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Change of Variables in Multiple Dimensions

  1. Jul 13, 2009 #1
    So, I've got a problem understanding the "algorithm" for changing variables in a more-than-one-dimensional integral. For the two-dimensional case, I've got a specific problem that I'm looking at:

    [tex]\int^{a}_{0}\left(\int^{2a-x}_{x}\frac{y-x}{4a^2+(y+x)^2}dy\right)dx[/tex]

    which I assume is an example of the more general case:

    [tex]\int^{b}_{a}\left(\int^{g(x)}_{f(x)}\phi(x,y)dy\right)dx[/tex].

    For the particular case, I am to make the given substitutions

    [tex]u=x+y[/tex]
    [tex]v=x-y[/tex]

    and evaluate the integral.

    I'm trying to figure out how to determine the new boundaries of integration. In the two dimensional example given above, it's easy enough to draw the region of integration and figure it out, but what if it's not such a simple situation? How does one juggle integration regions? Without drawing it, I'm left with the ugly inequalities

    [tex]\{x<y<2a-x\} \leftrightarrow \{u+v<u-v<4a-(u+v)\}[/tex]

    and

    [tex]\{0<x<a\} \leftrightarrow \{0 < u+v < 2a\}[/tex],

    which I can't seem to mentally sort through. In all of the examples I've found online and in textbooks, the region has been drawn. Is this just for illustrative (pun?) purposes, or is it because that's the only straightforward way to determine the regions of integration?

    Thanks!

    Some background about my understanding:

    I've had a basic introduction to the metric G, scale factors, and the Jacobian and the application to transformations.
     
  2. jcsd
  3. Jul 13, 2009 #2

    HallsofIvy

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    Okay, if u= x+ y and v= x- y, then the Jacobian, which you refer to below, is
    [tex]\left|\left|\begin{array}{cc}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right|\right|= \left|\left|\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right|\right|= 2[/tex] so dudv= 2dxdy and dxdy= (1/2)dudv.

    Adding u= x+y and v= x-y we get 2x= u+ v or x= u/2+ v/2. Putting that into u= x+y, u= u/2+ v/2+ y so y= u/2- v/2. The integrand is
    [tex]\frac{y- x}{4a^2+ (y+x)^2}= \frac{v}{4a^2+ u^2}[/tex]

    If you draw the lines x= 0, y= x, and y= a- x, you see that they form a triangle with one vertex at (a,a) on the vertical line x= a. Other than that, the line x= a is not really a boundary.

    The lower limit on the inner integral is y= x or y- x= v= 0. The upper limit is y= a- x or x+ y= u= a. Adding u= x+y and v= x-y gives u+ v= 2x or x= u/2+ v/2. The lower limit of the first limit is x= u/2+ v/2= 0 or u+ v= 0. The upper limit is x= u/2+ v/2= a or u+ v= 2a. We can, for example, write those as v= -u and v= -u+ 2a. Draw those four lines on a uv-plane which is the same as an xy-plane except that you label them u and v. v= 0 is the horizontal axis. u= a is a vertical line. u+ v= 0 or v= -u and u+ v= 2a or v= 2a- u are parallel lines at 45 degrees below the horizontal axis. That parallelogram (the original region in the xy-plane is a triangle) can be covered by taking u form 0 to a and v from -u up to a=0. The integral is
    [tex]\int_{u= 0}^a\int_{v= -u}^0\left(\frac{v}{4a^2+ u^2}\right)\left(\frac{1}{2}dudv}\right)[/tex]

    As you see from what I did above (and very likely it is what you did) how you change the limits of integration is heavily dependent on the geometry of the situation. I doubt there is any general method.
     
    Last edited: Jul 13, 2009
  4. Jul 13, 2009 #3
    I'm just uncomfortable with the "solution by drawing it on paper" strategy. What if you have a four-dimensional integral (it could happen...)? Maybe it's just one of those aspects of calculus that descends into numerical methods once the analysis becomes too bumpy.

    Anyways, thanks for the help!
     
  5. Jul 14, 2009 #4

    daniel_i_l

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    Gold Member

    If the area is a polygon and the transformation is linear, as in your case, you simply have to transform the verticies of the original polygon to get the verticies of the new polygon. This works in all dimensions.
    In the general case, if it's hard to integrate over the new point-set you probably didn't use the right transformation.
     
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