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Infinitesimal surface / volume

  1. Jun 21, 2011 #1
    When i develop integrals, changing the coordinates (cartesian-> polar for example), i always forget how to write infinitesimal surface or volume.
    Is there a sort of rule to derivate it?
    I mean, an intuitive way to remember it, not the mathematical derivation.

    (another thing: i've the same problem with the Prosthaphaeresis formulas)
     
    Last edited: Jun 21, 2011
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  3. Jun 21, 2011 #2

    HallsofIvy

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    What do you mean "infinitesmal surface or volume"? The differential of surface or volume?

    I believe the "rule" you are referring to is the Jacobian determinant.

    If u= f(x,y,z), v= g(x,y,z), w= h(x,y,z).

    Then the differential of volume, is given by
    [tex]dudvdw= \left|\begin{array}{ccc}\frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} & \frac{\partial w}{\partial x} \\ frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} & \frac{\partial w}{\partial y} \\ \frac{\partial u}{\partial z} & \frac{\partial v}{\partial z} & \frac{\partial w}\end{array}\right|dxdydz[/tex]
     
  4. Jun 21, 2011 #3

    micromass

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    Fixed LaTeX in Halls post...
     
  5. Jun 21, 2011 #4

    HallsofIvy

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    Thanks. For some reason couldn't get it to work. What was wrong?
     
  6. Jun 21, 2011 #5

    micromass

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    The very last \frac was of \frac{\partial w}. But you forgot a denominator :frown:
     
  7. Jun 21, 2011 #6

    I like Serena

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    The intuitive way to do it, is to look at the infinitesimal surface (or volume) and find out how long the sides of it are.
    Typically all regular coordinate systems have an orthonormal basis at any point, which makes the infinitesimal surface a rectangle (or cube).

    In cartesian an infinitesimal volume is:
    [tex]dV = dx \ dy \ dz[/tex]

    In spherical we have sides:
    [tex]dr, \ r d\theta, \ r \sin \theta \ d\phi[/tex]

    giving you an infinitesimal volume of:
    [tex]dV = r^2 \sin \theta \ dr \ d\theta \ d\phi[/tex]

    If you look at it like this, in general you don't need the Jacobian (which will yield exactly this result).

    The boundaries of the integral are always just the boundaries, such that you integrate over the entire surface (or volume).


    Note that the needed absolute value of the Jacobian in spherical coordinates is:
    [tex]|J| = |r^2 \sin \theta| = r^2 \sin \theta[/tex]
     
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