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Imaginary Center of Mass? What is that?

  1. Nov 19, 2011 #1
    Suppose I give you a curve

    [tex]f(x) = \sin^2 (x) + ln(x)[/tex]

    And suppose I tell you to rotate this curve about the x axis, we get disks. Now, I ask you, what is the center of mass of this object?

    Now immediately, you could say that [tex]\bar{y} = 0[/tex] because it is symmetric about the x-axis. I don't argue, and I say you are right. But what about [tex]\bar{x}[/tex]?

    It is actually NOT the same as the 2D- lamina. In fact

    [tex]\bar{x} = \frac{\int_{a}^{b} \pi x[f(x)]^2 dx}{\int_{a}^{b} \pi [f(x)]^2 dx}[/tex]

    Okay, so what am I trying get here?

    Evaluate the integral of the curve from x = 1 to x = 10

    [tex]\int_{1}^{10} \pi x (\sin^2 (x) + ln(x))^2 dx[/tex]

    Do this (do it on a computer, trust me on this) and you get some nonreal numbers. I asked my TA about it and he said when I expand [tex]f(x) = \sin^2 (x) + ln(x)[/tex], I get a term that I won't know how touch until I get to analysis.

    On Maple I got

    [PLAIN]http://img143.imageshack.us/img143/9805/comh.jpg [Broken]

    I asked my TA what it means physically and he said that he wasn't sure and maybe the complex/imaginary part meant it is significantly 0

    So what does having an imaginary part mean?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 19, 2011 #2
    You don't need Maple to carry out the integrals, you just apply integration by parts and obtain:

    int(ln(x) dx)=x ln(x)-x
    int(sin^2(x) dx)= (-sin(x) cos(x)+1)/2

    Now apply Barrow's rule

    My mistake. I didn't pay proper attention to the integrand
    Last edited: Nov 19, 2011
  4. Nov 20, 2011 #3
    how about this?

    Attached Files:

  5. Nov 20, 2011 #4


    Staff: Mentor

    That is just 0 to within your numerical precision.
  6. Nov 20, 2011 #5

    D H

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    It means Maple is stupid sometimes. The function is real everywhere in this interval, so there was no reason to introduce imaginary numbers. Yet it did.

    It's obvious how Maple solved that integral: They integrated a truncated infinite series. Had they used a infinite rather than truncated series the imaginary part would have gone to zero. They use truncated series because they want to give an answer in finite time. This, combined with its penchant to introduce complex numbers at the drop of a hat, can lead to complex results with tiny imaginary part (that should be zero).
  7. Nov 20, 2011 #6


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    Gold Member

    If you think about it, the quoted answer in post #1 is in an unhelpful form. To illustrate this, you could write it as
    Code (Text):

    -0.0000000004308080094 i
    where "?" denotes an unknown digit. If the first "?" is small enough to be ignored, the whole of the imaginary part can be ignored too. (Bear in mind that any numerical solution like this can only ever be an approximation to the exact answer.)
  8. Nov 20, 2011 #7
    never mind...


    That is odd, but Wolframalpha takes log(x) as ln(x) right?
  9. Nov 20, 2011 #8

    D H

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    Science Advisor

    Correct. Mathematica does things a bit better than does Maple in this regard. Maple is too quick to go to complex numbers. Mathematica tries to keep things real if it knows the function to be integrated is real throughout the integration interval. That, or it is "smart" enough to recognize that a non-zero imaginary part is bogus if it does use a complex formulation to calculate a result that is real.
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