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Integral of a function

  1. Mar 18, 2012 #1
    How do I compute the integral

    ∫z ∂u(z)/∂z dz

  2. jcsd
  3. Mar 18, 2012 #2


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    Hey crocomut and welcome to the forums.

    Are you aware of integration by parts?
  4. Mar 18, 2012 #3
    Yes, this is what I attempted and I got

    zu - ∫udz

    Does it make sense?
  5. Mar 18, 2012 #4


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    It does but I'm wondering what kind of context this question is in?

    For example are you asked to find it out in terms of the function u? Are you given u or the partial derivative? Are you given the integral of u with respect to z and thus have to find out the integral given only this?

    If you have this kind of information it would be helpful for us to know this since this context helps dictate what kind of answer you should provide.
  6. Mar 18, 2012 #5
    This is not a homework question or anything like that, I am just trying to learn about the log-law of the wall in fluid dynamics.

    So in reality what I have to do is solve for u in the following expression (there are constants in there but I removed for clarity):

    ∂u/∂z = 1/z

    The way this is solved in the book is u = ln(z) + C, so they integrate the left hand side and right hand side separately. However, if I solve it another way, i.e. if I put z on the left I get:

    z∂u/∂z - 1 = 0

    Integrating by parts I get zu - ∫udz - z + C = 0 from which I am not able to arrive at u = ln(z) + C. That is what is confusing me.

  7. Mar 18, 2012 #6


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    The thing you have to be careful about is the dimension of your u function.

    In your function you have indicated a partial derivative which implies that your function u is not just a function of z, but of possibly other variables. I'm assume since you are studying fluids, that it is at least a function of x and y as well.

    In other words you have to consider the total differential: this is written in the following form for a three dimensional function that maps to a one dimensional output (you can extend the idea to more dimensions if you want):

    du = ∂u/∂x dx + ∂u/∂y dy + ∂u/∂z dz, where du is the infinitesimal change in u and the others are as expected.

    From this it should appear the case that unless you assume your derivatives to be zero for the other variables, that you can't just get u from one partial derivative.

    In fact I would say that the way the author did it was also misleading as well. You said you took out the constants, but in this case they are probably very important.

    My guess is that the constants represent information about the other parameters like the x and y parameters if they exist. So if your solution is C1log(z) + C2 then C1 and C2 will involve your other variables if they are not just numeric constants.

    Also your answer for your transformation should give you the right answer as well because
    zu - ∫udz - z = zu - z(log(z) - 1 + C) - z = zu - zlog(z) + z - z - Cz = zu - zlog(z) = z(u - log(z)) = Cz so your extra z term cancels out and gives you u - log(z) = C which gives u = log(z) + C: Remember that you substitute u into the integral including the C term and you don't include an extra C term before you integrate. Keep in mind that constant of integration for u is actually the C you are given in u = log(z) + C: not being aware of this will confuse you into why I don't add an extra C term when I did my calculation above.
  8. Mar 18, 2012 #7
    Thank you for your time, I think I was correct in the way I was thinking and just needed confirmation. Your answer was great.
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