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Trying to integrate a non one-to-one parametric function

  1. May 23, 2013 #1
    Hi, I'm working on an independent research project - and am trying to integrate this (with respect to x between some arbitrary m and infinite).


    If you graph this as a parametric eqn (set r to 2 or 3), the problem is that it is not a one-to-one mapping. I want to find the area under the curve (and the part where there are two values of x, I want to include that twice.

    Is there any way I can do this?

    In the end I want to have the integral be a function of r (m is constant) that I can use elsewhere.

    I want to see how the area under the curve changes when I vary r

    Any suggestions?
    Last edited: May 23, 2013
  2. jcsd
  3. May 23, 2013 #2


    Staff: Mentor

    When we find the area "under a curve" it's implied that the lower boundary of the region is the horizontal axis. For your parametric curve, with r = 3, the graph is nearly a loop, so "under the curve" doesn't make much sense.
  4. May 24, 2013 #3
    First just get a bearing on it. This is what it looks to me:


    Ok, right off the bat, I'm going to plot it parametrically and since you want to integrate it, in the interest of just trying something even if it's wrong, I want to integrate it "twice" as you suggest where it's not one-to-one. That means I need to find where it's not. That's where the black dot is. Since [itex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/itex], then the point at which it has an infinite slope is when [itex]dx/dt=0[/itex]. Alright then, then I'll integrate the red to the t-value where [itex]\frac{dx}{dt}=0[/itex], and the blue from that spot to infinity:

    [tex]\mathop\int\limits_{\text{red}} y(x)dx=\mathop\int_{-\infty}^{\frac{dx}{dt}=0} y(t)\frac{dx}{dt} dt[/tex]
    [tex]\mathop\int\limits_{\text{blue}} y(x)dx=\mathop\int_{\frac{dx}{dt}=0}^{\infty} y(t)\frac{dx}{dt} dt[/tex]

    I don't know, is that right guys? I need to check.

    Attached Files:

    Last edited: May 24, 2013
  5. May 24, 2013 #4
    Hi !

    There is no difficulty to integrate on a closed area.
    In fact, the difficulty is that the antiderivative cannot be expressed in terms of a finite number of standard functions. So you cannot give the result on the form of a formula.
    The only way is a numerical integration, giving a numerical result for each numerical value of the parameter r.
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