Compute the following line integral along a path of your choice (Finding potential) Consider the following vector field:
$$\vec A(x,y,z)=(yz)\hat i+(xz)\hat j+(xy)\hat k$$
Compute the line integral of $A$ along a path of your choice connecting $(0,0,0)$ to $(1,1,1).$
I recognise that $\nabla\times A=0$ and so I can simply evaluate $V(1,1,1)-V(0,0,0)$ where $A=\nabla V$. 
I'm not sure how to find the function $V$ though. 
The solution says $V=xyz.$
 A: Assume V as a function of x,y and z.
Then $\nabla V=(\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y},\dfrac{\partial V}{\partial z})$
Compare components:
$$\dfrac{\partial V}{\partial x}=yz$$
$$\dfrac{\partial V}{\partial y}=xz$$
$$\dfrac{\partial V}{\partial z}=xy$$
Integrate:
$$V=xyz+f_1(y,z)$$
$$V=xyz+f_2(x,z)$$
$$V=xyz+f_3(x,y)$$
From here you can see that $V=xyz+const.$ by inspection. If you want to do it mathematically you have to subtract two equations from each other.
1-2: $f_1(y,z)=f_2(x,z)$ from that you can conclude $f_1(y,z)=f_1(z)$ and $f_2(x,z)=f_2(z)$, or you would have $y$ on the left but no $y$ on the right, and so forth. 
1-3: $f_1(z)=f_3(x,y)$.As one can see we have z on the left sid but no z on the right. Hence, $f_1$ is a constant function as it is not depending on $z$. if $f_1=const.$ $f_1=f_3(x,y)$ will imply that $f_3=f_1=const.$.  Same argument as before and we can conclude $f_1=f_3=const$. But if $f_1=const.$ than from 1-2 $f_2=const$.
With that we obtain $f_i=const$ for $i=1,2,3$.
At the end you will see, that $f_i=const$ for $i=1,2,3$.
A: $\nabla G = A$
what is G?
$G = \int yz \,dx + f(y,z) = \int xz \,dy + g(x,z) = \int xz \,dz + h(x,y)\\
G = xyz + f(y,z) = xyz + g(x,z) = xyz + h(x,y)$
$f(y,z), g(x,z), h(x,y)$ must all be constant functions, and they can be any constant function, so might as well make them $0.$
