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  1. #1
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    Question jacobian

    use the transformation T(u,v) = < u, u+v> to evaluate the iterated integral
    the integral from 1 to 2 the integral from x+3 to x+6 of 1/(sqrt(xy-x^2)) dy dx.

    please help if you know how to solve this question. thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kittycat View Post
    use the transformation T(u,v) = < u, u+v> to evaluate the iterated integral
    the integral from 1 to 2 the integral from x+3 to x+6 of 1/(sqrt(xy-x^2)) dy dx.

    please help if you know how to solve this question. thank you.
    I will attach the graphs later if i have time. i will show you how to set up the integral.

    Recall how we perform change of variables with Jacobians:

    Assuming all the conditions are fulfilled (look up these conditions in your text), we calculate the integral under the C^1 transformation T as follows:

    \iint_{R}f(x,y)~dA = \iint_{S}f(x(u,v),y(u,v)) \cdot \left| \frac {\partial (x,y)}{ \partial (u,v)} \right|~dudv

    where \frac {\partial (x,y)}{ \partial (u,v)} = \left|  \begin{array}{cc} \frac {\partial x}{\partial u} & \frac {\partial x}{\partial v} \\ & \\ \frac {\partial y}{\partial u} & \frac {\partial y}{\partial v} \end{array}\right|

    we are given:

    \boxed {x = u} and y = u + v \implies \boxed{v = y - x}

    So, we have: \frac {\partial (x,y)}{\partial (u,v)} = \left|  \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right| = 1


    Now, graph the region over which the original integral is being done, it will be a parallelogram. label the lines as i describe so you can follow, we have to transform each line under the directions of T. Label the line y = x + 3 as f_1, the line x = 2 as f_2, the line y = x + 6 as f_3, and the line x = 1 as f_4. note the intervals on which these lines are defined, as we have to change them under the transformation T as well. ( f_1: 1 \le x \le 2, f_2: 5 \le y \le 8, f_3: 1 \le x \le 2, and f_4: 4 \le y \le 7)

    Now, let's get this show on the road.

    For f_1:

    \begin{array}{c|c} y = x + 3 & x = u \\ \Rightarrow u + v = u + 3 & 1 \le x \le 2 \\ \Rightarrow \boxed {v = 3} & \boxed{1 \le u \le 2} \end{array}

    For f_2:

    \begin{array}{c|c} x = 2 & y = 2 + v \\ \Rightarrow \boxed{u = 2} & v = y - 2 \\ & 5 \le y \le 8 \\ & \boxed{3 \le v \le 6} \end{array}

    For f_3:

    \begin{array}{c|c} y = x + 6 & x = u \\ \Rightarrow u + v = u + 6 & 1 \le x \le 2 \\ \Rightarrow \boxed {v = 6} & \boxed{1 \le u \le 2} \end{array}

    For f_4:

    \begin{array}{c|c} x = 1 & y = 1 + v \\ \Rightarrow \boxed{u = 1} & v = y - 1 \\ & 4 \le y \le 7 \\ & \boxed{3 \le v \le 6} \end{array}

    Now we are ready. draw all those graphs between the given intervals on a new pair of axis. Let the vertical axis be v and the horizontal axis be u. you will notice the new region is a rectangle, which is a much simpler region to integrate over. not only that, but the integral gets a lot easier as well:

    We know: \iint_R f(x,y)~dA = \iint_S f(x(u,v),y(u,v)) \cdot \left| \frac {\partial (x,y)}{\partial (u,v)} \right|~dvdu

    = \int_{1}^{2} \int_{3}^{6} \frac 1{\sqrt{u^2 + uv - u^2}} |1|~dvdu

    = \int_{1}^{2} \int_{3}^{6} (uv)^{-\frac 12}~dvdu

    Now continue

    Check my computations, i only got 2 hours of sleep last night, so i am really tired and prone to silly mistakes right now


    EDIT: I attached the diagrams of the regions
    Attached Thumbnails Attached Thumbnails jacobian-first-region.jpg   jacobian-new-region.jpg  
    Last edited by Jhevon; October 1st 2007 at 07:56 PM.
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  3. #3
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    Hi Jhevon,

    Many thanks for your help. I should give you a hug for that.

    I have the final answer (2sqrt(6) -2sqrt(3)) *(2sqrt(2) -2) or approximately 1.189. Am I right?
    Last edited by kittycat; October 2nd 2007 at 07:51 AM.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kittycat View Post
    Hi Jhevon,

    Many thanks for your help. I should give you a hug for that.

    I have the final answer (2sqrt(6) -2sqrt(3)) *(2sqrt(2) -2) or approximately 1.189. Am I right?
    that's what i got. hopefully we're not both wrong

    i take it you followed all the steps ok?
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  5. #5
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    Just to confirm, I have the same answer
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