I am having trouble fully grasping the concept of setting a constant as a function of other variables: I would like to use a particular example where I could explain my thought process. Hopefully you could point out where I am going wrong!
Example: Decide whether the vector field $$F=i (x y-\sin (z))+j \left(\frac{x^2}{2}-\frac{e^y}{z}\right)+k \left(\frac{e^y}{z^2}-x \cos (z)\right)$$
is conservative in $D=[\{x,y,z\}):z\neq 0]$ and find a potential if it is.
(Please note that I am following a book example so I'm pretty sure what I am doing is correct, however I am going to point out when I do not understand what is going on and hopefully someone could comment and provide some guidance)
Solution and thought process:
As we can see from this definition the F is not defined when $z=0$. Lets define F1,F2,F3: $\text{F1}=x y-\sin (z);\text{F2}=\frac{x^2}{2}-\frac{e^y}{z};\text{F3}=\frac{e^y}{z^2}-x \cos (z);$
Hence:
$\frac{\partial \text{F1}}{\partial y}=x=\frac{\partial \text{F2}}{\partial x}$,
$\frac{\partial \text{F1}}{\partial z}=-\cos (z)=\frac{\partial \text{F3}}{\partial x}$
$\frac{\partial \text{F2}}{\partial z}=\frac{e^y}{z^2}=\frac{\partial \text{F3}}{\partial y}$
Which means that F might be conservative in domains not intersecting the xy-plane $z=0$.
What we must satisfy:
The potential $\phi$ must satisfy:
$$\frac{\partial \phi }{\partial x}=x y-\sin (z);\frac{\partial \phi }{\partial y}=\frac{x^2}{2}-\frac{e^y}{z};\frac{\partial \phi }{\partial z}=\frac{e^y}{z^2}-x \cos (z);$$
Lets start: $$\phi (x,y,z)=\int (x y-\sin (z)) \, dx=\text{c1}(y,z)+\frac{x^2 y}{2}-x \sin (z)$$
I understand(At least I believe I do) that since we are integrating with respect to one variable that this needs to be held constant.In other words we are allowed to let the constant depend on all the variables apart from the one we are integrating on( please reference if you have a good discussing about why this is the case)
It is now I get stuck:(Please comment on main question)
Because we have $c1(y,z)$, which we currently do not know what is: but what we do now is that:
$$\frac{x^2}{2}-\frac{e^y}{z}=\frac{\partial \phi }{\partial y}=\frac{\partial \text{c1}(y,z)}{\partial y}+\frac{x^2}{2}$$
Now it is this equation that makes me confused: In particular where do we get $$\frac{\partial \phi }{\partial y}=\frac{\partial \text{c1}(y,z)}{\partial y}+\frac{x^2}{2}$$
If we look at: End of example, and take the derivative of $\phi (x,y,z)$ with regards to x,y,z we clearly get the result we want. In other words I do not have a problem verifying that whats being done is correct. What I have a problem with is understanding the underlying logic of how it is derived.
Main question
If $\frac{x^2}{2}-\frac{e^y}{z}=\frac{\partial \phi }{\partial y}$
Then why are we not integrating the entire thing, such that:
$$\int \left(\frac{x^2}{2}-\frac{e^y}{z}\right) \, dy$$
My reasoning for this is that it feels like I am omitting a term by not integrating the entire thing. And I can't find the connection to why integrating the entire thing would be wrong( apart from the fact that when taking the derivative of final function it does not satisfy the conditions anymore)
End of example
You calculated $$\phi (x,y,z)=\int (x y-\sin (z)) \, dx=\text{c1}(y,z)+\frac{x^2 y}{2}-x \sin (z)$$
Now, since you are integrating with respect to $x$ only, you can have a function $c_1(y,z)$, which represents the constant of integration which usually appears. Oberve that in this case, this constant is a function, since $\frac{\partial c_1(y,z)}{\partial x}=0$ and therefore you could have any function, which depends exclusively on $y$ and $z$ that dissapears when you take the derivative with respect to $x$.
Now, going to your main question, you obtained
$$\phi (x,y,z)=\text{c1}(y,z)+\frac{x^2 y}{2}-x \sin (z)$$
Taking the partial derivative with respect to $y$, you have
$$\frac{\partial(\phi (x,y,z))}{\partial y}=\frac{\partial(\text{c1}(y,z)+\frac{x^2 y}{2}-x \sin (z))}{\partial y}=\frac{\partial(\text{c1}(y,z))}{\partial y}+\frac{\partial(\frac{x^2 y}{2})}{\partial y}+\frac{\partial(x \sin (z))}{\partial y}$$
Taking each partial derivative you obtain
$$\frac{\partial(\phi (x,y,z))}{\partial y}=\frac{\partial(\text{c1}(y,z))}{\partial y}+\frac{x^2}{2}$$
Since $\frac{\partial(x \sin (z))}{\partial y}=0$
You could also integrate with respect to $y$, but you would end up with a function $c_2(x,z)$, and following the same procedure, you would obtain it.