Surface of Revolution (parametrization)

Question

Question

For a curve $\gamma(t)=(x(t),y(t),z(t))$ what would be the parametrization of the surface that we get by the revolution of $\gamma$ in the zz' axis? What about a random line as an axis?

Attempt

I think that the answer for the first question is $X(t,u)=(x(t)cosu,y(t)sinu,z(t))$ but I'm not sure.

Answer 1

I'm assuming that with the "zz' axis" you mean the $z$-axis. A rotation with this axis keeps the $z$-coordinate of the points fixed and rotates the $(x,y)$-coordinates with the matrix $$\left[\matrix{\cos\theta&-\sin\theta \cr \sin\theta&\cos\theta\cr}\right]\qquad(\theta\in{\mathbb R})\ .$$ For your curve $$\gamma:\quad [a,b]\to{\mathbb R}^3,\qquad t\mapsto\bigl(u(t),v(t),w(t)\bigr)$$ we should assume that $t\mapsto w(t)$ is monotonic, say $w'(t)>0$. A parametrization of the resulting surface then would be $$X:\quad(\theta,t)\mapsto\left\{\eqalign{x(\theta, t)&= u(t)\cos\theta-v(t)\sin\theta \cr y(\theta,t)&= u(t)\sin\theta+ v(t)\cos\theta\cr z(\theta,t)&=w(t)\cr}\right.\quad\qquad(0\leq\theta<2\pi , \ a\leq t\leq b).$$

Answer 2

Your answer for the first question is almost right; $(x(t) \cos \theta, y(t) \sin \theta, x(t))$ for $0 \le \theta < 2\pi$ is not quite a valid parametrisation as you'll actually get ellipses. You'll need to rotate $(x(t),y(t))$ around the $z$-axis via a rotation matrix.

For an arbitrary axis, it would be best to rotate the $x, y, z$ axes such that the $z$-axis lines up to the axis via another rotation matrix. From here, you use the same method as previously outlined, and then rotate back.