Polar and Cartesian coordinates are two coordinate systems for describing the same points - so, for example, the point one unit above the origin is $(0,1)$ in Cartesian coordinates and $(\frac{\pi}{2},1)$ in polar (using the order $(\theta, r)$). Think of this as a different way of talking about the same point.
So when you "convert" an equation from Cartesian into polar, you shouldn't be changing the graph - you're just changing how you talk about the graph. For example, the equation $y = \sqrt{1 - x^2}$ in Cartesian coordinates can be rewritten as $r = 1$, $0 \leq \theta \leq \pi$ in polar. Notice that these equations look nothing alike - they produce the same graph, but the coordinate systems are so different that the equations have to be written very differently.
Likewise, $y = 6\sin{x} + 2$ can't be "converted" into polar by writing $r = 6\sin{\theta} + 2$ - a quick test with any graphing software you like should convince you that $r = 6\sin{\theta} + 2$ makes a big round shape, while $y = 6\sin{x} + 2$ makes a wiggly curve extending infinitely in each direction. Since those are different graph, this isn't a conversion.
But you can think of it as a transformation - the idea is that the act of "replacing" coordinates in one system with coordinates in the other system can be thought of as a function that turns points into other points. For example, the point $(0,1)$ in Cartesian coordinates could be transformed into the point $(0,1)$ in polar coordinates - which would mean $r = 1$ and $\theta = 0$, which would be the point with Cartesian coordinates $(1,0)$. Turning $y = 6\sin{x} + 2$ into $r = 6\sin{\theta} + 2$ is doing exactly this: you're transforming every point on the plane this way, and watching where the ones along your curve wound up.
Best Answer
$\vec v = v_r \hat r + v_\theta \hat \theta$, where $\hat r$ and $\hat \theta$ are the unit radial vector and unit rotational vector at $(r, \theta)$. Letting $\vec p = x\hat i + y \hat j$ be the position vector, and noting that $x = r\cos \theta, y = r\sin \theta$, we have $$\vec p = r\cos \theta \hat i + r\sin \theta\hat j$$. Now $\hat r$ is the direction that $\vec p$ changes when $r$ increases, and $\hat \theta$ is the direction that $\vec p$ changes then $\theta$ increases: $$\hat r = \frac{\frac{\partial\vec p}{\partial r}}{\left\|\frac{\partial\vec p}{\partial r}\right\|}\qquad\hat \theta = \frac{\frac{\partial\vec p}{\partial \theta}}{\left\|\frac{\partial\vec p}{\partial \theta}\right\|}$$
So $$\hat r = \cos \theta \hat i + \sin \theta\hat j\\\hat \theta = -\sin \theta \hat i + \cos \theta\hat j$$
Thus $$\begin{align}\vec v &= \pi r\hat \theta\\&=\pi(-r\sin \theta\hat i + r\cos \theta\hat j)\\&=-\pi y\hat i + \pi x \hat j\end{align}$$
Since you've not defined what $U$ and $V$ mean, I can't answer any further than that.