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Transform the region and compute the integral

  1. Jan 26, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    This is an exercise from Taylor & Mann page 468 Q5 :

    Use the transformation x=au and y=bv to map the region R defined by [itex]\frac{x^2}{a^2} + \frac{y^2}{b^2} ≤ 1[/itex] onto the uv plane.

    Evaluate : [itex]\int \int_R \frac{x^2}{a^2} + \frac{y^2}{b^2} dxdy[/itex]

    with the aid of this transofrm and polar coordinates.

    2. Relevant equations

    [itex]\int \int_R F(x,y) dxdy = \int \int_{R'} G(u,v)|J| dudv[/itex]

    Where |J| is the Jacobian.

    3. The attempt at a solution

    So if R is defined to be [itex]\frac{x^2}{a^2} + \frac{y^2}{b^2} ≤ 1[/itex], then using the transformation x=au and y=bv we define a new region R' by [itex]u^2 + v^2 ≤ 1[/itex]

    Now I can easily set up a Cartesian integral in terms of u and v, but the point is to use polars to simplify things.

    So let u = rcosθ and v = rsinθ and hence R' becomes [itex]r ≤ 1[/itex] since r>0 for [itex]0 ≤ θ ≤ 2π[/itex]

    The Jacobian of polars is just r, so J = r.

    Using all this information, our integral becomes :

    [itex]\int_{0}^{1} \int_{0}^{2π} r^3 dθdr = \pi/2[/itex]

    I'm getting the feeling I'm missing something here as the answer at the back of the book is πab/2 which sadly my integral almost evaluates to, but not quite.

    Any pointers?
     
  2. jcsd
  3. Jan 26, 2013 #2
    Just a slight careless mistake when you went from xy-space to uv-space:
    dx = a du
    dy = b dv
    This returns you the factor of ab that you are missing.
     
  4. Jan 26, 2013 #3

    Dick

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    What happened to the Jacobian of the transformation (x,y)->(u,v)?
     
  5. Jan 26, 2013 #4

    Zondrina

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    Ahhhhh thank you good sir :) I got the answer I wanted now.
     
  6. Jan 26, 2013 #5

    Zondrina

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    The Jacobian of polars is just r so yeah.
     
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