Let $f:\mathbb{R^3} \to \mathbb{R}$ $$f(x, y, z)=\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^2-4 R^{2}\left(x^{2}+y^{2}\right) $$ and $R,r \in \mathbb{R} , 0<r<R$
now define :
$Z=\{ (x,y,z) \in \mathbb{R^3} | f(x,y,z)=0\}$
what is shape of $Z$ in $\mathbb{R^3}$ ? how we can draw $Z$ in $\mathbb{R^3}$ ?
first of all we must compute $Z$ :
$\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^2-4 R^{2}\left(x^{2}+y^{2}\right)=0\Rightarrow $
$z^2=\pm2R\sqrt{x^2+y^2}-(\sqrt{x^2+y^2})^2-(R^{2}-r^{2})$
note that in the other hand we have $A \subset Z$ such that ,
$A=\{ (x,y,z) \in \mathbb{R^3} | x^2+y^2=R^2,z^2=r^2\}$
so $Z$ must include two circles with radios $R$ and $z=\pm r$. now for example how we can draw $Z$ for $R=2, r=1$ or what is graph of following equation ? is this torus ?
$$z^2=\pm2 \sqrt{x^2+y^2}-(\sqrt{x^2+y^2})^2-(2^{2}-1^{2})$$
It is a torus
Section with plane $y=0$ are the two circles $$x^2+z^2-2 R x=R^2-r^2 ;\; x^2+z^2+2 R x=R^2-r^2$$ the rotation of $2\pi$ of one of the two around $z$-axis gives the surface.
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