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    I am attaching a question and the solution I have written.

    The answer is correct.

    There is one point that I am unhappy about so I have fudge it sort of speak.

    The region is transformed from the triangle to the "trapezium" in the second page but I cannot justify analytically why the region "finishes" at u=0.

    For instance why can not be a straight line (or indeed a curve) which joins (0,1) to (1,0).

    (note there is also a singularity at x = 0 in the transformation equations)

    Any suggestions?


    jacobians_and_curvilinear_coodinates.pdf
    IMG_0016_NEW.pdf
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    I get the following:

    If u=x+y, v=\frac{y}{x} then:

    \frac{u}{x} = 1 +\frac{y}{x} = 1 + v \Rightarrow x = \frac{u}{1+v}

    \frac{u}{y} = \frac{x}{y} +1 = \frac{1}{v} + 1 = \frac{1+v}{v} \Rightarrow y = \frac{uv}{1+v}

    So in terms of limits:

    y=x \Rightarrow \frac{y}{x}=1 \Rightarrow v = 1

    x =1 \Rightarrow u = 1+v

    y=0 \Rightarrow \frac{uv}{1+v} = 0 \Rightarrow uv=0 \Rightarrow u=0 \text{ or } v = 0

    Does that help?
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    (Original post by atsruser)
    I get the following:

    If u=x+y, v=\frac{y}{x} then:

    \frac{u}{x} = 1 +\frac{y}{x} = 1 + v \Rightarrow x = \frac{u}{1+v}

    \frac{u}{y} = \frac{x}{y} +1 = \frac{1}{v} + 1 = \frac{1+v}{v} \Rightarrow y = \frac{uv}{1+v}

    So in terms of limits:

    y=x \Rightarrow \frac{y}{x}=1 \Rightarrow v = 1

    x =1 \Rightarrow u = 1+v

    y=0 \Rightarrow \frac{uv}{1+v} = 0 \Rightarrow uv=0 \Rightarrow u=0 \text{ or } v = 0

    Does that help?
    I am not in a "fit state" at present, but I think this does it.
    I will look at it tomorrow.
    many thanks.

    (PS will rep you as soon as I can rep you again)
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    (Original post by atsruser)
    I get the following:

    If u=x+y, v=\frac{y}{x} then:

    \frac{u}{x} = 1 +\frac{y}{x} = 1 + v \Rightarrow x = \frac{u}{1+v}

    \frac{u}{y} = \frac{x}{y} +1 = \frac{1}{v} + 1 = \frac{1+v}{v} \Rightarrow y = \frac{uv}{1+v}

    So in terms of limits:

    y=x \Rightarrow \frac{y}{x}=1 \Rightarrow v = 1

    x =1 \Rightarrow u = 1+v

    y=0 \Rightarrow \frac{uv}{1+v} = 0 \Rightarrow uv=0 \Rightarrow u=0 \text{ or } v = 0

    Does that help?
    I am ashamed but I have done something stupid (blame the vino).

    In the middle of the 1st page I have re-aranged incorrectly something that not even a GCSE student would have done.
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    (Original post by TeeEm)
    I am ashamed but I have done something stupid (blame the vino).

    In the middle of the 1st page I have re-aranged incorrectly something that not even a GCSE student would have done.
    Having thought about this, the transformation is a bit odd since the x-axis is apparently mapped to both the lines u=0 and v=0; it's not even a good function, if what I wrote above is correct.

    However, if we look at the original equations then if y=0 (i.e. for points on the x-axis) we have:

    v = y/x = 0 except at the origin
    u = x+y = x which isn't identically zero on the x-axis

    So I'm not sure what's going on. I think I've confused myself.
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    (Original post by atsruser)
    Having thought about this, the transformation is a bit odd since the x-axis is apparently mapped to both the lines u=0 and v=0; it's not even a good function, if what I wrote above is correct.

    However, if we look at the original equations then if y=0 (i.e. for points on the x-axis) we have:

    v = y/x = 0 except at the origin
    u = x+y = x which isn't identically zero on the x-axis

    So I'm not sure what's going on. I think I've confused myself.
    I am aware of the singularity but I am happy to ignore it since I know the integral converges.

    My problem was that I could not map the boundary correctly due to my stupidity.

    When you rearranged x=x(u,v) and y=y(u,v) then the penny dropped that my rearrangements (although I eliminated differently) was not the same!

    If you see my rearrangement for x = f(u,v) in the middle of the first page, you will laugh...

    Anyway, the problem is sorted

    (+5 already allocated this morning)
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    (Original post by TeeEm)
    I am aware of the singularity but I am happy to ignore it since I know the integral converges.
    My concern wasn't for the singularity though, but rather for the fact that 1) the mapping doesn't seem to be a function (it's multivalued on y=0, apparently) and 2) that I seem to get contradictory results depending on whether I play around with (x,y) or (u,v).

    Did you come up with that transform yourself or it is a standard thing?
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    (Original post by atsruser)
    My concern wasn't for the singularity though, but rather for the fact that 1) the mapping doesn't seem to be a function (it's multivalued on y=0, apparently) and 2) that I seem to get contradictory results depending on whether I play around with (x,y) or (u,v).

    Did you come up with that transform yourself or it is a standard thing?
    I made the question backwards, i.e. I started with ex over the trapezium and picked 2 simple transformation equations at random to create an "unrecognisable" integrand,( where naturally I know the transformation that will unlock it.)

    the integral has to exist because two different computer algebra systems will integrate it over the triangle to the answer I got.
 
 
 
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