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Calculus-integration over mass

  1. Jul 30, 2006 #1
    There are certain integrals which say integration over all m ,integration over all area,integration over line. I am confused with this. In calulus i am comfortable with integration with limits of an independent variable and the
    integration results in the area under the curve. But in doing the moment of inertia of a solid about the axis a small dm is present inside the integral
    and we say to integrate it for M. What is this integration about M exactly .

    Similarly what is integration over line and area. How can i visualize this?
     
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  3. Jul 30, 2006 #2

    ZapperZ

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    For an infinitesimal mass, what you have is

    [itex]dm = \rho dV[/itex]

    where [itex]\rho[/itex] is the mass density. So, in general, the mass of the object is the integral of dm over the volume that is occupied by that object, i.e.

    [itex]M = \int dm = \int \rho dV[/itex]

    If the density is uniform throughout the mass, then you can factor the density out of the integral and all you have, after doing the integral is

    [itex]M = \rho V[/itex]

    which is what you are familiar with as the mass of the object.

    The line and surface integral usually comes in in electrostatic. What you are doing actually is doing a line charge and surface charge integral. So instead of dm, you have [itex]d\lambda[/itex] or [itex]d\sigma[/itex] where [itex]\lambda[/itex] is the infinitesimal line charge defined as

    [itex]\lambda = Q dl[/itex]

    and [itex]d\sigma[/itex] is defined as

    [itex]d\sigma = Q dA[/itex].

    You'll notice that this is of similar form that we had for dm, where the "nature" of the quantity (i.e. mass M, or charge Q) is paired with the "dimension", i.e volume or length or area.

    Zz.
     
    Last edited: Jul 30, 2006
  4. Jul 31, 2006 #3

    quasar987

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    There is a thread about this going on in the math section. Here's the link,

    https://www.physicsforums.com/showthread.php?t=127425
     
  5. Jul 31, 2006 #4

    quasar987

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