# Density of a material in different dimensions

Hope this isn't something trivially wrong, I am a beginner in classical mechanics. But anyways, say there is a cube, of side length $a \ \text{cm}$. The volume of the cube is $a^3 \ \text{cm}^3$. If the total mass of the cube is $m \text{ kg}$, the density of the cube will be $\frac{m}{a^3} \text{kg/cm}^3$. Again, consider the cube as the combination of many square plates of negligible thickness. If the mass of each is $m_2 \ \text{kg}$, their density will be $\frac{m_2}{a^2} \ \text{kg/cm}^2$. Here's the question: we found the density in two ways, and their dimensions don't match. I think I know why this happens- we are finding the density of the cube by integrating the densities of each plate. But how would you answer the following question?

The density of a cube is $X \text{ kg/cm}^3$. Find the density of each square plate.

Again, I am sorry if this is something trivial, I am a novice. :)

• Is each square plate of different material? Oct 26 '13 at 16:28
• Density is an intensive property, so shouldn't it remain the same for each plate? Or are you trying to find the surface density, rather than the volume density? Oct 26 '13 at 16:38

What you found while finding density of plates of negligible thickness was surface mass density, but what you ade trying to find is volume mass density what is commonly called just density. When you tried to find the density(volume mass density) with the plates you would have to add both their thickness and mass of a cube thag yoh would make to find the density. Othewise if you just want density of negligible thickness plate you have to divide the mass with both area of plate and its negligible thickness.

For finding density of each square plate of density X assuming its side is a would be X x (da) here da is the thickness of the plate and what you get is the surface mass density of the plate. If you simply want volume density of the plate then it would still be X as density of a material ( all kinds ) is an intrinsic property and does not depend on shape and size of its individual particles if separated from parent body.

It sounds as if you are building up the cube by integrating a series of infinitesimally thin plates. In that case we would normally call the thickness of the plates $dx$, so the volume of each plate is $a^2 dx$ and the density is then $dm/(a^2dx)$. The units of the density are still $ML^{-3}$.

If:

$$dm = \rho a^2 dx$$

then to get the mass we integrate:

\begin{align} M &= \int_0^a \rho a^2 dx \\ &= \rho a^2 \int_0^a dx \\ &= \rho a^2 \left[ x \right]_0^a \\ &= \rho a^3 \end{align}

giving us your original result of $\rho = M/a^3$.