Shallow water equations in differential form and cylindrical coordinates I am ignoring bed height or any other extra terms. The version of SWE when written in cartesian is: 
$
\begin{pmatrix}
h  \\
hu  \\
hv  \\
\end{pmatrix}_t  + 
\begin{pmatrix}
hu  \\
hu^2 + \frac{g}{2} h^2  \\
huv  \\
\end{pmatrix}_x  +\begin{pmatrix}
hv  \\
huv \\
hv^2 + \frac{g}{2} h^2   \\
\end{pmatrix}_y = 0
$
First what is it in differential form? I think it is:
$\frac{\partial}{\partial t}h + \nabla \cdot (hv) = 0 \quad 
\frac{\partial}{\partial t}h + \nabla \cdot (hvv) = - \nabla \frac{h^2}{2} g  $
The pressure term at the end is what I'm really uncertain about. I think it must be a gradient as it comes from the pressure gradient term and it seems to fit but I couldn't derive it from Euler. What is it supposed to be?
Finally to write SWE in cylindrical coordinates, if I had the correct differential form can I just use the del in cylindrical coordinates formulas and $v = (u_r, u_\theta)$ and plug these in to get it? And can I expect it to be harder to solve in cylindrical as I seem to get forcing terms?
 A: Mass continuity equation. The first line of the system gives indeed the scalar conservation law
$$
\partial_t h + \mathrm{div} (h \boldsymbol{v}) = 0 .
$$
Using the vector calculus identity $\mathrm{div} (h \boldsymbol{v}) = h \,\mathrm{div}\, \boldsymbol{v} + \mathbf{grad}\, h\cdot \boldsymbol{v}$ and the definition of the material time derivative $D_t$, we have also $D_t h = -h \,\mathrm{div}\, \boldsymbol{v}$.
Momentum equation. The remaining lines give an equation on $h \boldsymbol v$. In the absence of fluid rotation and with a flat bottom surface, the conservation of momentum reads $D_t \boldsymbol{v} = -\mathbf{grad}(g h)$. After a bit of tensor calculus, one should arrive at
$$
 \partial_t (h \boldsymbol{v}) + \mathbf{div}(h \boldsymbol{v}\otimes \boldsymbol{v}) = -\mathbf{grad}(\tfrac12 gh^2) ,
$$
where $\otimes$ is the tensor product. Using the identity $\mathbf{grad}\psi = \mathbf{div}(\psi \boldsymbol{I})$ for the scalar field $\psi = \tfrac12 gh^2$ where $\boldsymbol{I}$ is the identity tensor, we can rewrite the above equation as a single conservation law. 
Of course, it is possible to solve this system in polar coordinates by setting
$x = r\cos\theta$, $y = r\sin\theta$. With $\boldsymbol{u} = h\boldsymbol{v}$ and $\boldsymbol{T} = \frac1h \boldsymbol{u}\otimes \boldsymbol{u} + \frac12 gh^2\boldsymbol{I} = \boldsymbol{T}^\top$, the above equations of motion lead to the system
\begin{aligned}
\partial_t h + \partial_r u_{r} + \partial_\theta (\tfrac{1}{r} u_{\theta}) &= -\tfrac{1}{r} u_r \\
\partial_t u_r + \partial_r T_{rr} + \partial_\theta (\tfrac{1}{r} T_{r\theta}) &= -\tfrac{1}{r} (T_{rr} - T_{\theta\theta}) \\
\partial_t u_\theta + \partial_r T_{r\theta} + \partial_\theta (\tfrac{1}{r}T_{\theta\theta})  &= -\tfrac{2}{r}T_{r\theta}
\end{aligned}
which may be written
\begin{aligned}
\partial_t h + \partial_r (hv_{r}) + \tfrac{1}{r} \partial_\theta ( hv_{\theta}) &= -\tfrac{h}{r} v_r \\
\partial_t (hv_r) + \partial_r (h{v_r}^2 + \tfrac12 gh^2) + \tfrac{1}{r}\partial_\theta ( hv_rv_\theta) &= -\tfrac{h}{r} ({v_r}^2 - {v_\theta}^2) \\
\partial_t (hv_\theta) + \partial_r (hv_rv_\theta) + \tfrac{1}{r} \partial_\theta (h{v_\theta}^2 + \tfrac12 gh^2)  &= -\tfrac{2 h}{r}v_r v_\theta
\end{aligned}
You may have a look at (1) for examples with cylindrical symmetry, i.e. where $\boldsymbol v$ is along the radial direction and where the flow is independent on $\theta$. In that case, the previous system becomes
$$
\begin{aligned}
\partial_t h + \partial_r u_r &= -\tfrac{1}{r} u_r \\
\partial_t u_r + \partial_r T_{rr} &= - \tfrac{1}{r h} {u_r}^2
\end{aligned}
$$
with $T_{rr} = \frac1h {u_r}^2 + \frac12 gh^2$. The speeds of sound of this $2\times 2$ hyperbolic system are $v_r \pm \sqrt{g h}$.
The previous case of cylindrical waves is pretty easy to solve numerically (cf. the mentioned article (1)), e.g. by using dedicated finite-volume methods. However, the full system in polar coordinates is not as easy to solve, mostly because of the $1/r$ factors in the flux derivatives w.r.t. $\theta$. These factors make this problem space dependent in a certain way.

(1) P. Glaister, "Shallow water flow with cylindrical symmetry", J. Hydraul. Res. 29:2 (1991), 219-227. doi:10.1080/00221689109499005
