Thread: 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?

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.