Consider the following parametrisation of a flower-like domain, where $R=0.4$ and $r=0.2$:
$\theta \mapsto [(R+r\cos(6\pi\theta))\cos(2\pi \theta), (R+r\cos(6\pi \theta))\sin(2\pi \theta)]$ which gives the following shape:
I'd like to get the three dimensional version of this: imagine the surface formed by rotating this figure and obtaining something which looks like a sphere but with many
Suppose we take the curve $\mathcal C$, parameterized by
$$\begin{cases} x(\theta) = (R+r\cos(16\pi\theta)) \cos(2\pi\theta) \\ y(\theta) = (R+r\cos(16\pi\theta)) \sin(2\pi\theta) \end{cases}$$
with $R=0.4$, $r=0.2$, and $\theta\in[0,1]$, and revolve it about the $x$-axis. Consider a cross section of the resulting surface taken perpendicular to the $x$-axis at $x=x_0\in\left[-\frac35,\frac35\right]$. It's easy to show that $\sqrt{x^2+y^2}\le \frac35$.
In the plane of the cross section, solve $x(\theta_0)=x_0$ for $\theta_0$ and plug the solution (there are multiple depending on your choice of $x_0$) into the parametric equations for $\mathcal C$. $x(\theta_0)$ and $y(\theta_0)$ define a circle in the cross sectional plane centered at $(x_0,0,0)$ and with radius $y(\theta_0)$. We can parameterize this circle by the 3D parametric curve,
$$\begin{cases} x'(\phi) = x(\theta_0) \\ y'(\phi) = y(\theta_0) \cos(2\pi\phi) \\ z'(\phi) = y(\theta_0) \sin(2\pi\phi) \end{cases}$$
with $\phi\in[0,1]$.
It follows that the complete surface is parameterized by
$$\begin{cases} x(\theta,\phi) = x(\theta) \\ y(\theta,\phi) = y(\theta) \cos(\pi \phi) \\ z(\theta,\phi) = y(\theta) \sin(\pi \phi) \end{cases}$$
with $(\theta,\phi)\in[0,1]^2$.
The figure shows $\mathcal C$ (red) plotted in the plane $z=0$, the surface of revolution, and one of the circles (green) traced out by revolving a point on $\mathcal C$ about the $x$-axis (rendered with $\theta_0\approx0.159$, i.e. in the plane $x=0.2$, which happens to correspond to $3$ distinct circles at $6$ distinct values of $\theta_0$).