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Thread: Change of Variables...final tomorrow!

  1. #1
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    Exclamation Change of Variables...final tomorrow!

    Use the transformation:
    x = (u+v)/2
    y = (v−u)/2
    double integral bounded by R of: (x − y)(x + y)dxdy

    where R is the region bounded by the lines y = x + 2, y = x, y = −x + 2 and
    y = −x + 4.


    Again...its the bounds of the integral that are killing me. Thanks ahead of time my book is just terrible.
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  2. #2
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    Note from your transformation that $\displaystyle u=x-y$ and $\displaystyle v=x+y$. Now, note that our region is a rectangle at 45 in the xy axes, and out bounding lines can be rewritten as: $\displaystyle x-y=-2$, $\displaystyle x-y=0$, $\displaystyle x+y=2$, and $\displaystyle x+y=4$, so we have bounds in uv of $\displaystyle u=-2$, $\displaystyle u=0$, $\displaystyle v=2$ and $\displaystyle v=4$. Does that help?

    --Kevin C.
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  3. #3
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    how did you find out what u and v were equal to
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  4. #4
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    I found u and v by the same methods one uses to solve a system of equations;
    since you have:
    $\displaystyle x=\frac{u+v}{2}=\tfrac{1}{2}u+\tfrac{1}{2}v$
    $\displaystyle y=\frac{v-u}{2}=-\tfrac{1}{2}u+\tfrac{1}{2}v$
    adding the two equations together gives:
    $\displaystyle x+y=\tfrac{1}{2}u+\tfrac{1}{2}v-\tfrac{1}{2}u+\tfrac{1}{2}v=v$
    and subtracting the second equation from the first gives:
    $\displaystyle x-y=\tfrac{1}{2}u+\tfrac{1}{2}v-(-\tfrac{1}{2}u+\tfrac{1}{2}v)=u$
    Does that make sense?

    --Kevin C.
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