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Change of Variables - Finding New Limits

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Integrate the following over the set E.

    [itex]\int_E \frac{2x+y}{x+3y} dA[/itex]

    Bounded by the lines:
    y = −x/3+1
    y = −x/3+2/3
    y = −2x
    y = −2x + 1

    2. Relevant equations


    3. The attempt at a solution

    I can up to the same point everytime, but always get stuck on finding the new limits. Here is where I get to.

    Let u=2x+y and v=x+3y

    I take the Jacobian to find the distortion factor (or whatever you want to call it) and invert it. Please excuse my poor attempt at a matrix.

    [itex]\left| 2 \ 1 \right| \\
    \left|1 \ 3 \right| [/itex]

    This is 5, which inverted it 1/5. This gives me the following:

    [itex]\frac{1}{5} \int_E \frac{u}{v} du \ dv[/itex]

    Now the part I don't get. My new limits for integration. On the solution page for this problem, there is another Jacobian presented that doens't seem to match anything, except slightly matching the new limits.

    So back to the question at hand - how does one figure out those new limits?

    And as a secondary question - how do you choose u and v? In this case it is easy. But in other cases can you just choose whatever you want (within reason of course) and run it through the same steps to get your new limits?

    I've seen some examples where an easy shape is drawn onto x-y co-ords and then mapped onto the u v plane.

    For example take a unit square located at the origin. (0,0), (0,1), (1,0), (1,1). Is a correct method then to put those values into u and v and get new points?
  2. jcsd
  3. Feb 6, 2014 #2


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    Solve for ##x## and ##y## in terms of ##u## and ##v## and then plug them into the expressions for the boundaries.
  4. Feb 6, 2014 #3
    Okay, this should be right.

    I have x=(3u-v)/5 and y=(2v-u)/5

    But I think I am doing something wrong as this is just getting messier and messier the further I go.

    Using the first boundary y=-(x/3)+1 and my two equations there I end up with v=(5-2u)/3.

    Which I am pretty sure is wrong because in the answer the limits are single digit numbers....
    Last edited: Feb 6, 2014
  5. Feb 6, 2014 #4


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    You have u = 2x + y and v = x + 3y.

    The given boundaries are these:
    There are straightforward arithmetic operations which will turn the first two into "v = x + 3y = ..." and the second two into "u = 2x + y = ...".
  6. Feb 6, 2014 #5
    Ooooh yes. I see now. That is very straightforward. :)

    Now I have:
    2 ≤ v ≤ 3
    0 ≤ u ≤ 1

    I am guessing you choose which for v and which for u based on plotting the graph and seeing which corresponds to which axis? I tried swapping them and it wasn't working so well (I got results with x instead of values).

    Now for some practice problems.
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