The helicoid is the surface swept out by a rotating horizontal line as it rises
along the $z$-axis (see the figure below). It can be described by the parameterized surface $x : \mathbb{R}^2 \to \mathbb{R}^3$, $x(u, v) = (av\cos u, av\sin u, bu)$, where $a$, $b$ are positive constants. Show that $x$ is a regular surface by computing $|x_u \times x_v|$. (Notation: $x_u =
\frac{\partial x}{\partial u}$, etc.)
Okay, I'm not sure of the steps I'm supposed to take to compute this. I THOUGHT that I was supposed to differentiate with respect to $u$ and then do the same for $v$, and then take the cross product of those two derivatives; but how does that show the surface is regular. Do I even have the right idea as to what I'm supposed to do?
Best Answer
You have to show that at it each point $p=x(u,v)$ of the surface $S=x(\mathbb{R}^2)$ there is a tangent plane $T_pS$. Since the tangent plane is generated by the two vectors $\displaystyle \dfrac{\partial x}{\partial u}$ and $\dfrac{\partial x}{\partial v}$, its existence is therefore equivalent to the normal vector $$ n\equiv\dfrac{\partial x}{\partial u}\times \dfrac{\partial x}{\partial v} $$ being non-zero.
Now $$ \dfrac{\partial x}{\partial u}=(-av\sin u,av\cos u,b),\quad \dfrac{\partial x}{\partial v}=(a\cos u,a\sin u,0), $$ and $$ n=(-ab\sin u,ab\cos u,-a^2v)=a(-b\sin u,b\cos u,-av). $$ It follows that $$ \lVert n\rVert=a\sqrt{a^2v^2+b^2} \ne 0. $$