Surfaces of revolutions are other type of surfaces constructed by curves. In the previous entry, a cylinder is noting a planar curve $\alpha$ moved along a fix direction $a$, that is, we translate $\alpha$ along a direction. If $a$ is the given direction, a translation in this direction is $T_t(x)=x+t a$, $x\in{\mathbb R}^3$ and $t\in{\mathbb R}$. Then the cylinder on basis $\alpha$ is $$\cup_{t\in {\mathbb R}}T_t(\alpha(s)):s\in I\}.$$
In order to define a surface of revolution, we consider a curve $\alpha$ contained in a plane $P$ and we rotate $\alpha$ about a line $L$ contained in the plane $P$. We know that the parametrization is $X(s,\theta)=(f(t)\cos\theta,f(t)\sin\theta,g(t))$, where $\alpha(t)=(f(t),0,g(t))$. Again, the difficulties appear when we prove that $X$ is an embedding. For this reason we assume again that $\alpha$ is an embedding or a simple closed curve.
With the curve $\alpha(t)=( \sin(t),0,1+\cos(t)\cos(2t))$, with $t\in (0.5,2.5)$, we observe that there is a self-intersection, so it does not define a surface.
Other example is the torus generated by the circle $\alpha(t)=(1+2\cos(t),0,2\sin(t))$ because it intersects the $z$-axis.
Definitively, we impose that the curve $\alpha$ is an embedding or a simple closed curve. In the next picture, the surface is generated by the simple closed curve $\alpha(t)=(2+\sin(t),0,\cos(t)+\cos(2t))$.
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