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Math Help - Jacobian determinant

  1. #1
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    Jacobian determinant

    I'm going to rush through most of this stuff, as I feel I've got it correct. It's just the answer to the last question that's bugging me.
    Question
    Let R be the region in the xy-plane enclosed by the lines:
    x + y = 0, x + y = 1, x y = 1, x y = 4.
    (a) Draw a picture of the region R.
    answer:
    y= -x, y = 1-x, y=x-1 & y = x-4. It makes a rectangle with the corners (0.5,-0.5), (1,0), (2,-2), & (2.5,-1.5).

    (b) Define a change-of-variables transformation by the equations
    u = x + y & v = x y.
    What is the image S in the uv-plane of R under this transformation?
    answer:
    solving for x and y:
    x=1/2(u+v) & y=1/2(u-v)
    Putting these into the above equations I get the lines u=0, u=1, v=1/2 and v=2. It makes a rectangle with the corners (0,1/2), (0,2), (1,1/2) & (1,2)

    (c) Solve the transformation equations for x and y in terms of u and v, and hence find the determinant of the Jacobian matrix.
    answer:
    \tfrac{\partial x}{\partial u} = \tfrac{1}{2}, \tfrac{\partial x}{\partial v} = \tfrac{1}{2}, \tfrac{\partial y}{\partial u} = \tfrac{1}{2}, \tfrac{\partial y}{\partial v} = \tfrac{\text{-}1}{2}
    \frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix}\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{\text{-}1}{2}\end{vmatrix}=\frac{\text{-}1}{2}

    (d)Use this change of variables to evaluate
    \int \int_R (x - y)e^{x^2-y^2}dA
    answer:
    simplifying
    x-y = \tfrac{1}{2}(u+v)-\tfrac{1}{2}(u-v) = v
    x^2-y^2 = (\tfrac{1}{2}(u+v))^2 - \tfrac{1}{2}(u-v))^2 = \frac{uv}{2}
    makes the equation:
     \int_{\frac{1}{2}}^{2}\int_{0}^{1}v.e^{\frac{vu}{2  }}(\tfrac{\text{-}1}{2}). \partial u\partial v
    inner integral:
     \int_{\frac{1}{2}}^{2}\int_{0}^{1}v.e^{\frac{vu}{2  }}(\tfrac{\text{-}1}{2}). \partial u\partial v = -e^{\frac{uv}{2}}\bigg|_{0}^{1} = \int_{\frac{1}{2}}^{2}\text{-}e^{\frac{v}{2}}+1. \partial v
    outer integral:
    \int_{\frac{1}{2}}^{2}\text{-}e^{\frac{v}{2}}+1. \partial v = v-2e^{\frac{v}{2}}\bigg|_{\frac{1}{2}}^{2} = (2 - 2e^{1})-(\tfrac{1}{2} - 2e^{\frac{1}{4}})= 1\tfrac{1}{2} - 2(e^{1}-e^{\frac{1}{4}})

    So is my answer to (d) correct? It's seems so unwieldy and unpleasant an answer considering how nice and easy the rest of the Question was. If anyone could check my work and affirm/negate my answers, please do!
    cheers
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  2. #2
    MHF Contributor matheagle's Avatar
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    1 you need not find the inverse transformation

    I would obtain {\partial(u,v)\over\partial(x,y)}=-2 instead, so our jacobian is absolute
    value of the reciprocal of -2, i.e., +.5.

    2 clearly x^2-y^2=(x+y)(x-y)=uv not uv/2

    3 you need the absolute value around the jacobian, so it's plus 1/2, not -1/2.

    4 The region is 0<u<1 and 1<v<4.

     .5\int_1^4\int_0^1  ve^{uv}dudv =.5\int_1^4(e^v-1)dv
    Last edited by matheagle; May 6th 2010 at 11:40 PM.
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  3. #3
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    thanks for that. Too many stupid small mistakes on my part add up to a nasty final answer!
    One thing: why do you have the limits for v: 1<v<4 and not 1/2<v<2?
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  4. #4
    MHF Contributor matheagle's Avatar
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    just insert u and v into....

    x + y = 0, x + y = 1, x y = 1, x y = 4.

    you need not even draw this, it's too simple
    somehow you have that 1/2 you incorrectly obtained via bad algebra
    check that inverse transformation, I bet it's wrong
    YOU don't need to obtain the inverse transform.
    I get the upside down jacobian and take the reciprocal
    Lots of books don't mention that you can do that
    Plus you do need the absolute values too.
    Is this from Stewart by the way?
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