I just wondering what the implicit equation would be if an ellipse with major axis a and minor axis b, rotating about the Z axis with a distance of $R_0$. The $R_0$>a and $R_0$>b which means the rotation will result in a non-degenerate torus. My aim is to determine if some points are inside the toroidal surface. The surface is shown in the image found online.
I'll suppose that, as in the diagram, the minor axis $b$ is in the radial direction.
This is a surface of revolution generated, for example, by the ellipse $$\frac{(x - R_0)^2}{b^2} + \frac{z^2}{a^2} = 1$$ in the $xz$-plane. We can produce the formula for the surface of revolution generated by rotating this ellipse about the $z$-axis by replacing $x$ with the radial variable $\sqrt{x^2 + y^2}$: $$\frac{\left(\sqrt{x^2 + y^2} - R_0\right)^2}{b^2} + \frac{z^2}{a^2} = 1.$$
A point $(x, y, z)$ is inside the elliptical torus iff the LHS of the previous display is $< 1$.
You can obtain this as follows.
If you start with a slice where $y = 0$, you begin with the equation $$ \frac{z^2}{a^2} + \frac{(x - R_0)^2}{b^2} - 1 = 0 $$ However, this doesn't give you the rotated version; to rotate it about the $z$-axis, simply replace the $x$ by $\sqrt{x^2 + y^2}$, yielding $$ \frac{z^2}{a^2} + \frac{\Big(\sqrt{x^2 + y^2} - R_0\Big)^2}{b^2} - 1 = 0 $$ This is a little unsatisfying though, since polynomials are much nicer than radicals. However, a little bit of manipulation yields $$ \frac{z^2}{a^2} + \frac{x^2 + y^2 + R_0^2}{b^2} - \frac{2R_0}{b^2}\sqrt{x^2 + y^2} - 1 = 0 $$ which, if you isolate the radical and square both sides yields $$ \bigg(\frac{z^2}{a^2} + \frac{x^2 + y^2 + R_0^2}{b^2} - 1\bigg)^2 - \frac{4R_0^2}{b^4}(x^2 + y^2) = 0 $$ Voila, a polynomial!
