Orbit space of $\partial D\times D$ under the action of $S^1$ by multiplication

Let $D$ be the unit disk, $D=\{z: |z|\leq 1\}$. Let $S^1$ act on $\partial D\times D$ by pointwise multiplication. What is the orbit space of this action?

I think that since $\partial D=S^1$, this factor will vanish in the orbit space. Is it true that $\frac{\partial D\times D}{S^1}=\frac{\partial D}{S^1}\times \frac{D}{S^1}?$ Is the orbit space of $D$ under $S^1$ is just $D$? the orbits can be identified with radiused $0\leq r\leq 1$.

Thanks!

Define $p : S^1 \times D^2 \to D^2, p(z_1, z_2) = z_2/z_1$. This is a continuous surjective map. Therefore it is an identification map because domain and range are compact. The orbit of $(z_1,z_2) \in S^1 \times D^2$ consists of all $(zz_1,zz_2)$ with $z \in S^1$. Hence $p$ maps each orbit to a single point of $D^2$. Conversely, $p(z_1,z_2) = p(z_1',z_2')$ means $z_2/z_1 = z_2'/z_1'$. With $z = z_1'/z_1 \in S^1$ we obtain $(z_1',z_2') = (zz_1,zz_2)$ so that $(z_1,z_2),(z_1',z_2')$ belong to same orbit. This shows that $p$ induces a unique homeomorphism $\hat{p} : (S^1 \times D^2) / S^1 \to D^2$ such that $\hat{p} \circ \pi = p$, where $\pi : S^1 \times D^2 \to (S^1 \times D^2) / S^1$ denote the quotient map.
(1) This shows that if you have continuous actions of a topological group $G$ on spaces $X, Y$, then in general $(X \times Y)/G \ne X/G \times Y/G$. This is similar to the situation in a field: You cannot expect $\frac{ab}{c} = \frac{a}{c}\frac{b}{c}$.
(2) If you have a continuous action of a topological group $G$ on a space $X$, you get a continuous action of $G$ on $G \times X$. Defining $p : G \times X \to X, p(g,x) = g^{-1} \cdot x$, you can easily show that $p$ induces a homeomorphism $\hat{p} : (G \times X) / G \to X$.