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Verify the equation of integration

  • Thread starter yungman
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  • #1
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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:

Answers and Replies

  • #2
verty
Homework Helper
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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.
 
  • #3
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Thanks, it's a strange question.
 

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