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Math Help - Path Integral

  1. #1
    Senior Member vincisonfire's Avatar
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    Path Integral

    I have to compute the arc length of  r = 1 + cos(\theta)
    I found the formula  \int_0^{2\pi} \sqrt{r^2+(\frac{dr}{d\theta})^2} d\theta =  \int_0^{2\pi} \sqrt{2+2cos(\theta)} d\theta *
    I don't know how to do this integral.
    Can you please help me?
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  2. #2
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    Dear vincisonfire,

    one of tricks: cos(theta) = cos(2*(theta/2)) and
    cos(2x) = cos^2(x)-sin^2(x).
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  3. #3
    Senior Member vincisonfire's Avatar
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    I would have *\int \sqrt{1+cos(\theta)} = \int \sqrt{1+cos^2(\frac{\theta}{2})-sin^2(\frac{\theta}{2})} = \int \sqrt{2cos^2(\frac{\theta}{2})}
    Back to my problem *
     2 \int_0^{2\pi} cos(\frac{\theta}{2}) d\theta= 4 [sin(\frac{\theta}{2})]_0^{2\pi} = 0
    I should have 8 no ?
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  4. #4
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    khm...khm...
    \sqrt{x^2} \neq x
    such in x = -3
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  5. #5
    Eater of Worlds
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    Integrate from 0 to Pi and multiply by 2.

    8sin(\frac{\theta}{2})|_{0}^{\pi}=8
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  6. #6
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    Hello, vincisonfire!

    I have to compute the arc length of  r \:=\: 1 + cos(\theta)

    I found the formula: .  \int_0^{2\pi} \sqrt{r^2+(\tfrac{dr}{d\theta})^2} d\theta \;\;=  \int_0^{2\pi} \sqrt{2+2\cos\theta}\, d\theta

    We're expected to know this identity: . \cos^2\tfrac{\theta}{2} \:=\:\frac{1+\cos\theta}{2} \quad\Rightarrow\quad 1 + \cos\theta \:=\:2\cos^2\!\tfrac{\theta}{2}

    We have: . \sqrt{2(1 + \cos\theta)} \;=\;\sqrt{2\left(2\cos^2\!\tfrac{\theta}{2}\right  )} \;=\;2\cos\tfrac{\theta}{2}


    Therefore: . L \;=\;2\int^{2\pi}_o\!\!\cos\tfrac{\theta}{2}\,d\th  eta\quad\hdots . Got it?

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  7. #7
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    It is not true!

    See above!
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  8. #8
    Math Engineering Student
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    Quote Originally Posted by Soroban View Post

    We have: . \sqrt{2(1 + \cos\theta)} \;=\;\sqrt{2\left(2\cos^2\!\tfrac{\theta}{2}\right  )} \;=\;2\cos\tfrac{\theta}{2}


    Therefore: . L \;=\;2\int^{2\pi}_o\!\!\cos\tfrac{\theta}{2}\,d\th  eta\quad\hdots . Got it?
    Here's where the mistake is, that should be \int_{0}^{2\pi }{\left| \cos \frac{\theta }{2} \right|\,d\theta }, hence,

    \begin{aligned}<br />
   \int_{0}^{2\pi }{\left| \cos \frac{\theta }{2} \right|\,d\theta }&=2\int_{0}^{\pi }{\left| \cos \theta  \right|\,d\theta } \\ <br />
 & =2\left( \int_{0}^{\pi /2}{\cos (\theta )\,d\theta }-\int_{\pi /2}^{\pi }{\cos (\theta )\,d\theta } \right) \\ <br />
 & =2(1+1)=4.<br />
\end{aligned}
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