Very confused on fluid flow question I am very confused on how to solve the following problem:
If $2 \pi m$ represents the volume that is ejected per unit of time per unit length of the $z$ axis, obtain the velocity $v(r)$, ie the line source strength of m, also obtain the potential function for the line source.
Next, suppose that two line sources parallel to 0z have the strength m and -m, and say we cut the $xy$ plane into $(L/2,0)$ and $(-L/2,0)$ then find the velocity potential for this configuration and show that as L approaches zero and m approaches infinity the potential becomes $\frac{-\mu cos(\theta)}{r}$, obtain the streamlines and the equipotential and show they form mutually orthogonal family of curves.
I don't even know how to begin. The question is part of practice among others involving line integrals etc, which I have been more successful with. This one I don't even know how to begin. I know extremely little about physics if that matters.
Please help me
 A: The volumetric flow rate (per unit length in the z-direction) through the surface of a cylinder of radius $r$ and axis coinciding with the z-axis is
$$Q = \int_C\mathbb{v} \cdot \mathbb{n}\, dl= \int_0^{2\pi}v_r rd \theta= 2\pi m.$$
By symmetry the velocity field has a nonzero radial component $v_r$ that depends only on $r$.
Hence,
$$\int_0^{2\pi}v_r rd \theta= 2 \pi r v_r=2\pi m,$$
and
$$v_r = \frac{m}{r}.$$
For incompressible, irrotational flow, the velocity potential is defined by $\nabla \phi = \mathbb{v}$ and satisfies Laplace's equation $\nabla^2 \phi = 0$.
With only a nonzero radial component this reduces to
$$\frac{\partial \phi}{\partial r}= v_r = \frac{m}{r}.$$ 
Integrating we find that the potential function (up to an integration constant that can be arbitrarily set to $0$) is given by
$$\phi(r) = m \ln r = m \ln \left(\sqrt{x^2 + y^2}\right).$$
Since Laplace's equation is linear, the velocity potential due to multiple sources is a linear combination of the potentials corresponding to the individual sources. 
In the second part of the problem, you have a source located at $x = L/2, y = 0$ ejecting fluid and a second source (actually a sink) located at $x = -L/2, y = 0$ absorbing the fluid at the same rate.
The total potential is obtained by adding the potentials for these two line sources. The potential of  each individual source is obtained by translating the potential derived for a line source at $x = 0, y=0$ to the actual location of the source. Hence,
$$\phi = m \ln \left(\sqrt{(x-L/2)^2 + y^2}\right) - m \ln \left(\sqrt{(x+L/2)^2 + y^2}\right).$$
