= Solution
Work away from the <line source>, so $r>0$. In <polar coordinates>, $u_r=Q/(2\pi r)$ and $u_\theta=0$. The supplied <divergence> and <curl> formulas give
$$
\nabla\cdot\mathbf u=\frac1r\partial_r\left(\frac Q{2\pi}\right)=0,
\qquad
(\nabla\times\mathbf u)_z=\frac1r\left[\partial_r(ru_\theta)-\partial_\theta u_r\right]=0.
$$
Hence \b[the field defines an <incompressible flow> and an <irrotational flow> for $r>0$]. The restriction matters: the outward <volume flux> through any enclosing <circle> is
$$
\int_0^{2\pi}\frac{Q}{2\pi r}r\,d\theta=Q.
$$
The origin is the excluded injection point. Applying the <divergence theorem> to a disk through that <singularity> would violate its smoothness hypothesis.
Back to article page
