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Help with change of variables in multivariable calculus/analysis

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##H## be the parallelogram in ##\mathbb{R}^2## whose vertices are ##(1,1), (3,2), (4,5), (2,4).## Find the affine map ##T## which sense ##(0,0)## to ##(1,1), (1,0)## to ##(3,2), (0,1)## to ##(2,4)##. Show that ##J_T=5## (the Jacobian). Use ##T## to convert the integral
    $$
    \alpha = \int_H e^{x-y}dxdy
    $$
    to an integral over ##I^2## (unit square) and thus compute ##\alpha##.


    2. Relevant equations



    3. The attempt at a solution

    So I'm not sure why I'm having some trouble with this. I've done everything up to solving the integral. The affine map I found is ##T(x,y) = (1,1) + (2x+y,x+3y)##, and I've checked that it is correct. However, in my mind, it makes sense that in changing the integral to be over the square, I take ##T^{-1}(H) = I^2##, that is, obtaining:
    $$
    \alpha = \int_H e^{x-y}dxdy = \int_{I^2} f(T^{-1}(x,y))|J_{T^{-1}}|dxdy
    $$

    However, the integral is supposed to be ##f(T(x,y))|J_T|##, not of ##T^{-1}##. Can someone explain why this is so? I think I'm just forgetting something from lower division calculus, as it's been a while. It made sense back then, but for some reason this problem is screwing me up. Thanks in advance.
     
  2. jcsd
  3. Mar 13, 2014 #2

    Mark44

    Staff: Mentor

    I'm pretty rusty on this stuff, but since no one else has answered, I'll take a shot at it. Your original integral is over the parallelogram H. You want the new integral to be over the unit square, so you need to transform both the H region and the integrand to get a new integral with the same value as the old. IOW, you want to apply your transformation T to both H and the function in the integrand. You don't want to apply T to the region and T-1 to the function's inputs.

    The idea is similar to integration by substitution where you start with one integral:
    $$\int_a^b f(x)dx$$

    and use a substitution u = u(x), du = u'(x)dx
    to rewrite the integral and the limits of integration in a form that is more easily solvable...
    $$ \int_{u(a)}^{u(b)} f(u(x))du$$
     
  4. Mar 13, 2014 #3
    Thanks Mark44. I mean, I get you want to apply it to BOTH region and integrand, but I'm not sure why applying T to H would give me the integral over the unit square, as T is from the unit square to the parallelogram.
    In my lower division calculus, when we learned about this, our transformations were always from the "harder" region to the "easier" region. However, in this case, our T maps the "easier" region (unit square) to the "harder" region (parallelogram H), and for some reason it's just not clicking in my head. I should be getting this if my past class is any , but I'm not sure why it isn't now. Perhaps I'm just recalling the definition incorrectly...

    Oh a quick edit, on your comment about applying ##T## to the region and ##T^{-1}## to the function inputs, I applied ##T^{-1}## to both the region AND the inputs
     
    Last edited: Mar 13, 2014
  5. Mar 13, 2014 #4

    Mark44

    Staff: Mentor

    I misread (or actually, didn't read closely enough) your first post, thinking that T transformed H to the unit square.

    Did applying T-1 to the region and the inputs work out for you? I would think that's what you need to do.
     
  6. Mar 13, 2014 #5
    Well the thing is, nothing went wrong per say, but there was a dispute in office hours about whether it was applying ##T## or ##T^{-1}## to the region/integrand, and the majority was on the side of applying ##T##, but it just didn't make sense to me as to why. So, applying ##T^{-1}## is correct then?
     
  7. Mar 13, 2014 #6

    Mark44

    Staff: Mentor

    You want the integral to be evaluated over the unit square. T sends the unit square to the parallelogram - you want to go the other way, from H to the unit square.

    BTW, it's per se, Latin for "in itself".
     
  8. Mar 13, 2014 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Not sure which you mean by T, so I'll use R instead: that is, we can change variables from (x,y) to (u,v), where
    [tex] \pmatrix{x\\y} = R \pmatrix{u\\v} [/tex]
    Then
    [tex] \alpha = \int_H f(x,y) \, dx \, dy = \int_{[0,1]^2} f(x(u,v),y(u,v))
    \left| \frac{ \partial(x,y)}{\partial(u,v)}\right| \, du \, dv [/tex]
     
  9. Mar 13, 2014 #8
    Alright, so then your R goes from H to the unit square?
     
  10. Mar 13, 2014 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The way I wrote it, R takes (u,v) into (x,y), so takes the unit square in (u,v)-space into the quadrilateral in (x,y)-space. Never mind using canned formulas; think about the concepts and meanings first.
     
  11. Mar 13, 2014 #10

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    I think Zalculus is getting confused by his own notation. The parallelogram is given in ##(x,y)## space and is being transformed into the unit square in the ##(u,v)## space. The transformation that is requested in the original problem is to find the transformation that takes the unit square in the ##(u,v)## space to the parallelogram in the ##(x,y)## space. So instead of writing$$
    T(x,y) = (1,1) + (2x+y,x+3y)$$he should have written$$
    (x,y) = T(u,v) = (1,1) + (2u+v,u+3v)$$
     
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