How to mathematically color the regions bounded by a parametric curve? Usually, if an implicit equation $F(x, y) = 0$ defines a curve (or curves) on the x-y plane, then we can use the inequalities $F(x, y) < 0$ or $F(x, y) > 0$ to color the regions bounded by the curve (or curves). In this way, we can make interesting pictures.
Suppose $(x, y) = (f(t), g(t))$ defines a parametric curve (an example picture) on the plane. How to color the regions (an example picture) bounded by the curve without converting the parametric equation to an implicit equation?
 A: Given a closed curve $\gamma: t\mapsto{\bf z}(t)$ $\ (0\leq t\leq T)$  in the plane under mild technical conditions for each point $c=(a,b)\notin\{{\bf z}(t)\ |\ t\in [0,T]\}$ the winding number $n_\gamma(c)$ is defined by
$$n_\gamma(c):={1\over 2\pi}\int_0^T {(y(t)-b)\dot x(t)-(x(t)-a)\dot y(t)\over (x(t)-a)^2+(y(t)-b)^2}\ dt ={1\over 2\pi i}\int_\gamma{dz\over z-c}\ .$$
This number is always an integer (this is a miracle). Color  the point $c$ green if $n_\gamma(c)$ is odd and white otherwise.
A: You can use the winding number to discriminate among the regions.  If we use complex numbers to represent the plane, the winding number of the closed parametric curve $z = \gamma(t)$, $0 \le t \le 1$, around a point $a$ not on the curve is 
$$n(\gamma;a) = \frac{1}{2\pi i} \oint_\gamma \frac{dz}{z-a} = \frac{1}{2\pi i} \int_0^1 \frac{\gamma'(t)\ dt}{\gamma(t)-a}$$
which should take integer values (so you can use numerical methods to calculate this approximately and round to the nearest integer).
