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Math Help - Double Integrals

  1. #1
    Member Abu-Khalil's Avatar
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    Double Integrals

    Evaluate \int\int_Re^{\frac{x-y}{x+y}} where R=\{(x,y)\in\mathbb{R}^2:0\leq x,0\leq y,x+y=1\} using x-y=u,x+y=v.

    I got \frac{1}{2}\int_0^2\int_u^1e^{\frac{u}{v}}dvdu but maple says im wrong.
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  2. #2
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    Quote Originally Posted by Abu-Khalil View Post
    Evaluate \int\int_Re^{\frac{x-y}{x+y}} where R=\{(x,y)\in\mathbb{R}^2:0\leq x,0\leq y,x+y=1\} using x-y=u,x+y=v.

    I got \frac{1}{2}\int_0^2\int_u^1e^{\frac{u}{v}}dvdu but maple says im wrong.
    Your limits of integration are incorrect see the diagram

    Double Integrals-capture.jpg

    So u=-v..v;v=0..1

    I hope this helps
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  3. #3
    Member Abu-Khalil's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Your limits of integration are incorrect see the diagram

    Click image for larger version. 

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    So u=-v..v;v=0..1

    I hope this helps
    Not really >.<

    I did x=\frac{u+v}{2},y=\frac{v-u}{2} and then

    y=0\iff u=v
    y=1-x\iff =\frac{v-u}{2}=1-\frac{u+v}{2}\iff v=1
    x=0\iff u=0
    x=1\iff u=2

    could you do it in the same way?
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  4. #4
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    Quote Originally Posted by Abu-Khalil View Post
    Not really >.<

    I did x=\frac{u+v}{2},y=\frac{v-u}{2} and then

    y=0\iff u=v
    y=1-x\iff =\frac{v-u}{2}=1-\frac{u+v}{2}\iff v=1
    x=0\iff u=0
    x=1\iff u=2

    could you do it in the same way?
    yes really

    On the bottom of the triangle y=0 and x=0..1

    using this we get in the u-v plance

    u=x-y \iff u=x-0 \iff u=x and
    v=x+y \iff v=x+0 \iff v=x so as x goes from 0 to 1

    u and v go to 1. so we get the equation u=v


    Now on the hypotenuse y=-x+1 so we get

    u=x-y \iff u=x-(-x+1) \iff u=2x-1 and
    v=x+y \iff v=x+(-x+1) \iff v=1 so as x goes from 1 to 0

    So we get the horizontal line segment v=2 as u goes from -1 to 1

    Finally on the last part we get x=0 y =0..1


    u=x-y \iff u=0-y \iff u=-y and
    v=x+y \iff v=y \iff v=y so as y goes from 1 to 0

    we get the line segment  u=-v

    So we get the triangle in my above post.
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  5. #5
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    always a sketch will help to clarify things.

    for the (x,y) region you have a triangle whose coordinates are (0,0),\,(0,1),\,(1,0), thus under the transformation it's (0,0),\,(-1,1),\,(1,1) for the (u,v) region.

    from there things are a bit clear to take the new region where the double integral is taken.

    of course, don't forget the Jacobian.
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  6. #6
    Member Abu-Khalil's Avatar
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    Thanks you both
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