Question about potential function of a vector field

Question

Lets say we have conservative vector field $\mathbf{F}(x,y,z)=y\sin{z}\mathbf{i}+x\sin{z}\mathbf{j}+xy\cos{z}\mathbf{k}$

I found it's potential function like this $$\begin{cases} \frac{\partial{\phi}}{\partial{x}} = y\sin{z} \\ \frac{\partial{\phi}}{\partial{y}} = x\sin{z} \\ \frac{\partial{\phi}}{\partial{z}} = xy\cos{z} \end{cases}\int \to \begin{cases} \phi(x,y,z) = xy\sin{z}+C_1(y,z) \\ \phi(x,y,z) = xy\sin{z}+C_2(x,z) \\ \phi(x,y,z) = xy\sin{z}+C_3(x,y) \end{cases}$$

Stupid question but is the constant of the potential function $C(x,y,z)$ or C. Because in my books solution its C and I think it has to be $C(x,y,z)$.

Answer 1

It seems that it has to be constant. From the first coordinate of vector field you obtain that $\phi (x,y,z)$ must be of form $\phi(x,y,z) = xy\sin z + C_1(y, z)$. Now you have to compare the vector field which is generated by this potential with your initial vector field. They agree on the first component, but you have to check 2nd and 3rd yourself. It's necessary that $x\sin z = \frac{\partial \phi}{\partial y} = x\sin z + \frac{\partial C_1(y, z)}{\partial y}$, so $\frac{\partial C_1(y, z)}{\partial y} \equiv 0$ and $C_1(y,z)$ has no dependence on $y$. Then you check third coordinate $xy \cos z = \frac{\partial \phi}{\partial z} = xy\cos z + \frac{\partial C_1(z)}{\partial z}$. From here it follows that $\frac{\partial C_1(z)}{\partial z} \equiv 0$ and $C_1(x,y,z)$ is constant.

Answer 2

Not a stupid question at all. The problem is one of notation. What you really know after the first step is that $\phi$ is $xy \sin z$ plus some function of just $y$ and $z$. Perhaps best to call that $C_1(y, z)$. Similarly for the other two...but the three "C"s are not all the same. In the case you've looked at, you can pick all three to be zero. And in general, you can add a constant to all of them, of course. But the problem becomes one of "how can I glue together the three different solutions that I've found, adding a function of only $y$ and $z$ to the first, or $x$ and $z$ to the second, etc., to get a single answer that's consistent with all three constraints?"

Answer 3

Consider that the function $\phi $ is just one and only function, not three. So in fact, $\ C_1 = C_2 = C_3 $. Since you are demanding it not to be a function of x, nor a function of y or z, then the only choice is that it must be just a number. The $\ C_i $ functions are intermediate steps that follow from the integration with respect to one variable, but get defined when you differentiate wrt to the next one, and compare with the partial derivative you already know (although maybe you have got this just by guessing). Try to differentiate $\phi $ and see that $\frac{\partial{C}}{\partial{x}} = \frac{\partial{C}}{\partial{y}} = \frac{\partial{C}}{\partial{z}} = 0$