# Verify the equation of integration

1. Sep 7, 2013

### yungman

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
$$\frac{1}{2\pi}\int_0^{2\pi} \cos(1+\cos t)\cosh(2+\sin t)\;dt$$

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

Am I correct?

Thanks

Last edited: Sep 7, 2013
2. Sep 7, 2013

### verty

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. Sep 7, 2013

### yungman

Thanks, it's a strange question.