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Math Help - line integrals

  1. #1
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    line integrals



    may i know how to solve this question? i tried it , as seen in the picture, but got stuck
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by alexandrabel90 View Post

    may i know how to solve this question? i tried it , as seen in the picture, but got stuck
    \frac{d}{dt}\frac{1}{\sqrt{x^2+y^2+z^2}}=-\frac{1}{2\cdot (x^2+y^2+z^2)^{3/2}}\cdot (2x \frac{dx}{dt}+2y\frac{dy}{dt}+2z\frac{dz}{dt})=\ld  ots

    So \frac{1}{\sqrt{x^2+y^2+z^2}} is an anti-derivative of the integrand in question...
    Last edited by Failure; April 12th 2010 at 10:52 PM.
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  3. #3
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    how were you able to deduce from this integration that the equation that you wrote is the anti-derivative of the integrand in question?

    thanks!
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    how were you able to deduce from this integration that the equation that you wrote is the anti-derivative of the integrand in question?

    thanks!
    To be quite honest with you it was primarily because this is a famous example of a "central force" being conservative (gravitation), and I knew that in the one-dimensional case, it is quite obvious that this amounts to \int -\frac{1}{r^3}\cdot r\, \frac{dr}{dt}\, dt=-\int\frac{1}{r^2}\,dr = \frac{1}{r}+C.
    The three-dimensional case is a little less obvious than this, but it also amounts to a simple substitution of r := \sqrt{x^2+y^2+z^2}, checking that dt can be gotten rid of and then the integral really does turn out to have a value that is dependent on r only, and therefore only on the location of the end-points of the path.

    Moral of the story: Nothing beats memory when it comes to problem-solving . I am told that all experts mainly rely on memory (aka. experience), and use reasoning only when they absolutely have to.
    Last edited by Failure; April 12th 2010 at 11:43 PM.
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