What is the Difference Between ∫X.dY and ∫Y.dX in Physical Applications?

[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?

 

[mfb]

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

I can't even think of what ∫v.dp means

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).

 

[AlephZero]

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.

Correct.

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.

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".

I can't even think of what ∫v.dp means. What does it mean?

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

 

[transparent]

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".

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?

 

[pmacias]

I can understand your confusion and frustration with understanding the difference between ∫X.dY and ∫Y.dX in the physical world. In pure mathematics, these integrals may represent different areas, but in the physical world, they have different physical interpretations and applications.

∫X.dY represents the work done by a force F as it moves through a displacement in the Y direction. This can be thought of as the area under the force-displacement curve. On the other hand, ∫Y.dX represents the work done by a force F as it moves through a displacement in the X direction, which can be thought of as the area under the force-displacement curve in the X direction.

In physical situations, the direction of the force and the direction of the displacement are crucial in determining the work done. This is why ∫X.dY and ∫Y.dX may give different results. The first integral is calculating the work done in the Y direction, while the second is calculating the work done in the X direction. This is why only the first is called the work done, as it is the work done in the direction of the displacement.

In the case of pressure-volume work, ∫p.dv represents the work done by a gas as it expands or contracts. This can be thought of as the area under the pressure-volume curve. On the other hand, ∫v.dp does not have a physical interpretation as it does not represent a physical quantity. This is because volume cannot be directly related to pressure in the same way that force is related to displacement.

In summary, while ∫X.dY and ∫Y.dX may seem similar in pure mathematics, in the physical world they have different interpretations and applications. It is important to consider the direction of the force and displacement when interpreting these integrals in physical situations.