# Differentials in mechanics

1. Sep 14, 2014

### txdw16

Hey all,

I just started a fluid mechanics class and I'm having trouble interpreting the physical meaning behind differentials in this free body diagram.

For example, γδxδyδz. I know gamma is the specific weight of the block of fluid. And I know δ is the differential length in x, y, or z directions. But together as an expression what does this mean?

My thought process is:

λ=ρg

and it makes sense that the weight of the "block" of fluid is

λ*Volume

But wouldnt that look something like this:

∫∫∫λδxδyδz

If you're not integrating, then what does it mean when they're just hanging there?
Also, is there a difference between the notation δx and dx?

Thanks!

2. Sep 14, 2014

### Matterwave

The δ notation usually means you are considering a very small element. So the integral is not needed because the element is so small (probably you will take the size to 0 in some specific way at the end of the calculation) that you don't need the integral since γ will not vary inside this small volume.

3. Sep 14, 2014

### txdw16

Okay thanks for the response that makes sense. But as far as the geometric meaning goes, what does γδxδyδz mean?

In my head I think "The change in specific weight in any direction".

But that phrase makes me think more along the lines of:
$\frac{\partial{\gamma}}{\partial{x}}\hat{i} + \frac{\partial{\gamma}}{\partial{y}}\hat{j} + \frac{\partial{\gamma}}{\partial{z}}\hat{k}$

I'm probably just confusing myself. Any insight would be appreciated

4. Sep 14, 2014

### voko

This is an example of "physics math" that is meant to be more intuitive than rigorous math required to analyse the situation would have been, but it is only intuitive after you have grasped its ways. $\delta x$ means a "small change in $x$", and it may also mean $dx$ whenever that is convenient. $\delta x \delta y \delta z$ is the volume of a (small) box with (small) sides $\delta x, \ \delta y, \ \delta z$. And because things are as small as we need them to be, $\lambda$ does not change significantly within the box, the weight of the box is just $\lambda \delta x \delta y \delta z$ We ignore all of the inaccuracies of this description and say that things are equal when they are approximately equal.

5. Sep 14, 2014

### SteamKing

Staff Emeritus
You had the correct insight in the OP, and then you started thinking too much.

The volume of the small element shown in the picture is indeed δxδyδz. We assume that because the dimensions of this element are so small, the specific weight γ can be treated as a constant. Hence, the weight of this tiny element is expressed as δw = γδxδyδz.