I have two systems of differential equations given in polar coordinates and I am supposed to find the $\omega$- and $\alpha$-limit sets for each, and I'm not sure if its the polar coordinates giving me trouble but I can't get a good visualization of the problem and I'm stuck. I know the $\omega$-limit set is the set of all points y such that as $n\to \inf t_n\to \inf$ and the flow goes to y and the $\alpha$-limit set is where $t_n \to -\inf$ and again the flow goes to y but I'm not sure how to show this without a phase portrait or really how to draw the portraits.
the first system is this: $$r'=r^3-3r^2+2r, \theta'=1$$ I solved this by seperation of variables to get $r(t)=\frac{\pm\sqrt{e^{2c+2t}+1}+e^{2c+2t}+1}{e^{2c+2t}+1}$ and $\theta(t)=t+\theta_0$ so $\theta(t)$ is inceasing when $t \gt \theta_0$ and I graphed the solution of r(t) and noticed that r=1 or 2 and kind of zips down from 2 to 1 depending on c, and for the negative part it zips up from -2 to -1 but I don't know why or what that means for the limit sets, but it seemed important to note. also, solving for c in terms of $r_0$ got really messy an I'm not sure if I even needed to do that.
the second system is: $$r'=sin(r), \theta'=-1$$ the solutions here were $r(t)=2cot^{-1}(e^{c-t})$ and $\theta(t)=-t+\theta_0$ and again the graph of r(t) does the zippering between 0 and $\pi$ but I'm not sure what that means for the limit sets
For an autonomous d.e. $r' = f(r)$, the equilibrium points are the solutions of $f(r) = 0$, and these are the only possible $\alpha$ or $\omega$ limit sets. In the polar system $r' = f(r),\; \theta' = c$ (with $c$ a nonzero constant, and $f(r) = 0$ so nothing funny happens at the origin), the $\alpha$ and $\omega$ limit sets are the circles $r = $constant, where the constant is a solution of $f(r) = 0$ (or the origin itself in the case $r=0$).