Show that the circular cylinder S = {(x, y, z) ∈R^3 : y^2+z^2 = 1}can be covered by a single regular surface patch, and hence is a surface.
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Expert's answer
2020-10-19T16:01:14-0400
Solution
A surface patch can be given by (x,y)↦(rx,ry,tan(r−π/2)).
However we must be careful on how we define the open set that this is defined over
Here {x,y∣x2+y2∈(0,π)}.
Also, .r=+x2+y2.
Clearly, the map is smooth since r=0. Also r<π.
Hence tan(r−π/2) tan is well defined.
For surjection
Let (x,y,z) be a given point on the cylinder. Since tan is continuous and injective on the range (−π/2,π/2) and unbounded its surjective, and hence we get r such that tan(r−π/2)=z. So take a=yr,b=zr. Now y2+z2=1⇒y,z≤1y2+z2=1⇒y,z≤1 . Hence a,b≤r<π. Hence the pre-image lies in our given domain.
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