How do I solve $F = \nabla\times G$ for $G$? Given the vector valued function $F(x,y,z) = (xz,-yz,y)$ find $G$ such that $F = \nabla\times G$
I let $G(x,y,z) = (G_1,G_2,G_3)$ and expanded $\nabla \times G$ then equated the components to $F$ but I don't know what to do from here.
How would I go about solving these equations? or is there a better way of doing this? Thanks for any help.
 A: The problem amounts to finding a vector potential $\mathbf G$ such that $\mathbf F = \nabla \times \mathbf G$. The existence of a vector potential for $\mathbf F$ is equivalent to saying that $\mathbf F$ is divergence free:
$$\nabla \cdot \mathbf F = \nabla \cdot (\nabla \times \mathbf G) = 0.$$
In our case we have $\nabla \cdot \mathbf F = 0$, so there indeed exists a vector potential $\mathbf G$.
Now note that the choice for $\mathbf G$ is far from unique. Let $f$ be any differentiable scalar function on $\mathbb R^3$. We then have
$$\nabla \times(\mathbf G + \nabla f) = \nabla \times \mathbf G .$$
Thus $\mathbf G' = \mathbf G + \nabla f$ is also a vector potential for $\mathbf F$. We can exploit this non-uniqueness to simplify the set of equations $$ \partial_y G_z - \partial_z G_y = F_1, \quad \partial_z G_x - \partial_x G_z = F_2, \quad \partial_x G_y - \partial_y G_x = F_3 \; ,$$
by choosing our $f$ such that $\partial_z f = -G_z$. This simplifies the set of equations to
$$-\partial_z G'_y = F_1, \quad \partial_z G'_x = F_2, \quad \partial_x G'_y - \partial_y G'_x = F_3 \;.$$
This choice of $f$ is purely to start off with a simple set of equations, and does not need calculation.
The problem now amounts to finding $\mathbf G'$. Integrating the first two equations yields $G'_y$ and $G'_x$ with integration constants $C_1(x,y)$ and $C_2(x,y)$ respectively. In the last equation we have to make a choice for these constants.
