How is the surface integral of $\vec J$ carried out in Ampere's law?

Question

In Ampere's law, the right hand side consists of the surface integral $\int_s\vec j\cdot ds$ which is supposed to represent the current passing through the surface of the enclosed volume. However, $\vec j$ leaving the surface will contribute a positive value to the integral while that entering will contribute a negative value. Hence the integral should give a zero value, rather than the current $I$

Answer 1

The right integral is: $$i_{\Sigma}(t)=-\iint_{\Sigma}{\mathbf j (\mathbf r,t)\bullet \mathbf n(\mathbf r) \ d\Sigma}$$ where $\mathbf n$ is the versor of the surface. If you have a closed surface: $$i_{\Sigma}(t)=-\iint_{\Sigma}{\mathbf j (\mathbf r,t)\bullet \mathbf n(\mathbf r) \ d\Sigma}=-\iiint_{V}div \mathbf j(\mathbf r, t) \ dv$$