# Exact differentials

1. Mar 1, 2016

### Happiness

Why can't (1.39) be put in the form of an exact differential? Seems like I could and the solution to the first equation is

$x-a\phi\sin\theta=c$, where $c$ is an arbitrary constant.

Let $x-a\phi\sin\theta$ be $h$.

By considering $dh=\frac{\partial h}{\partial x}dx+\frac{\partial h}{\partial \phi}d\phi=0$, we get the first equation of (1.39). So it must be a solution. Isn't it?

Derivation 4:

2. Mar 1, 2016

### TeethWhitener

I don't know for sure, but I think it's because $h = h(x,y,\theta,\phi)$, so
$$dh=\frac{\partial x}{\partial h}dx+\frac{\partial y}{\partial h}dy+\frac{\partial \theta}{\partial h}d\theta+\frac{\partial \phi}{\partial h}d\phi$$
and $\frac{\partial \theta}{\partial h}$ is nonzero, so you don't actually get back equation 1.39.