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Thread: Dispersed Volume Integral

  1. #1
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    Dispersed Volume Integral

    A represents a volume that is directly proportional to the value (0.5w - d) and inversely proportional to how the volume is distributed from a central point in space (c^2). Would the integral of c^2 represent the distributed volume?

    Dispersed Volume Integral-ac.jpg
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  2. #2
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    Re: Dispersed Volume Integral

    Hey rickS.

    You have a value for the volume in proportion (meaning that A = cV where A is area, V is value and c is some non-zero constant) and inverse proportionality in terms of how its distributed.

    This means the distribution of volume (i.e. the density) is p = d*1/|(X-C)| where p is density, C is some point and X is an arbitrary point with d as a constant (again - non-zero).

    You need to clarify what you mean by distributed volume because that term by itself makes absolutely no sense.
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