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Homework Help: Change of Variables in Multple Integrals

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Let D be the region y [tex]\leq[/tex] x [tex]\leq[/tex] 1 and 0 [tex]\leq[/tex] y [tex]\leq[/tex] 1.

    Find the boundaries of the integration for the change of variables u and v, if:

    x = u + v and y = u-v.


    2. Relevant equations

    None


    3. The attempt at a solution

    I solved for u and for v in terms of x and y.

    u = [tex]\frac{x+y}{2}[/tex]

    v = [tex]\frac{x-y}{2}[/tex]

    Then I got stuck.

    The book answer's is:

    v [tex]\leq[/tex] u [tex]\leq[/tex] 1-v and 0 [tex]\leq[/tex] v [tex]\leq[/tex] 1/2


    Can anyone show me how to figure this problem out without any reference to graphing (just

    pure algebraic)?
     
  2. jcsd
  3. Jul 11, 2010 #2

    vela

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    Why do you guys always like to make things more difficult for yourself by refusing to draw a picture?

    Just rewrite the boundaries of D in terms of u and v. For example,

    [tex]y \le x \Rightarrow u-v \le u+v \Rightarrow 0 \le 2v \Rightarrow v \ge 0[/tex]
     
  4. Jul 11, 2010 #3
    Well, I did that part already. However, how is it possible to get v [tex]\leq[/tex] 1/2 (the

    book's answer). For example,

    x [tex]\leq[/tex] 1 => u + v [tex]\leq[/tex] 1 => v [tex]\leq[/tex] 1 - u

    Which is supposedly not equivalent to v [tex]\leq[/tex] 1/2

    How would you go about solving the other ones (v [tex]\leq[/tex] u [tex]\leq[/tex] v-1) ?
     
  5. Jul 11, 2010 #4

    vela

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    You need to do the third side of D, y≥0. When you combine it with what you got for x≤1, you can get that upper limit for v.
     
  6. Jul 11, 2010 #5
    Okay, can you explain it a bit more. Sorry about being unknown about what you mean, but I

    think if I finally understand what you are trying to say, I can finally get this problem correct.
     
  7. Jul 11, 2010 #6

    vela

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    y≥0 translates to u≥v, and x≤1 gave you u≤1-v. Combining those, you get v≤u≤1-v. (You almost had this above, but you flipped a sign.) You can't have v=1, for instance, because you'd get 1≤u≤0. What restriction on v must exist for those inequalities to be satisfied?
     
  8. Jul 11, 2010 #7
    Now, I think I am starting to get it. I think I am having a conceptual misunderstanding of the

    situation at hand! I am elated that you pointed out how I flipped the sign. Here is the reason

    why I flipped the sign:

    y [tex]\leq[/tex] 1 => u - v [tex]\leq[/tex] 1 => u [tex]\leq[/tex] 1 + v (correct answer: u [tex]\leq[/tex] 1 - v)

    Can you tell me why this would not work?
     
  9. Jul 11, 2010 #8

    vela

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    That is valid, but it doesn't help much. Another constraint gave you v≥0, so if u≤1-v holds, u≤1+v automatically holds as well.
     
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