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Integration in two different coordinate systems

  1. Oct 19, 2012 #1
    Hi all. I am very puzzled by the following.

    Let [itex]x_1[/itex] and [itex]x_2[/itex] be two coordinate systems related by [itex]x_1=1-x_2[/itex].

    Now if [itex]y(x_1) = x_1[/itex] and [itex]z(x_2) = 1-x_2[/itex], then clearly [itex]y(x_1)=z(x_2)[/itex].

    Now integrating the function in each coordinate system gives

    [tex]Y(x_1) = \int y(x_1) dx_1 = \int x_1 dx_1 = \frac{x_1^2}{2} + C [/tex]

    [tex]Z(x_2) = \int z(x_2) dx_2 = \int (1-x_2) dx_2 = -\frac{x_2^2}{2} + x_2 + D [/tex]

    Now, however,

    [tex]Y(x_1) = Y(1-x_2) = \frac{(1-x_2)^2}{2} + C = \frac{1}{2} - x_2 + \frac{x_2^2}{2} + C \neq Z(x_2) [/tex]

    In words, [itex]Y(x_1) \neq Z(x_2)[/itex] regardless of the values of [itex]C[/itex] and [itex]D[/itex]. One would expect, however that [itex]Y(x_1) = Z(x_2)[/itex] - but then where was my mistake? Thanks!
  2. jcsd
  3. Oct 19, 2012 #2


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    Since any difference in the constant term is explained away by 'constant of integration', the remaining puzzle is the reversal of the sign. This is simply because the integral wrt x2 ran in the opposite direction, right-to-left, so to speak. So the integration range was reversed.
  4. Oct 20, 2012 #3
    Let me try to highlight the problem. Say that you know boundary condition [itex]Y(x_1=0) = 0[/itex] (and hence [itex]Y(x_2=1)=0[/itex]). This means that [itex]C=0[/itex] and [itex]D=-\frac{1}{2}[/itex].

    Now let us say we want to evaluate the integral of [itex]y[/itex] at [itex]x_1=1[/itex] ([itex]x_2=0[/itex]). We should be able to do this in either coordinate system, but

    [tex] Y(x_1=1) = \frac{1}{2} [/tex]
    [tex] Z(x_2=0) = -\frac{0}{2} + 0 + \frac{-1}{2} = -\frac{1}{2} \neq Y(x_1=1)[/tex]
  5. Oct 20, 2012 #4


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    It makes sense, if you think about how the functions relate. Just some example:

    Let x_1 = 0, x_2 = 1, x_1 = 1, x_2 = 0, x_1 = 2,x_2 = -1, and so on and so forth.

    What this tells me is that you actually need to put a negative sign in front of the integral when you switch from one to another due to the fact that the integration range reverses. Hopes this help.
  6. Oct 20, 2012 #5
    The error is a result of the implied assumption that the limits are the same from the constant of integration.. this isn't true. Limits must change with the coordinate system. The constant of integration accounts for more than you would think.

    Imagine integrating over [itex] x_1 \in (a,b) [/itex]

    the integrals are equivalent...

    [itex] \int_a^bx_1dx_1 \Leftrightarrow \int_{1-a}^{1-b}(1-x_2)dx_2 [/itex]
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