Why does h = h(x,y,θ,φ) prevent (1.39) from being an exact differential?

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The discussion centers on why the expression \( h = h(x,y,\theta,\phi) \) prevents equation (1.39) from being an exact differential. The key point is that when deriving \( dh \), the term \( \frac{\partial \theta}{\partial h} \) is nonzero, which disrupts the conditions necessary for an exact differential. Consequently, this leads to the conclusion that the solution cannot be expressed in the required exact form, despite initial appearances suggesting otherwise.

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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?

Screen Shot 2016-03-02 at 12.04.16 am.png

Derivation 4:
Screen Shot 2016-03-02 at 12.04.37 am.png
 
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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.
 
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