# Integration in real world.

1. Nov 16, 2013

### transparent

What is the difference between ∫X.dY and ∫Y.dX in the physical world? I know what the difference is in pure mathematics. ∫X.dY represents the are bounded by the curve and the Y axis while ∫Y.dX represents the area bounded by the curve and the X axis. But I am unable to translate this into physical situations.

For example when we are doing ∫F.dX, we are calculating force multiplied by displacement for very small displacements and then adding them all up. Similarly ∫X.dF should mean that we are multiplying the displacement of the particle for a very small change in F with that small change. Should they not mean the same thing and give the same result? Yet only the first is called the work done. One other example that I can think of is that of pressure-volume work. ∫p.dv is the work done. I can't even think of what ∫v.dp means. What does it mean?

2. Nov 16, 2013

### Staff: Mentor

X and Y are arbitrary variables. If you do not define them, the integrals are meaningless.

Integrate the volume for a pressure change. If volume is a function of pressure, this is possible (I am not sure if there is a useful physical interpretation for that).

3. Nov 16, 2013

### AlephZero

Correct.

No. we are multiplying "the displacement of the particle", not "the displacement of the particle for a very small change in F". In other words, "the displacement from origin", not "the small change in displacement".

If you write down a mathematical formula "at random", usually it doesn't mean anything physically.

4. Nov 16, 2013

### transparent

That makes sense. But ∫v.dp should mean something since dH=dU+pdV+Vdp. Here dU is the change in internal energy and pdV is the work done. What is Vdp?