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Integral with infinitesimal under root

  1. Jun 15, 2014 #1
    I want compute the following integral:

    $$\\ \int f(x,y) \sqrt{dx^2+dy^2}$$
    Is correct this pass-by-pass:
    $$\\ \sqrt{\left( \int f(x,y) \right)^2} \sqrt{dx^2+dy^2} = \sqrt{\left( \int f(x,y) \right)^2 (dx^2+dy^2)} = \sqrt{\left( \int f(x,y) \right)^2 dx^2 + \left( \int f(x,y) \right)^2 dy^2} = \sqrt{\left( \int f(x,y) dx \right)^2 + \left( \int f(x,y) dy\right)^2}$$
    ?
     
  2. jcsd
  3. Jun 15, 2014 #2

    micromass

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    Please define what this means. Or at least, please tell us where you saw this.
     
  4. Jun 15, 2014 #3
    The only time that I have ever seen notation like this is when ##x## and ##y## can be parameterized. Then This is interpreted as
    ##\displaystyle\int f(x,y)\sqrt{\left(\frac{dx}{dt}\right)^2dt^2+\left(\frac{dy}{dt}\right)^2dt^2}=\int f(x,y)\sqrt{x'(t)^2+y'(t)^2}dt##
    which may be hard to compute but should make sense. Passing the integral under the square root does not make sense. An integral is a limit of sums which can't be passed under the square root.

    The integral property ##\left|\int fdx\right|\leq\int\left|\ fdx\right|## should convince you that your first step is not guaranteed to work.
     
  5. Jun 15, 2014 #4

    verty

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    Converting to polar coordinates is an option, that ugly square-root will become ##dr##.
     
  6. Jun 15, 2014 #5
    That's true too. It depends on whether this is a surface or line integral.
     
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