Circular cylinder $S=\{ (x,y,z) : x^2+y^2=1 \}$ can be covered with a single surface patch.

Question

I somewhere found that we can take $U$ an annulus instead of a disc where $U=\{ (u,v): 0 < u^2+v^2 < π \}$. Can anyone please explain me that how a cylinder can be covered with a single surface patch from $U$?

Answer 1

Any point in the annulus $U$ is uniquely of the form $(t \cos \theta, t \sin \theta)$ for some real $t \in (0,\sqrt{\pi}), \theta \in [0,2\pi).$ Map this point to the point of the cylinder $(x,y,z)=(\cos \theta, \sin \theta, \cot t^2).$ This is clearly a subset of the cylinder as it satisfies $x^2+y^2=1.$ Also, because $\theta$ ranges in $[0,2\pi),$ for any fixed $z$ the entire slice of the cylinder at that $z$ level gets covered. Finally, because the cotangent of $t^2$ for $t \in (0,\sqrt{\pi})$ takes on every real value, every level $z$ indeed gets hit, showing the result of mapping the annulus as above covers the whole cylinder.

Answer 2

Parametrization

EDIT1:

For Cylinder

$$ a=1 ; $$

$$ (a \cos u, a \sin u, v),(u,0, 2 \pi),(v,v1,v2) $$

where z coordinate is depth or height of cylinder between two limits.

For Annulus

$$ a=1 ; $$

$$ (a v \cos u, a v \sin u, 0 ),(u,0, 2 \pi),(v,v1,v2) $$