Given parametrization of torus is equal to level surface

Question

I need to show that the torus $T=\pmatrix{(R+r\cos\phi)\cos \theta\\(R+r\cos\phi)\sin\theta\\r\sin\phi}$ is equal to the surface given implicitly by $(\sqrt{x^2+y^2}-R)^2+z^2-r^2=0$. I already got from T to the level surface, but how do I show that the level surface doesn't contain any other points than those in T?

Answer 1

Set $u = \sqrt{x^2 + y^2} - R$. Then the level surface is given by $u^2 + z^2 = r^2$ which is the equation of a circle of radius $r$ in $(u,z)$. Hence, we can find $\varphi$ such that $u = r \cos(\phi)$ and $z = r \sin \phi$. Then

$$ u + R = \sqrt{x^2 + y^2} \implies (u+R)^2 = x^2 + y^2 $$

which is the equation of a circle of radius $u + R$ in $(x,y)$ and so we can find $\theta$ such that $x = (u + R) \cos \theta$ and $y = (u + R) \sin \theta$.