(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Identify integral as the mean value of a harmonic function at a point and evaluate the integral:

[tex]\frac{1}{2\pi} \int_0^{2\pi} \; cos(1+cos(t)) cosh(2+sin(t)) \; dt[/tex]

Using:

[tex] u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt[/tex]

2. Relevant equations

[tex] u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt[/tex]

This equation show if [itex]u[/tex] is harmonic function in a region, then the value of u at any point [itex](x_0, y_0)[/itex] equal to the average of u on the line integral of a circle inside the region centered at [itex] (x_0, y_0) [/itex].

3. The attempt at a solution

[tex]\frac{1}{2\pi} \int_0^{2\pi} \; cos( 1+cos(t) ) cosh( 2+sin(t)) \; dt[/tex]

[tex] x(t) = 1+cos(t) \hbox { and } y(t) = 2+sin(t) [/tex]

[itex]\nabla^2 u=0 [/tex] and has continuous 1st and 2nd partial derivatives implies [itex]u[/tex] is harmonic function which implies:

[tex] u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt[/tex]

[tex] x(t) = 1+cos(t) \hbox { and } y(t) = 2+sin(t) \;\; \Rightarrow\;\; x_0=1 \;\hbox{ and }\; y_0=2\hbox{ Therefore } u(x_0, y_0)=u(1,2) \; \hbox{ and } r=1[/tex]

Therefore the line integral is a circle center at (1,2).

[tex]u(1,2)= cos(1)cosh(2) = \frac{1}{2\pi} \int_0^{2\pi} \; cos(1+cos(t)) cosh(2+sin(t)) \; dt[/tex]

Which the answer is [itex]cos(1)cosh(2)[/itex].

But the answer of the book is [itex] 2\pi u(0,0) = 2\pi cos(1) cosh(2)[/itex] which implies the center of the circle is (0,0). I don't see what I done wrong. Please help.

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