Given $\Phi((0, \infty) \times (0, 2\pi) \times \Bbb R) \rightarrow \Bbb R^3$,
$\Phi (r, \phi , z) := \begin{pmatrix} r \cos(\phi) \\ r \sin (\phi) \\ z \\ \end{pmatrix}$,
I would like to evaluate $\Bbb R^3 \setminus \Phi((0, \infty) \times (0, 2\pi))$.
My guess is the following:
First, we note that $z$ remains the same under $\Phi$, so we should be able to cut this out for a moment and imagine that we actually want to evaluate $\Bbb R^2 \setminus \theta((0, \infty) \times (0, 2\pi))$ with
$\theta(r, \phi) := \begin{pmatrix} r \cos(\phi) \\ r \sin (\phi) \\ \end{pmatrix}$.
$r$ and $\phi$ are the polar coordinates of a pair $(x, y) \in \Bbb R^2 \setminus \{(0, 0)\}$, their images under $\Phi$ represent their transformation into cartesian coordinates. The domain of definiton should look something like this: $\Bbb R^2 \setminus (0, x)$ with $x \in \Bbb R$. For all of these points, we can find such polar coordinates and transform them into cartesian coordinates. Therefore, we would get that
$\Bbb R^2 \setminus \theta((0, \infty) \times (0, 2\pi)) = \Bbb R^2 \setminus \{(0, x) : x \in \Bbb R\}$.
As I said, $z$ remains the same under $\Phi$, and cutting out $\Bbb R$ just cuts out the whole $z$-axis, so we would move one dimension lower, and with the application of the reasoning above, we get that the set we are looking is just
$\Bbb R^2 \setminus \{(0, x) : x \in \Bbb R\}$,
so we're working in the $2$-dimensional real space without this "tunnel" that is spanned by $(0, x)$.