Question about determining whether vector field is conservative and about determining a potential function for said vector field. So, I've received this question to solve. Can anyone help me? I do not understand how to show that the vector field is conservative in this case. A detailed solution would help to understand what I've been doing wrong so far. (I seem to get answers that indicate that the vector field in question is not conservative, but it should be.)
The question is listed below:
Show that the vector field
F$(x, y, z) = (2x + y)$i$ + (z\cos(yz) + x)$j$ + y\cos(yz)$k
is conservative and determine a potential function.
 A: If we manage to find a potential function $U$ for $\mathbf{F}$, we are done so let's try to do that. We need:
$$ \frac{\partial U}{\partial x} = 2x + y \implies U(x,y,z) = \int (2x + y) \, dx + G(y,z) = x^2 + yx + G(y,z) $$
for some function $G = G(y,z)$ which depends only on $y,z$. Next,
$$ \frac{\partial U}{\partial y} = z \cos(yz) + x \implies x + \frac{\partial G}{\partial y} = x + z \cos(yz) \implies \\
G(y,z) = \int z \cos(y z) \, dy + H(z) = \sin(yz) + H(z)$$
for some function $H = H(z)$ that depends only on $z$.
Finally,
$$ \frac{\partial U}{\partial z} = y \cos(yz) \implies y \cos(yz) + \frac{\partial H}{\partial z} = y \cos(yz) \implies H \equiv C $$
so a potential for $\mathbf{F}$ is given by
$$ U(x,y,z) = x^2 + yx + \sin(yz) + C $$
where $C \in \mathbb{R}$ is arbitrary.
A: Hint:
consider the potential function
$$
f(x,y,z)=x^2+xy+\sin(yz)
$$
can you see how I've found it starting from:
$$
\frac{\partial f}{\partial x}= 2x+y
$$
$$
\frac{\partial f}{\partial y}= x+z\cos (yz)
$$
$$
\frac{\partial f}{\partial z}= y\cos(yz)
$$
A: I think levap pretty much covers it (on how to find a potential function), but the only thing I wanted to add is another way to determine if a vector field is conservative. 
Another way to check if $F$ is conservative is to calculate curl$F$ = curl$\nabla f$ = $\bigg(\dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z}\bigg)i + \bigg(\dfrac{\partial F_x}{\partial z} - \dfrac{\partial F_z}{\partial x}\bigg)j + \bigg(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\bigg)k = 0$. If the curl is not $0$, then $F$ cannot be conservative. 
But if the curl does end up being $0$, you also have to check that $F$ is defined over all of $\mathbb{R}^3$ (or, for all $(x, y, z) \in \mathbb{R}^3$, $F(x, y, z)$ exists), and that all its component functions have continuous partials. In your case, since all the component functions are products and sums of continuous functions that are defined on the entire domain of $\mathbb{R}^3$, you don't have to worry about this condition, but I would just state it on the side.  
Indeed, 

$\bigg(\dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z}\bigg) = cos(yz) - yzsin(yz) - cos(yz) + yzsin(yz) = 0;$ 
$\bigg(\dfrac{\partial F_x}{\partial z} - \dfrac{\partial F_z}{\partial x}\bigg) = 0 - 0 = 0;$ and
$\bigg(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\bigg) = 1 - 1 = 0$.

Then $F$ is indeed conservative. 
I would also refer to this previous post: Question on conservative fields, which gives detailed information on equivalent definitions of conservative. The previous solutions use property 3, there exists $f$ st $\nabla f = F$. 
A: First, check that $\operatorname{curl}\mathbf F=0$. I’ll leave the details of that to you. It is, so $\mathbf F$ is irrotational. Its domain is all of $\mathbb R^3$, which is simply connected, so by Poincaré’s theorem, $\mathbf F$ is also conservative.  
A scalar potential function $U(x,y,z)$ for $\mathbf F$ is given by the line integral $\int_\Gamma\mathbf F\cdot d\mathbf r$, with $\Gamma$ any convenient smooth curve that lies within the domain of $\mathbf F$ joining the origin to the point $P(x,y,z)$. The domain here is all of $\mathbb R^3$, so we can take $\Gamma$ to be the line segment parameterized as $\mathbf r(t)=(tx,ty,tz)$, $t\in[0,1]$. We then have $$\begin{align} U(x,y,z) &= \int_0^1\mathbf F(tx,ty,tz)\cdot(x,y,z)\,dt \\ &= \int_0^1x(2tx+ty)+y(tz\cos(t^2yz)+tx)+z(ty\cos(t^2yz))\,dt \\ &= x^2+xy+\sin(yz).\end{align}$$
