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The volume current density $ J$ is defined as $ \frac {dI}{da_{\perp}}$ where $dI$ is a small current segment flowing in the volume with cross section $da_{\perp}$ where $da_{\perp}$ is a small area element that is perpendicular to the current flow.

So in the next simple example, a uniform volume current density flowing parallel to the axis of a simple cylinder, the cross section is circular, so our intuition tells us that $J$ should be equal to $ \frac{I}{\pi r^2}$ , we did this without integrals, but now if I want to do it as an integral,

$ I = \int_S \vec J . \vec da$ where the surface $S$ is the cylinder which contains three parts, the curved surface and both circular ends, so $S = S_1 +S_2 + S_3$ respectively.

Computing the integral, the $ \vec J$ vector is parallel to the axis while the normal vector of the curved surface is radial, so the contribution from $S_1$ vanishes, this leaves us with the 2 circular ends,

$ I = \int_{S_2} \vec J . \vec da + \int_{S_3} \vec J . \vec da$

But in that case, $ \vec J$ is parallel to one of the $ \vec da$'s and antiparallel to the other, so both integrals cancel and that leaves us with $ I = 0$ , so how do we use the definition of $ I = \int_S \vec J . \vec da$ ?

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  • $\begingroup$ Notice there is a problem (that doesn't really affect your question) in your problem statement. You first define $J$ to be a scalar, and then you start using it as a vector quantity $\vec{J}$. $\endgroup$ – The Photon Aug 6 at 4:45
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But in that case, $ \vec J$ is parallel to one of the $ \vec da$'s and antiparallel to the other, so both integrals cancel and that leaves us with $ I = 0$ , so how do we use the definition of $ I = \int_S \vec J . \vec da$ ?

You just calculated the net current out of a closed surface.

As Kirchhoff's current law should lead you to expect, this is zero (for steady state conditions).

If you want the current in the cylinder, don't use a closed surface. Just use a surface that sections the cylinder.

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