1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted