Find the Pre-Image of $\ q\ $, where $\ q^{-1}(S)\ $, where $\ S=f^{-1}([0,+\infty))$ 
Let $\ f:\mathbb{R^2}\rightarrow\mathbb{R}\ $ be given by $\ f(x,y)=y(x-y)\ $ and let $\ q:\mathbb{R^2}\rightarrow\mathbb{R^2}\ $ be given by $\ q(r,\theta)=(r \ \text{cos}(\theta),r \ \text{sin}(\theta))$. Find the pre-image $\ q^{-1}(S)\ $, where $\ S=f^{-1}([0,+\infty))$.

I have found that the pre-image of $\ f\ $ is the set $\ f^{-1}=\{(x,y)\in\mathbb{R^2}: y(x-y)\geq 0\}$. Is it also true that $S$ is a closed?
To find $\ q^{-1}(S)\ $, I thought this would correspond to where $\ (r,\theta)\geq S$. But I am very confused on how to proceed. I don't have the intuition of what to do next.

By definition:
$$q^{-1}(S)=\{(r,\theta)\,\colon q(r,\theta)\in S\}$$
Then, 
$$q(r,\theta)\in S\iff q(r,\theta)\in f^{-1}([0,+\infty))\iff f(q(r,\theta))\geq 0.$$
This yields 
$$ r\sin(\theta)(r\cos(\theta)-r\sin(\theta))\geq 0$$
Calculating, I obtained that this happens if and only if
$$r^2(\sin(\theta)\cos(\theta)-\sin^2(\theta))\geq 0$$
Maybe you should apply some trigonometric identity to ''simplify'' this result
