So I have come across this previous post, that reads:
Deriving Biot-Savart Law from Maxwell's Equations
"As an exercise, I've been trying to derive the Biot-Savart law from the second set of Maxwell's equations for steady-state current
$$\begin{align}&\nabla\cdot\mathbf{B}=0&&\nabla\times\mathbf{B}=\mu_0\mathbf{J}\end{align}$$
I've been able to do this using the fact that an incompressible field has a vector potential $\mathbf{A}$, allowing me to rewrite the second equation as
$$\nabla^2\mathbf{A}=-\mu_0\mathbf{J}$$
which can be solved by components using the Green's function for the Laplacian, yielding
$$\mathbf{A}(\mathbf{x})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\,d^{3}\mathbf{x}'$$
and since $\nabla\times\left(\psi\mathbf{J}\right)=\psi\nabla\times\mathbf{J}+\nabla\psi\times\mathbf{J}$, $$\nabla\times\mathbf{A}=\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}\times(\mathbf{x}-\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|^3}\,d^{3}\mathbf{x}'$$
as desired. However, if instead I take the curl of both sides of Ampere's Law, and use the identity $\nabla \times \left( \nabla \times \mathbf{B} \right) = \nabla(\nabla \cdot \mathbf{B}) - \nabla^{2}\mathbf{B}$ initially, I find that
$$\nabla(\nabla \cdot \mathbf{B}) - \nabla^{2}\mathbf{B}=0-\nabla^2\mathbf{B}=\mu_0\nabla\times\mathbf{J}$$
which I can again solve like Poisson's equation, yielding
$$\mathbf{B}(\mathbf{x})=-\frac{\mu_0}{4\pi}\int\frac{\nabla'\times\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\,d^3\mathbf{x}'$$
which can be simplified using the identity $\psi(\nabla\times\mathbf{J})=-\nabla\psi\times\mathbf{J}+\nabla\times\left(\psi\mathbf{J}\right)$, giving
$$\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{x}')\times(\mathbf{x}-\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|^3}\,d^3\mathbf{x}'-\frac{\mu_0}{4\pi}\int\nabla'\times\left(\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\right)\,d^3\mathbf{x}'$$
The first integral is precisely the Biot Savart law, but I have no idea how to make the second integral vanish. I've exhausted any obvious vector calculus identities, and Stokes theorem doesn't help much. I'm clearly missing an obvious mistake, but I haven't been able to locate it. This is similar to other questions that have been asked before, but I have a specific question about a step in the derivation which is not answered elsewhere."
The solution in the comments are basically saying that the reason this last integral is zero, is because we are integrating over all of space, and they can show that this can be turned into a surface integral that vanishes when the surface is being taken to infinity.
But what if we choose a volume that doesnt go to infinity, yet STILL encloses all of the current density? would this integral vanish? and how do I prove this
I have a feeling that I would need to plug this solution back into amperes law as I am taking the curl of the curl of amperes law which loses information about the function J, Can someone here prove biot savart in the last way described by the post, that can be generalised to any volume enclosing charge, and also checking that the solution satisfied maxwells equations :)
