Finding Potential for a conservative vector field using Line Integral The vector field $\textbf{A} = (yz , xz , xy ) $ in cartesian coordinates is irrotational, or has zero curl. This means there should exist a function $\phi : \mathbb{R}^3 \rightarrow \mathbb{R} $ with $ \nabla \phi  = \textbf{A} $. To find $\phi$ I could  look at the partial derivatives like $\frac{\partial \phi }{\partial x } = yz $ and end up getting $\phi = xyz + c $, where $c \in \mathbb{R}$.
However I want to do it using a line integral of $\textbf{A}$. In one dimension, if $g(x) = f'(x)$ then $f(x) = \int_{x_0}^{x} g(s) ds$ ,  so analogously this should mean that $\phi = \int_{\textbf{r}_0}^{\textbf{r}} \textbf{A} \cdot d\textbf{r}$. So using this line of reasoning I get that $\phi = \int_{\textbf{r}_0}^{(x,y,z)}( YZ \,  dX + XZ \,  dY + XY \, dZ ) = 3xyz + c, c \in \mathbb{R}$, but the gradient of this function is $3\textbf{A}$.
Where Have I gone wrong? Do I need to specify a path, despite line integrals of conservative vector fields being independent of path? If so, why?
Edit: Here is how I integrated. Let $\textbf{r}_0 = (a,b,c )$ be arbitrary. $ \phi = \int_{\textbf{r}_0}^{(x,y,z)}( YZ \,  dX + XZ \,  dY + XY \, dZ ) =  YZX  + XZY + XYZ  \, \, |_{\textbf{r}_0}^{(x,y,z)} = 3XYZ  \, \, |_{\textbf{r}_0}^{(x,y,z)} = 3xyz -3abc = 3xyz + c $ where $c$ is the constant of integration.
 A: Ok that's definitely not how line integrals are defined. You can't just pull out the $YZ$ from the integral and then say $\int dX=X$, and then multiply to get $YZX$ and then evaluate the difference at the endpoints (likewise for the other two terms). This is the problem one falls into when not careful with definitions.
The first way of solving the problem, which is the obvious way, is to observe that the differential form
\begin{align}
\omega= YZ\,dX+XZ\,dY+XY\,dZ=d(XYZ)
\end{align}
is equal to the $d$ of some function. Hence, we have a potential function, determined up to a constant as $\phi(x,y,z)=xyz$.
Since you want to avoid doing this, we can try another approach. If you somehow already know that the integral is path-independent, then the function $\phi:\Bbb{R}^3\to\Bbb{R}$ defined as
\begin{align}
\phi(x,y,z):=\int_{\gamma_{(x,y,z)}}\omega,
\end{align}
where $\gamma_{(x,y,z)}$ is ANY path joining say the origin and $(x,y,z)$, will be such that $\omega=d\phi$.
Of course, to actually calculate this integral explicitly, one has to work carefully from the definitions. If you already know the integral is path-independent, then to evaluate it, you may as well choose a simple path, such as the straight-line segment joining the two points: given any point $(x,y,z)$, let us consider the simple path $\gamma_{(x,y,z)}:[0,1]\to\Bbb{R}^3$ defined as $\gamma_{(x,y,z)}(t)=(tx,ty,tz)$. With this, the calculation proceeds as follows:
\begin{align}
\phi(x,y,z)&=\int_{\gamma_{(x,y,z)}}YZ\,dX+XZ\,dY+XY\,dZ\\
&=\int_0^1(ty)(tz)(x\,dt)+(tx)(tz)(y\,dt)+(tx)(ty)(z\,dt)\\
&=\int_0^13t^2xyz\,dt\\
&=xyz \int_0^13t^2\,dt\\
&=xyz
\end{align}
Of course, now it is easily double-checked that $\omega=d\phi$, so everything works out.
A: $\textbf{A} = (yz, xz, xy)$. As you mentioned the curl is zero. While the potential function is obvious in this case, you can follow the below steps in general.
$\displaystyle \frac{\partial \phi}{\partial x} = yz, \frac{\partial \phi}{\partial y} = xz, \frac{\partial \phi}{\partial z} = xy$
$\displaystyle \phi(x, y, z) = \int A_x \ dx = xyz + g(y, z)$
Taking partial derivative wrt $y, \displaystyle \frac{\partial \phi}{\partial y} = xz + \frac{\partial g}{\partial y}$. Comparing with $j$ component of $\textbf{A}$, $\displaystyle \frac{\partial g}{\partial y}$ is zero. In other words, $g$ is not a function of $y$.
Then take partial derivative wrt $z, \displaystyle \frac{\partial \phi}{\partial z} = xy + \frac{\partial g}{\partial z}$. Comparing with $k$ component of $\textbf{A}$, $\displaystyle \frac{\partial g}{\partial z} = 0$.
So $\phi(x, y, z) = xyz$.
