Given $$\alpha : I \rightarrow \mathbb R^2, \ \ \ \alpha(t)=(\alpha_1(t),\alpha_2(t)) \text.$$ It is a regular space and also homomorphic and $\alpha_2(t)>0$ for all $t \in I$.
Let $S$ be the surface in $x$-$y$-$z$-space provided by the rotating the curve $$\{(x ,y ,0 ) \mid x =\alpha_1(t) ,\ y=\alpha_2(t),\ t\in I \}$$ around the $x$-axis.
How can I prove that $S$ is a regular surface?
All these symbols just got me confused. If you help me to solve this I would very much appreciate it.