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Change of Variables in Multiple Integrals

  1. Aug 14, 2011 #1
    The problem is:

    R is the parallelogram bounded by the lines x+y=2, x+y=4, 2x-y=1, and 2x-y=4. Use the transformation u=x+y and v=2x-y to find the area of R.


    I am not sure how to complete this problem. My first issue is that I don't know how to convert the transformation functions into functions of u and v, to find the Jacobian.

    Second, the region R is a diamond shape in the xy plane. I don't understand how to map this figure into the uv plane. Could someone please provide some advice on this problem?

    Thanks for your help.
     
  2. jcsd
  3. Aug 14, 2011 #2

    dynamicsolo

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    You don't need to convert anything. The transformation functions were chosen to match the terms in the equations for the lines. So one transformed line is just u = 2 , and similarly for the rest. You will end up with a rectangle in the uv-plane, for which you can find an area. You will then need the Jacobian in order to find the factor you must multiply this by in order to obtain the area in the xy-plane.
     
  4. Aug 14, 2011 #3
    The general formula for the Jacobian is |d(x,y) / d(u,v)|... so in this case, would I find |d(u,v) / d(x,y)|, since I can't find the first unless the transformation functions are not in terms of u and v?

    Just to confirm, after finding the Jacobian, to find the area, I will compute:

    int( int(Jacobian) dv) du)

    since the area is found by integrating 1 over the region, correct?


    I see what you mean now... u=x+y and x+y=2,4 so this will translate to u=2,4 on the uv plane. And similarly for v.
     
    Last edited: Aug 14, 2011
  5. Aug 14, 2011 #4
    [itex] |d(u,v) / d(x,y)| [/itex] is the notation for the Jacobian. Does [itex] d(x,y) [/itex] mean anything to you, other than the top part of the symbol for the Jacobian?

    The "formula" or definition of the Jacobian of a transformation [itex] f: R^n \rightarrow R^n [/itex] is as follows. The Jacobian is the determinant of the matrix [itex]Df [/itex] where [itex]Df_{i,j} = \frac{\partial f_i}{\partial x_j} [/itex]

    Your problem was stated using different notation. Translating between your notation and my notation gives: [itex] u(x,y) = f_1(x_1,x_2); v(x,y) = f_2(x_1,x_2); x=x_1,y=x_2. [/itex]
     
  6. Aug 14, 2011 #5

    vela

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    It's basic algebra. You have
    \begin{align*}
    x+y &=u \\
    2x-y &= v
    \end{align*}
    Compare to, say,
    \begin{align*}
    x+y &=1 \\
    2x-y &=2
    \end{align*}
    How would you solve for x and y in the latter case?
     
  7. Aug 14, 2011 #6
    I understand that the Jacobian is the determinant. What this means to me is that I should find:

    | du/dx du/dy |
    | |
    | dv/dx dv/dy |

    From what I understand, the Jacobian of the transformation given by x=g(u,v) and y=g(u,v) is:

    [itex] |d(x,y) / d(u,v)| [/itex]

    But because my transformation functions are functions of x,y, then I will find:

    [itex] |d(u,v) / d(x,y)| [/itex]

    Is this interpretation correct?
     
  8. Aug 14, 2011 #7
    I think I understand that already, see above... my problem was that I copied the functions, on paper, in terms of y so that I could plot the functions. I overlooked the relationship because of this (whoops, really tired lol).
     
  9. Aug 14, 2011 #8

    dynamicsolo

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    I believe there is a way to use the Jacobian written the other way if the functions can be inverted, but they were nice to you in this problem by giving you linear functions. You can solve the system of two equations for u and v as functions of x and y . (Add the two to get u + v = 3x , and so forth.)

    Now you can use |d(x,y) / d(u,v)| . You'll have linear functions which are simple to differentiate, giving you a Jacobian which has a constant value. (Keep in mind that we use the absolute value of the determinant.)

    I believe that is correct. In this problem, you won't really have to integrate, since geometry will give you the area of your region in the uv-plane.


    On a quick search, here is a slightly more complicated example which illustrates the method: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/change/change.html" [Broken]
     
    Last edited by a moderator: May 5, 2017
  10. Aug 14, 2011 #9
    Yes, that's right. I'm sorry, I brought it up. It's a common mistake for people to think that the slash in the symbol for derivatives or the Jacobian is a division sign. So, when you called the symbol for the Jacobian a formula, I thought you were taking that mistake. Sorry for jumping to conclusions.
     
  11. Aug 14, 2011 #10

    vela

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    I think you missed the point I was making which was it's straightforward to invert the transformation functions to find x=x(u,v) and y=y(u,v).
     
  12. Aug 14, 2011 #11
    No problem, I am sure that information might be of use to someone who comes across this thread in the future.

    So what I have pulled together is the following...

    Using lin alg to convert the transformation functions:

    x=u/3 + v/3
    y=2u/3 - v/3

    J =
    |d(x,y)/d(u,v)|=
    |1/3 1/3|
    |2/3 -1/3|
    =-1/3

    so:

    int( int( 1/3 dv) du) u=2..4 v=1..4

    or based on simple geometry, 2*3=6




    If all of the above is correct, my only question that still remains is that of the use of the Jacobian when changing variables. When calculating

    |d(u,v)/d(x,y)| = -3

    and

    |d(x,y)/d(u,v)| = -1/3

    so does this relationship imply (as a general rule) that the correct Jacobian for integration is the inverse of that calculated for the transformation function given?
     
    Last edited: Aug 14, 2011
  13. Aug 14, 2011 #12
    See the post immediately above this one.

    Are you just implying that I should use linear algebra, subtracting / adding equations to find the transformation funcs as functions of u,v?
     
  14. Aug 14, 2011 #13

    dynamicsolo

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    J is the absolute value of the determinant, or 1/3 .

    Isn't that (1/3) * 2 * 3 ?


    I hunted around a bit and found that your statement is correct as a general rule. I think the limitation would be that you may not always be able to invert x and y to write u and v . (Presumably, it's true anyway for the implicit inverse functions -- you just wouldn't be able to write the expressions...)

    In the end, for this problem, you want to have the value |d(x,y)/d(u,v)| = 1/3 , since we have to go back from the uv-plane to the xy-plane to give the area for the original parallelogram.
     
    Last edited: Aug 14, 2011
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