Anti-curl operator

It is known that if a vector field $\vec{B}$ is divergence-free, and defined on $\mathbb R^3$ then it can be shown as $\vec{B} = \nabla\times\vec{A}$ for some vector field $A$.

Is there a way to find $A$ that would satisfy this equation? (I know there are many possibilities for $A$)

Note: I want to find it without using the explicit formula for $B_x(x,y,z), B_y(x,y,z), B_z(x,y,z)$, but maybe with a formula involving surface/curve integrals. For example, I've found that in the 2D case (if $B_z=0$ and $\vec{B}=\vec{B}(x,y)$) then $A$ can be shown as:

$$\vec{A}(x,y)=\hat{z}\int_{\vec{R_0}}^{\vec{r}} (\hat{z}\times\vec{B})\cdot\vec{dl}$$

I am looking for something similar in the general case.

This is called Poincare's Lemma. I will write the standard version, in a neighborhood around the origin. The usual phrasing is to say that a closed form is locally exact, the fact that this is not globally true is the stuff of cohomology. This is from pages 94-96 of Calculus on Manifolds by Michael Spivak.

Given your divergence-free vector field $(F_1(x,y,z), \; F_2(x,y,z), \; F_3(x,y,z)),$ the $x$-coordinate of the new vector field $G$ is $$ G_1(x,y,z) = \int_0^1 \; \left( \; t z F_2(tx, ty,tz) - t y F_3(tx, ty,tz) \; \right) \; dt, $$ the $y$-coordinate is $$ G_2(x,y,z) = \int_0^1 \; \left( \; t x F_3(tx, ty,tz) - t z F_1(tx, ty,tz) \; \right) \; dt, $$
with $z$-coordinate $$ G_3(x,y,z) = \int_0^1 \; \left( \; t y F_1(tx, ty,tz) - t x F_2(tx, ty,tz) \; \right) \; dt. $$

Note that fractions tend to show up if you have any exponents. I did a test run with a random field, $$ H = (xyz, \; x y^2 z^3, \; x y^3 z^5).$$ I then took the curl to get $$ F = \nabla \times H = ( 3 x y^2 z^5 - 3 x y^2 z^2, \; x y - y^3 z^5, \; y^2 z^3 - x z).$$ The three components are what I am calling $F_1,F_2,F_3.$ We know that $F$ is a curl, by construction, and we know it is divergence free (check!). Going through Poincare's recipe, after fixing a few of my bookkeeping errors, gave instead $$ G_1 = \frac{1}{2} x y z - \frac{1}{10} y^3 z^6 - \frac{1}{7} y^3 z^3,$$ $$ G_2 = \frac{4}{7} x y^2 z^3 - \frac{1}{4} x^2 z - \frac{3}{10} x y^2 z^6,$$ $$ G_3 = \frac{4}{10} x y^3 z^5 - \frac{1}{4} x^2 y - \frac{3}{7} x y^3 z^2.$$ This has a bit of a different appearance from $H.$ That is fine. As $H,G$ have the same curl, it follows merely that $(G-H)$ is the gradient of some function.

On that note, if you have a curl-free field $W = (W_1, W_2, W_3),$ it is the gradient of a function $f$ given by $$ f(x,y,z) = \int_0^1 \; \left( \; x W_1(tx, ty,tz) + y W_2(tx, ty,tz) + z W_3(tx, ty,tz) \; \right) dt.$$

I'm not sure if this is what you meant by excluding the 'explicit formula', but such a vector field is constructed in the proof of Helmholtz' theorem; $\mathbf{A}(\mathbf{r})$ is given by

$$\mathbf{A}(\mathbf{r}) = \frac{1}{4\pi} \int_{\mathbb{R}^3} \frac{\nabla \times \mathbf{B}(\mathbf{r}')}{\mathbf{|r - r'|}}$$

where $\mathbf{r'}$ is the variable you're integrating over.

To see why this works, you need to take the curl of the above equation; however, you'll need some delta function identities, especially

$$\nabla^2(1/\mathbf{|r - r'|}) = -4 \pi \delta(\mathbf{r - r'}).$$

If you're at ease with those, you should be able to finish the proof on your own. If you're not sure, just ask over here and I'll be glad to provide details.

I do NOT (repeat: NOT) own the rights to this webpage, but I think this might be more along the lines of about what the OP was asking http://galileo.math.siu.edu/Courses/251/S12/vpot.pdf