{k}{x^2+ y^2+ z^2} and
    2) \rho(1, 0, 1)= \frac{k}{1^2+ 0^2+ 1^2}= 1
    You can use that to find k and so find the full density function.

    Of course, the mass is just the density function integrated over the curve.
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  3. #3
    Junior Member
    Joined
    Jun 2009
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    thanks,

    i did mean inversely, it's just first time i have to translate that expression
    to english.

    in any case, my problem was that i didn't know about that k...
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