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Thread: Double Integrals

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

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

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

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

    Double Integrals-capture.jpg

    So $\displaystyle u=-v..v;v=0..1$

    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 $\displaystyle u=-v..v;v=0..1$

    I hope this helps
    Not really >.<

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

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

    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 $\displaystyle x=\frac{u+v}{2},y=\frac{v-u}{2}$ and then

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

    could you do it in the same way?
    yes really

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

    using this we get in the u-v plance

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

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


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

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

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

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


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

    we get the line segment$\displaystyle u=-v$

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