- #1

yungman

- 5,618

- 225

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

[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: