Question

Q. Show that the circular cylinder S={(x,y,z)∈R^3 |x^2+y^2=1} can be covered by a single surface patch and so a surface.

Answer 1

Answer on Question #75562 – Math – Differential Geometry | Topology

Question

Show that the circular cylinder $S = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 = 1\}$ can be covered by a single surface patch and so a surface.

Solution

We can take $U$ an annulus instead of a disc, where $U = \{(u, v) \colon 0 < u^2 + v^2 < \pi\}$. Any point in the annulus $U$ is uniquely of the form $(t \cos \vartheta, t \sin \vartheta)$ for some real $t \in (0, \sqrt{\pi})$, $\vartheta \in [0, 2\pi)$. Map this point to the point of the cylinder $(x, y, z) = (\cos \vartheta, \sin \vartheta, \cot(t^2))$. This is clearly a subset of the cylinder as it satisfies $x^2 + y^2 = 1$. Also, because $\vartheta$ 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 a hit, showing the result of mapping the annulus as above covers the whole cylinder.

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