It may take 5 seconds for fieldlines to fully computeThe problem is simply phrased. If we are given $\vec B:\R\rightarrow \R$, can we find $\vec A:\R \rightarrow \R$ s.t. $\nabla \times\vec A = \vec B$, and if so, what is it?
One may have learned in vector calc that the necessary condition for $\vec A$ to exist is that $\nabla \cdot \vec B$ is 0. Here I wish to take a differential geometry approach to find answers to these questions.
If $\vec A$ were to exist, then for a given connected surface $S$ we have the following relations by 3D Stoke's theorem:
$$\iint_S \vec B \cdot d\vec s = \iint_S \nabla \times \vec A \cdot d\vec s = \oint _{\partial S} \vec A\cdot d\vec l.$$
It is immediately recognized that $\vec B = \nabla \times\vec A$ acts on surfaces and $\vec A$ acts on lines. Hence it should be natural to identify $\vec B$ with a 2-form and $\vec A$ with a 1-form.
Formally, we shall take $\omega$ to be a 2-form such that when acting on vector bundles $\vec \xi_1, \vec \xi_2 \in \R^*$ in spatial coordinate $\vec x \in \R$,
$$\omega_{\vec x}(\vec \xi_1, \vec \xi_2) = \vec B(\vec x)\cdot(\vec \xi_1\times\vec\xi_2). \tag{1}$$
Then because we want $\Omega$ to be defined satisfying $d\Omega = \omega$ s.t.
$$\oint _{\partial S} \vec A\cdot d\vec l=\iint_S \vec B \cdot d\vec s=\iint_S \omega = \int_{\partial S} \Omega , $$
we shall for any $\vec \xi\in \R^*$, $\vec x \in \R$ define $$\Omega_{\vec x}(\vec \xi) = \vec A(\vec x) \cdot \vec \xi. \tag{2}$$
Finding $\Omega$
The problem is now rephrased: If $\omega$ is a 2-form in $\R^3$, can we find $\Omega$ a 1-form such that $d\Omega = \omega$?
We shall immediately realize that $\R^3$ is a vector space (and also null-homotopic), so by Poincarè Lemma we know that the necessary and sufficient condition for $\Omega$ to exist is that $\omega$ is exact.
That is, $d\omega = 0$.
Since for any connected volume $V$ in $\R^3$, we have:
$$\int_V d\omega = \int_{\partial V} \omega= \int_{\partial V} \vec B \cdot d\vec s, $$
taking the limit of $V\rightarrow 0$ gives $d\omega(\Pi) = \nabla \cdot \vec B |\Pi|$, where $\Pi$ is any parallelogram and $|\cdot|$ is the volume operator in $\R^3$.
So $d\omega = 0 \Rightarrow \nabla\cdot \vec B=0$.
We can also find the explicit analytical expression of $\vec A$ in terms of $\vec B$ using the cocone operator:
\begin{align}
\Omega &= p\omega\\
\Rightarrow \Omega_{\vec x}(\vec \xi) &= \int_0^1 \omega_{\vec x t}(\vec x, \vec \xi)t dt\\
&= \int_0^1(\vec x\times \vec \xi)\cdot \vec B(\vec x t) t dt\\
&=\int_0^1(t\vec B(\vec x t)\times\vec x)\cdot \vec \xi dt\\
&=\left(\int_0^1 t \vec B(\vec x t)\times \vec x dt\right) \cdot \vec \xi
\end{align}
Now as $ \Omega_{\vec x}(\vec \xi) \equiv \vec A(\vec x) \cdot \vec \xi$,
$$\vec A(\vec x) = \int_0^1 t\vec B(\vec x t)\times \vec x dt.$$
Of course $\vec A$ is non-unique by choice of a gauge, so in general we have
$$\vec A(\vec x) = \left(\int_0^1 t\vec B(\vec x t) dt\right)\times \vec x + \nabla \Phi(\vec x).$$
Applications
We shall interest ourselves first with the vector field $\vec J = (y, z, x)$:
The divergence of this field is obviously $0$. We can also check to find its curl:
$$\nabla \times \vec J = \left(\pd{x}{y}-\pd{z}{z}, \pd{y}{z}-\pd{x}{x}, \pd{z}{x}-\pd{y}{y}\right)=-(1,1,1).$$
Now based on our derivation, we can find its anti-curl by:
\begin{align}
\vec B_0 = p\{\vec J\}&= \int_0^1 t(yt,zt,xt) dt\times \vec x\\
&=\int_0^1 t^2 dt(y,z,x)\times \vec x\\
&=\frac{(y,z,x)\times(x,y,z)}{3}. \tag{3}
\end{align}
\frac{((y,z,x)\times(x,y,z))}{3}Field lines of $\vec B_0.$
One may be tempted to ask further what is the 'anti-curl' of $\vec B_0 = \frac{(y,z,x)\times (x,y,z)}{3}$?
Now its divergence is
\begin{align}
\nabla\cdot \vec B_0&= \frac{(x,y,z)\cdot(\nabla\times(y,z,x))-(y,z,x)\cdot(\nabla\times(x,y,z))}{3}\\
&=\frac{(x,y,z)\cdot(-1,-1,-1)}{3}\\
&=-\frac{x+y+z}{3}.
\end{align}
If we ignore the fact that $\vec B_0=\frac{(y,z,x)\times(x,y,z)}{3}$ is not exact, we can still try to integrate it using our derived formula:
\begin{align}
\vec A_0 = p\left\{\vec B_0\right\}&=\int_0^1 \vec x\times \frac{(yt,zt,xt)\times(xt,yt,zt)}{3} t dt\\
&=\frac{(y,z,x)\times (x,y,z)}{3} \int_0^1 t^3 dt\times \vec x\\
&= \frac{((y,z,x)\times (x,y,z))\times (x,y,z)}{12}.
\end{align}
It's important to remember that cross products are not associative, so $((y,z,x)\times(x,y,z))\times(x,y,z)$ does not cancel to everywhere 0. The traces of this field looks like this :
\frac{((y,z,x)\times(x,y,z))\times(x,y,z)}{12}Field lines of $\vec A_0.$
Comparing it to the curl-field, we can see that there are many places where the field lines of $\frac{((y,z,x)\times(x,y,z))\times(x,y,z)}{12}$ are not "curling" around those of $\frac{(y,z,x)\times(x,y,z)}{3}$.
\frac{((y,z,x)\times(x,y,z))\times(x,y,z)}{12}
\frac{(y,z,x)\times(x,y,z)}{3}Field lines of $\vec A_0$ (purple), and field lines of $\vec B_0$ (red).
That is, we are hoping to see the fields of $\vec B_0$ to be perpendicular to those planes in which if we consider the plane in which the field lines of $\vec A_0$ are "curling". But this is not the case here (look at the poles of the "field-sphere" in the above graph).
Algebraically: $$\nabla \times \vec A_0 = \left(\frac{4z^2+x^2-3xy+xz}{12}, \frac{4x^2+xy+y^2-3yz}{12}, \frac{4y^2+z^2-3xy-yz}{12}\right),$$
which is not equal to: $$\vec B_0=\frac{(x,y,z)\times(y,z,x)}{3} = \left(\frac{z^2-xy}{3}, \frac{x^2-yz}{3},\frac{y^2-zx}{3}\right)$$.
This is of course due to the non-zero divergence of the curl field $\vec B_0$. Thinking in the terms of the derivations of Poincre Lemma, we know that depending on the surface, the net flux over the surfaces $S_1$ and $S_2$ will not be equal even if they have the same boundary.
That is: $$\int_{S_1} \frac{(y,z,x)\times (x,y,z)}{3}\cdot d\vec s \neq \int_{S_2} \frac{(y,z,x)\times (x,y,z)}{3}\cdot d\vec s$$ for different choices of $\partial S_1 = \partial S_2$.
To fix that, we need to subtract it by a function with the same divergence to make it exact.
With $\nabla \cdot \frac{(y,z,x)\times (x,y,z)}{3} = -\frac{x+y+z}{3}$, we can easily think of $\nabla \cdot \frac{(x^2, y^2, z^2)}{6} = \frac{x+y+z}{3}$ and $\nabla \times \frac{(x^2, y^2, z^2)}{6}=0$ s.t. if we let $$\vec B(x,y,z) =\frac{(y,z,x)\times (x,y,z)}{3}+\frac{(x^2, y^2, z^2)}{6}, \tag{4}$$
$\frac{(y,z,x)\times (x,y,z)}{3}+\frac{(x^2, y^2, z^2)}{6}$ with no divergence and same curl as $\vec J$.
then \begin{align}
\vec A(x,y,z)&=\int_0^1 \left(\frac{(yt,zt, xt)\times(xt,yt,zt)}{3}+\frac{\left(x^2t^2, y^2t^2, z^2t^2\right)}{6}\right) \times \vec x t dt\\
&=\left(\frac{(y,z,x)\times(x,y,z)}{3}+\frac{\left(x^2, y^2, z^2\right)}{6}\right)\times (x,y,z)\int_0^1 t^3 dt\\
&= \left(\frac{(y,z,x)\times(x,y,z)}{12}+\frac{\left(x^2, y^2, z^2\right)}{24}\right)\times (x,y,z). \tag{5}
\end{align}
$\vec A$ with curl equal to $\vec B$.
Putting them all together, $(y,z,x)$, anti-curl of $(y,z,x)$:
anti-curl of $(y,z,x)$, anti-curl of anti-curl of $(y,z,x)$:
Now the divergence of this field $\vec A$ is
$$\nabla \cdot \vec A = \frac{xy+yz+zx}{4}.$$
We can easily think of a curl free field with the same divergence, which is $\frac{xyz(1,1,1)}{4}$ to cancel out its divergence and maintain its curl:
$$\vec A = \left(\frac{(y,z,x)\times(x,y,z)}{12}+\frac{\left(x^2, y^2, z^2\right)}{24}\right)\times (x,y,z)- \frac{xyz(1,1,1)}{4}.$$
A divergence free $\vec A$ (purple) satisfying $\nabla \times \vec A = \vec B$ (red).
