# Clarify definite integration of differentials in physics problems

The notation is misleading - while $$\mathrm{d}U_p$$ really is the differential of a state function $$U_p$$, there is no state function $$W$$ whose differential $$\mathrm{d}W$$ could be. That is, it is an "inexact differential" or inexact 1-form, which you can evaluate along paths but for which no potential function exists. The "$$W$$" we usually write on the l.h.s. of your equation should be thought of as $$W[\gamma]$$, where $$\gamma$$ is the path you're integrating along, i.e. "work" is a functional on paths whose value you get by integrating the inexact differential $$\mathrm{d}W$$. Some people are careful to make the inexactness visible by writing inexact differentials as $$\delta W$$, but there seems to be no consensus on this.

See also this answer by Joshphysics for a formal proof of the fact that the existence of the path functional "work" $$W$$ is equivalent to the existence of a 1-form $$\mathrm{d}W$$. Physically, this still has the meaning of $$\mathrm{d}W$$ being the "infinitesimal version" of $$W$$, but as said, the crucial difference to something like potential energy is that $$W$$ is not a function on spatial points, but a function on the paths, hence the $$\mathrm{d}$$ in $$\mathrm{d}W$$ does not denote ordinary differentiation.

The force from pressure in your second example is the same, just in one dimension higher: The $$P\mathrm{d}A$$ is an inexact 2-form that can be integrated over 2-dimensional objects (=surfaces), and this produces a functional on surfaces that we can call $$F$$ that assigns to any surface that integral.

In general, we note that the viewpoint that unifies both exact and inexact differentials is the notion of differential p-forms that can be integrated over p-dimensional objects. The differentials $$\mathrm{d}U_p, \mathrm{d}W, \mathrm{d}F$$ appearing here are all examples of such forms. The $$\mathrm{d}U_p$$ is special, because it is the (exterior) derivative of a 0-form (a function) $$U_p$$, while the others are no such derivatives. In full generality, if you have a $$p$$-form $$\omega$$ that is the exterior derivative of a $$p-1$$-form, you can use a general version of Stokes' theorem to reduce an integral of $$\omega$$ over a $$p$$-dimensional object to the integral of $$\sigma$$ over the $$p-1$$-dimensional boundary of that object.

Since you're wondering in a comment how to tell whether any given form (or "differential" is a derivative or not: This is answered by Poincaré's lemma: On nice (contractible) regions, it is necessary and sufficient for the (exterior) derivative of a form to vanish in order for it to have a $$p-1$$-form that is its antiderivative.