Flux of a vector field with an integrable tangential singularity

Consider the vector field $\mathbf{F}=(p(x),q(x),r(z))$ in the open unit cube, where $p$ and $r$ extend to continuously differentiable functions on $[0,1]$, and $q$ is continuous on $(0,1]$ with ${\int }_0^1|q(x)|dx<\infty$. Even if $q$ is unbounded as $x\downarrow 0$, the normal component on an $x$-face has the finite trace $\pm p(x)$. The two $y$-faces have finite improper integral fluxes $\pm {\int }_0^{1}q(x)dx$ and therefore cancel. The remaining face contributions giveTo justify this use of the divergence theorem, first apply it to $[\epsilon ,1]\times [0,1{]}^2$, on which the field is regular. Its divergence is $p^{\prime }(x)+r^{\prime }(z)$, independent of $q$. Its outward flux is $p(1)-p(\epsilon )+(1-\epsilon )[r(1)-r(0)]$. Taking $\epsilon \downarrow 0$ yields the displayed answer, and absolute convergence makes the two $y$-face limits finite separately. For example, $q(x)=1/[x(1+(\ln x{)}^2)]$ diverges at zero butThis argument defines the normal traces and the face integrals explicitly; it does not assert that the whole vector field has a finite value on the singular face.