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Verify the equation of integration

  1. Sep 7, 2013 #1
    I want to verify the the value of ##x_0## and ##y_0## of the given integral according to the formula of Mean Value of Harmonic function
    [tex]\frac{1}{2\pi}\int_0^{2\pi} \cos(1+\cos t)\cosh(2+\sin t)\;dt[/tex]


    Mean Value of Harmonic function on a disk ##\Omega## given:
    [tex]u(x_0,y_0)=\frac {1}{2\pi}\int_{\Omega}u[(x-x_0),(y-y_0)] d\Omega[/tex]
    [tex]\Rightarrow\;u[(x-x_0),(y-y_0)]=\cos(1+\cos t)\cosh(2+\sin t)[/tex]
    [tex]\Rightarrow\;(x-x_0)=1+\cos t,\;(y-y_0)=2+\sin t[/tex]
    Using Polar coordinates, ##x=r\cos t,\;y=r\sin t## where ##r=1## in this case.
    [tex](x-x_0)=1+\cos t\;\Rightarrow\; x_0=-1\;\hbox{ and }\;(y-y_0)=2+\sin t\;\Rightarrow\;y_0=-2[/tex]

    Am I correct?

    Thanks
     
    Last edited: Sep 7, 2013
  2. jcsd
  3. Sep 7, 2013 #2

    verty

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    Homework Helper

    I don't know this subject but the smallest |x - x_0| is when t = ##π \over 2##, then x = x_0, so x_0 = -1, x = - cos(t). And for t = 3/4 π, |y - y_0| = 1 is the smallest, so I think, when y = -1, y - y_0 = 1, so y_0 = -2, y = sin(t). So I agree with your calculations. x - x_0 is a sum of two terms, one independent of t and one dependent, and the same goes for y - y_0, so I think this must be right.
     
  4. Sep 7, 2013 #3
    Thanks, it's a strange question.
     
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