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yungman
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Homework Statement
Identify integral as the mean value of a harmonic function at a point and evaluate the integral:
\frac{1}{2\pi} \int_0^{2\pi} \; cos(1+cos(t)) cosh(2+sin(t)) \; dt
Using:
u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt
Homework Equations
u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt
This equation show if u[/tex] is harmonic function in a region, then the value of u at any point (x_0, y_0) equal to the average of u on the line integral of a circle inside the region centered at (x_0, y_0).<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> \frac{1}{2\pi} \int_0^{2\pi} \; cos( 1+cos(t) ) cosh( 2+sin(t)) \; dt<br /> <br /> x(t) = 1+cos(t) \hbox { and } y(t) = 2+sin(t)<br /> <br /> \nabla^2 u=0 [/tex] and has continuous 1st and 2nd partial derivatives implies u[/tex] is harmonic function which implies:&lt;br /&gt; &lt;br /&gt; u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt&lt;br /&gt; &lt;br /&gt; 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&lt;br /&gt; &lt;br /&gt; Therefore the line integral is a circle center at (1,2).&lt;br /&gt; &lt;br /&gt; u(1,2)= cos(1)cosh(2) = \frac{1}{2\pi} \int_0^{2\pi} \; cos(1+cos(t)) cosh(2+sin(t)) \; dt&lt;br /&gt; &lt;br /&gt; Which the answer is cos(1)cosh(2).&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; But the answer of the book is 2\pi u(0,0) = 2\pi cos(1) cosh(2) which implies the center of the circle is (0,0). I don&amp;#039;t see what I done wrong. Please help.
