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Double integral using transformations

  1. Jan 9, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s2.ipicture.ru/uploads/20120109/dT4m6rNG.jpg

    The attempt at a solution

    [tex]x=\frac{u}{1+v}[/tex] and [tex]y=\frac{uv}{1+v}[/tex]
    Transforming the integrand: [tex]\frac{x+y}{x^2}e^{x+y}=\frac{(1+v)^2 e^u}{u}[/tex]
    [tex]dxdy=J.dudv[/tex]
    [tex]J=\frac{v(1+v)^2 +1+uv}{(1+v)^3}[/tex]
    The double integral becomes: [tex]\int\int \frac{e^u[v(1+v)^2+1+uv]}{u(1+v)}.dudv[/tex]
    This is the part where i'm having some trouble as i broke the large integrand into smaller ones:
    [tex]\int\frac{e^u[v(1+v)^2+1+uv]}{u(1+v)}.du=\int\frac{e^u v (1+v)}{u}.du+\int\frac{e^u}{u(1+v)}.du+\int\frac{e^u v}{(1+v)}.du[/tex]
    The integral of [itex]\frac{e^u}{u}[/itex] is what's blocking my progress through this problem. I did an online search and it appears that it can't be solved unless i use some special function that mathematicians invented called exponential integral. So, i think i might have made an error somewhere and might have inadvertently overstepped into unknown territory. I haven't started working on the limits yet, as i wouldn't be able to integrate at this point.
     
    Last edited: Jan 9, 2012
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  3. Jan 9, 2012 #2

    ehild

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    Check J.


    ehild
     
  4. Jan 9, 2012 #3

    sharks

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    [tex]\frac{\partial x}{\partial u}=\frac{1}{1+v}[/tex]
    [tex]\frac{\partial x}{\partial v}=\frac{-u}{(1+v)^2}[/tex]
    [tex]\frac{\partial y}{\partial u}=\frac{v}{1+v}[/tex]
    [tex]\frac{\partial y}{\partial v}=\frac{u}{(1+v)^2}[/tex]
    I had made a mistake in [itex]\frac{\partial y}{\partial v}[/itex]
    [tex]J=\frac{u+uv}{(1+v)^3}[/tex]
    The u-v graph is:
    http://s2.ipicture.ru/uploads/20120109/JPuR66Sn.jpg
    The double integral becomes:
    [tex]\int^1_0\int^2_{2v} e^u .dudv=e^2-\frac{e^2}{2}+\frac{1}{2}[/tex]
    The answer is incorrect. I'm supposed to get [itex]e^2-1[/itex]
     
    Last edited: Jan 9, 2012
  5. Jan 9, 2012 #4

    ehild

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    Check the boundary u=2v. It is not correct.

    ehild
     
  6. Jan 9, 2012 #5

    sharks

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    I'm not sure what went wrong, as my conversion from the Cartesian coordinates to uv coordinates appear correct, so here is my Cartesian graph:
    http://s2.ipicture.ru/uploads/20120110/rOnUC6d7.jpg
    The graph i drew is just a sketch for describing the boundaries and region area, so it's not graphically very accurate.

    Converting the xy coordinates to uv coordinates (i took the x and y coordinate from each set and plugged them into the original transformations from the problem):
    (x,y) --> (u,v)
    (0,0) --> (0,0)
    (1,1)--> (2,1)
    (2,0) --> (2,0)
     
    Last edited: Jan 9, 2012
  7. Jan 10, 2012 #6

    ehild

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    Transform the lines instead of the points.

    From the integration domain you have the relations y≤x≤2-y and 0≤y≤1.

    As x, y both are non-negative, so are u and v : u≥0, v≥0.

    y≤x →v≤1
    x≤2-y→x+y≤2→u≤2

    y≥0 is true when u≥0, v≥0

    y≤1 →u≤1/v+1 as u≥0, v≥0. As v≤1, 1/v+1≥2, and this is true when u≤2.

    So you have the domain for u,v u≥0, v≥0, v≤1, u≤2.

    ehild
     
    Last edited: Jan 10, 2012
  8. Jan 10, 2012 #7

    sharks

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    Transforming the lines:
    For the line y=1, [itex]v=\frac{1}{u-1}[/itex]
    For the line y=x, v=1
    For the line y=2-x, [itex]v=\frac{2-u}{u-2}[/itex]
    For the line y=0, v=0

    I plotted the graph for these transformed lines (quite complicated):
    http://s1.ipicture.ru/uploads/20120110/TUT1Skrk.png
    In the graph above, i used an online graphing calculator, and i still don't understand how [itex]v=\frac{2-u}{u-2}[/itex] can give a single straight line v=-1. For example, at the point where u=2, v=0, which is not found on the line v=-1. So i tried to make u the subject of formula and i got u=2, which makes this even more confusing. So, [itex]v=\frac{2-u}{u-2}[/itex] has actually 2 lines??

    Then, i deduced the points of intersection on the uv plot based on the lines which intersect on the xy plot:
    http://s1.ipicture.ru/uploads/20120110/MARBBCEA.png
     
    Last edited: Jan 10, 2012
  9. Jan 10, 2012 #8

    ehild

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    y=2-x is the same as x+y=2. x+y=u, so u=2.
    Include also the GT LT relations.

    ehild
     
  10. Jan 10, 2012 #9

    sharks

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    Not sure what you mean.:confused:

    I drew the graph of u-v again and included the line u=2 and all the points of intersection:
    http://s1.ipicture.ru/uploads/20120110/c1y1BkLX.png
    This is probably the most complex transformation i've dealt with, yet. Is there another point of intersection where the lines v=-1 and u=2 cross each other? I can't get the correct triangular shape if i simply join all the points outwards, which result in a parallelogram. However, if i ignore the part below the line v=0, then it becomes a triangle with coordinates: (1,0), (2,0), (2,1).

    The limits would then be:
    For v fixed, u varies from u=v+1 to u=2
    v varies from v=0 to v=1

    This is wrong, as after integration, i get the answer: e

    OK, i'm not ready to give up, so i'm going to revise the transformations:
    For the line y=1, [itex]v=\frac{1}{u-1}[/itex]
    For the line y=x, v=1
    For the line y=2-x, [itex]v=\frac{2-u}{u-2}[/itex]⇒u = 2 and v=-1
    For the line y=0, v=0 and u=0

    I get a triangle with coordinates: (0,1), (2,0), (2,1)
    The integration gives me: [itex]\frac{1}{2}(e^2 +1)[/itex] which is wrong again!
     
    Last edited: Jan 10, 2012
  11. Jan 10, 2012 #10

    ehild

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    It is not a triangle, as v=0 and u=0 corresponds to two lines.

    Investigate if u and v can be negative or not when 0<y<1 and y<x<2-y.

    You are right, this is a tricky transformation. But the integral is very easy after you have found the correct boundaries. Do not give up.

    ehild
     
  12. Jan 10, 2012 #11

    sharks

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    I am quite sure that the new point of intersection is at (0,1) but i didn't transform the line y=1 by making u the subject of formula, which would be: [itex]u=\frac{1+v}{v}[/itex]
    But i suppose that the graph of [itex]v=\frac{1}{u-1}[/itex] must be the same as [itex]u=\frac{1+v}{v}[/itex]? (i plugged in a few coordinates and it appears to be equivalent).
     
    Last edited: Jan 10, 2012
  13. Jan 10, 2012 #12

    ehild

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    [itex]u=\frac{1+v}{v}[/itex]and [itex]v=\frac{1}{u-1}[/itex] are the same. And you have a point of intersection at (0,1).


    ehild
     
  14. Jan 10, 2012 #13

    sharks

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    I've encircled the relevant points of intersection in the graph below:
    http://s1.ipicture.ru/uploads/20120110/ML97PRfn.png
    If i join all the points of intersection, it shows a pentagon, however, since x and y are both greater or equal to zero, u and v are likewise. This means that i get a rectangular shape, with coordinates: (0,0), (0,1), (2,0), (2,1). I'm not sure though as the shape of the x-y plot is triangular and i'm supposed to always get the same basic shape after transformation, right?
     
  15. Jan 10, 2012 #14

    ehild

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    That triangle is a rectangle with one side of length zero. It is the same basic shape if you can convert to the other one by stretching and compressing.


    ehild
     
  16. Jan 10, 2012 #15

    sharks

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    The limits would then be:
    For v fixed, u varies from u=0 to u=2
    v varies from v=0 to v=1
    The integration then gives: [itex]e^2 -1[/itex] finally!

    I seriously doubt that i will have enough time and insight to do this type of problem in the exams. I might require an internet-enabled device to search for assistance here, but my request will most probably be rejected by the examiners. :redface: Thank you for your help and patience, ehild.
     
    Last edited: Jan 10, 2012
  17. Jan 10, 2012 #16

    ehild

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    You are welcome. I suggest to read my post #6 again.

    ehild
     
  18. Jan 10, 2012 #17

    sharks

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    From your post #6, here is what i've been able to understand and deduce:

    Based on the given boundaries, y≤x≤2-y and 0≤y≤1, i break them down and transform each boundary to u and/or v:

    Since both boundaries for x and y are ≥0 (i based this conclusion from the x-y graph), the boundaries for u and v are also ≥0.
    Hence, u≥0 and v≥0.

    y≤x → uv≤u → v≤1
    x≤2-y → x+y≤2 → u≤2
    Therefore, 0≤u≤2 and 0≤v≤1

    y≥0 → u≥0 or v≥0
    y≤1 → v≤1/(u-1) or u≤(1+v)/v → u≤1/v+1
    Therefore, 0≤u≤1/v+1 and 0≤v≤1/(u-1)

    So, now, i ended up with 4 boundaries in u-v. I don't know what's next. I should probably plot these individually or should i compare the boundaries?

    For comparison of the boundaries, i'm assuming that the u's and v's all describe the same region. Is that a correct assumption?

    If i compare the 2 boundaries for u, then i get:
    2=1/v+1
    which gives, v=1
    This would mean 0≤v≤1

    If i compare the 2 boundaries for v, then i get:
    1=1/(u-1)
    which gives, u=2
    This would mean 0≤u≤2

    The comparison results seem to indicate that there was no need to do any comparisons in the first place, as the results were already obtained after the conversion from x-y to u-v coordinates earlier. The boundaries are correct as the integration gives the correct answer, but i'm not sure about this method. It's too easy compared to the graphical method that i've done previously.:smile: I'll have to reproduce this simpler method on other similar problems to verify its validity.
     
    Last edited: Jan 10, 2012
  19. Jan 10, 2012 #18

    ehild

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    You can apply the method together with plotting the boundaries.
    This problem was quite challenging. I had to work with it quite much.

    ehild
     
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