0
$\begingroup$

Is the area/surface density just the volume density with one height dimension in the volume calculation set to 1?

$\endgroup$
0
$\begingroup$

No, that would still give you a 3-dimensional volume. You need to think in just 2 dimensions. You can just ignore the 3rd dimension to get an area. If someone talks about buying property they use the units of square meters. They don't even consider the 3rd dimension, height. If you assume a height of 1 the number may not change but the units will.

$\endgroup$
3
  • $\begingroup$ what if you have an integral with a height of 1 , integrate from 0 to 2 . you get a length. not a square length. You can assign any units you wish for that extra dimension. Can't the same be said of my example which uses the x dimension as a number representing area? I see your reasoning , it is true in multiplication but this is integration....or am I missing something ...? $\endgroup$
    – user86411
    Oct 3 '17 at 15:45
  • $\begingroup$ you can't assign any unit you wish. the unit will be a length but it will mulitple the surface area previously determined so you get a surface area multiplied by another length giving a volume. $\endgroup$
    – Natsfan
    Oct 3 '17 at 17:34
  • $\begingroup$ That is correct and normally I would agree with you ...the reason I am hedge hawing is that after reading your response I recalled the triple integral using dx dy dz in order . Each step uses two dimensions and each step calculates the quadrature , the essence of integration, by assuming the y direction is 1 then the dx integrates to x and there pops out the volume, since you will end up with x, y and z , all is well. But if you assign it another dimension of units as you suggest you get into trouble. otherwise I would agree with you 100% $\endgroup$
    – user86411
    Oct 3 '17 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy