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Integrating over unit circle

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Express

    f(x,y) = 1/sqrt(x^2 + y^2) . (y/sqrt(x^2 + y^2)) .exp(-2sqrt(x^2 + y^2))

    in terms of polar coordinates [tex]\rho[/tex] and [tex]\varphi[/tex] then evaluate the integral over a circle of radius 1, centered at the origin.

    2. Relevant equations

    x = [tex]\rho[/tex]cos[tex]\varphi[/tex]
    y = [tex]\rho[/tex]sin[tex]\varphi[/tex]

    sin^2[tex]\varphi[/tex] + cos^2[tex]\varphi[/tex] = 1

    3. The attempt at a solution

    ok so here's my effort

    after rearranging and substituting: f([tex]\rho[/tex],[tex]\varphi[/tex]) = sin[tex]\varphi[/tex]exp(-2[tex]\rho[/tex])

    now let's integrate!
    limits are 0 [tex]\leq[/tex] [tex]\rho[/tex] [tex]\leq[/tex]1
    and 0 [tex]\leq[/tex] [tex]\varphi[/tex] [tex]\leq[/tex] 2[tex]\pi[/tex]

    [tex]\int[/tex][tex]\int[/tex] sin[tex]\varphi[/tex]exp(-2[tex]\rho[/tex]) d(fi) d(rho)

    the problem is sin becomes -cos so, -cos(2pi) - -cos(0) = 0

    giving a final answer of zero doesn't make much sense, does it? so what arent i getting?
     
    Last edited: Jan 25, 2010
  2. jcsd
  3. Jan 25, 2010 #2
    Didn't you drop a [tex]\rho[/tex] in the integrand?

    AB
     
  4. Jan 25, 2010 #3

    vela

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    You made a mistake when converting f to polar coordinates, and you made another one in writing down the integral. It turns out they cancel each other, so you got the right answer, which is 0.

    Note that the original integrand is an odd function of y. Since the unit circle is symmetric about the y-axis, the integral turns out to be 0.
     
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