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Math Help - Help with an integral

  1. #1
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    Help with an integral

    Hi.

    Can anybody help me with the solution of the next integral:

    \int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi

    where \theta=\theta(\phi).

    The problem is that don't know the relation between \theta and \phi.
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  2. #2
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    Quote Originally Posted by guest View Post
    Hi.

    Can anybody help me with the solution of the next integral:

    \int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi

    where \theta=\theta(\phi).

    The problem is that don't know the relation between \theta and \phi.
    looks somewhat like an arc length integral for a polar function ... in what context was this integral set up?
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  3. #3
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    lets try this: call the value of the definate integral f, square it and evaluate it as a double integral. the square root of the double integral should be f also lets pretend that theta and phi are independant since we havent got a relationship between them

    f=\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi
    f^2=(\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi)^2
    f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta(\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2) (d \phi)^2 f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta\sin^2 \thetad\phi^2+\int_{\phi_0}^{\phi}\int_{\phi_0}^{\  phi}\cos^2\thetad\theta^2
    f^2=\int_{\phi_0}^{\phi}(\phi-\phi_0)\cos^2\theta\sin^2\theta*d\phi+\int_{\phi_0  }^{\phi}\frac{2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0}{4}d\theta
    f^2=(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}
    finally
    f= \sqrt{(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}}

    i cannot guarentee this is the answer youre looking for, however this is the only way i was able to integrat the function e^{-x^2}
    good luck!!
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  4. #4
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    Quote Originally Posted by skeeter View Post
    looks somewhat like an arc length integral for a polar function ... in what context was this integral set up?
    This is a part of the formula for the speed down an unknown path on a sphere with friction.
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  5. #5
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    Quote Originally Posted by JeffN12345 View Post
    f=\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi
    f^2=(\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi)^2
    f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta(\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2) (d \phi)^2 f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta\sin^2 \theta d\phi^2+\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi}\c  os^2\theta d\theta^2
    f^2=\int_{\phi_0}^{\phi}(\phi-\phi_0)\cos^2\theta\sin^2\theta*d\phi+\int_{\phi_0  }^{\phi}\frac{2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0}{4}d\theta
    f^2=(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}
    finally
    f= \sqrt{(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}}
    Your intgrating would be okay if \theta \neq \theta(\phi). I think that because of not knowing the relation \theta = \theta(\phi) this integral can't be solved annalyticaly.
    Last edited by guest; September 3rd 2009 at 01:43 AM.
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