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player1_1_1
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Homework Statement
I have very funny thing to do; I must count average distance between two points in circle of R radius:D (hahaha)
Homework Equations
Defined integrals, parametric equations, etc
The Attempt at a Solution
I am describing the circle in two parametres:
[tex]\begin{cases}x(t)=R\cos t\\ y(t)=R\sin t\end{cases}[/tex]
I can choose any two points by choosing any parameters [tex]t_1,t_2[/tex] in [tex]\langle 0;2\pi\rangle[/tex] limits. Two points [tex]A_1,A_2[/tex] will be described by coordinates
[tex]A_1=\left(R\cos t_1,R\sin t_1\right),A_2=\left(R\cos t_2,R\sin t_2\right)[/tex]
and distance between them in [tex]t_1,t_2[/tex] function
[tex]s\left(t_1,t_2\right)=\sqrt{\left(\cos t_1-\cos t_2\right)^2+\left(\sin t_1-\sin t_2\right)^2}[/tex]
now I am making double integral in limits in which [tex]t_1,t_2[/tex] changes, tzn [tex]t_1,t_2\in\langle 0;2\pi\rangle[/tex] divided by the field where the function is defined
[tex]\frac{\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{\left(\cos t_1-\cos t_2\right)^2+\left(\sin t_1-\sin t_2\right)^2}\mbox{d}t_1\mbox{d}t_2}{\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\mbox{d}t_1\mbox{d}t_2}[/tex]
is it correct answer for average value of this function?
now I am counting integral in this way, bottom integral is [tex]4\pi^2[/tex] value, so I have
[tex]\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{\left(\cos t_1-\cos t_2\right)^2+\left(\sin t_1-\sin t_2\right)^2}\mbox{d}t_1\mbox{d}t_2=[/tex]
[tex]=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{\cos^2t_1-2\cos t_1\cos t_2+\cos^2t_2+\sin^2t_1-2\sin t_1\sin t_2+\sin^2t_2}\mbox{d}t_1\mbox{d}t_2=[/tex]
[tex]=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{2-2\cos t_1\cos t_2-2\sin t_1\sin t_2}\mbox{d}t_1\mbox{d}t_2=[/tex]
[tex]=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{2\left(1-\cos\left(t_2-t_1\right)\right)}\mbox{d}t_1\mbox{d}t_2=[/tex]
[tex]=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{4\sin^2\left(\frac{t_2-t_1}{2}\right)}\mbox{d}t_1\mbox{d}t_2=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}2\left|\sin\left(\frac{t_2-t_1}{2}\right)\right|\mbox{d}t_1\mbox{d}t_2[/tex]
is it good way of thinking? how can I count this double integral (i mean what to do with this abs)? thanks for help!