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Double integrals and change of variables

  1. Jan 26, 2008 #1
    Hi, everyone!

    I have a problem in understading the change of variables in double integrals. Here is an example
    [tex]\int\int x^2+y^2dx dy=\int \frac{x^3}{3}+y^2x dy=\frac{x^3y}{3}+\frac{y^3x}{3}+C_1[/tex]
    but if I first do a change in poral coordinates I get
    [tex]\int\int r^2 r drd\theta=\int\frac{r^4}{4}d\theta=\frac{r^4\theta}{4}=\frac{1}{4}(x^2+y^2)^2 arctan(\frac{y}{x})+C_2[/tex]
    which is not the first answer. A more simple example is
    [tex]\int\int dxdy=xy+C_1[/tex]
    and in poral
    [tex]\int\int rdrd\theta=\int\frac{r^2}{2}d\theta=\frac{r^2\theta}{2}=\frac{1}{2}(x^2+y^2) arctan(\frac{y}{x})+C_2[/tex]

    Can someone explain what I do wrong?

    Thanks in advance!
     
    Last edited: Jan 26, 2008
  2. jcsd
  3. Jan 26, 2008 #2
    you have to pick a region over which to integrate and then transform that region. for example let's integrate your first integral over the quarter of the unit circle in the first quadrant in the xy plane.

    [tex]\int^{1}_{0}\int^{\sqrt{1-x^2}}_{0} x^2+y^2~dydx[/tex]

    in polar coordinates your integral will look like

    [tex]\int^{\frac{\pi}{2}}_{0}\int^{1}_{0} r^3 dr d\theta[/tex]
     
    Last edited: Jan 26, 2008
  4. Jan 26, 2008 #3
    Why should I choose a region? When I have to evaluate a indefinite double integral, I can't apply change of variables?
     
  5. Jan 26, 2008 #4
    I think this is where the problem is, because you seem to think that when you evaluate an indefinite single variable integral that you aren't integrating over a "region" when in fact you are, the segment of the real line [a,x], so it is similar the in the multiple variable case that you still would need to chose an arbitrary region over which to integrate.
     
  6. Jan 26, 2008 #5
    Ahaa! So my integrals over [tex]dx dy[/tex] are correct by the ones over [tex]dr d\theta[/tex] are not?
     
  7. Jan 27, 2008 #6

    Defennder

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    Actually no, since even if you're doing a double integral wrt dxdy, and not cylindrical or spherical coordinates, you'll still need to specify a region over which the integration is done.
     
  8. Jan 27, 2008 #7
    I was referring to d_leet's post. I mean the integrals over [tex]dx dy[/tex] are correct in sense that I can evalute them and then substitude the values a<x<b, c<y<d for some region. But for integrals over [tex]dr d\theta[/tex] I can not first evalute them and then substitude the values r1<r<r2, [tex]\theta_1<\theta<\theta_2[/tex].

    Does it makes any sense, or I am talking nonsenses?? :smile:
     
  9. Jan 27, 2008 #8

    Defennder

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    It's possible to integrate in both cylindrical and cartesian coordinates without the limits, but it's not possible to change variables without first knowing the limits. Or at least that's what I feel.
     
  10. Apr 14, 2010 #9
    I have a general question about double integrals. Since acceleration is the second derivative of the position function, can one not use a double integral notation to express the position function as a double integral of the acceleration function, like x(t) = [tex]\int a(t) dt dt[/tex]? (I can't seem to make two integral symbols back to back, but that is supposed to be two integrals before a(t).)
     
  11. Apr 15, 2010 #10

    HallsofIvy

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    1) Please don't "hijack" other people's threads for new questions.

    2) A "double integral" is with respect to two different variables. When you go from acceleration to position, you are integrating, twice, with respect to time.
     
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