Question #138613
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.
1
Expert's answer
2020-10-19T16:01:14-0400
SolutionSolution

A surface patch can be given by (x,y)(xr,yr,tan(rπ/2)).(x,y)\mapsto (\frac{x}{r},\frac{y}{r}, tan (r-\pi/2)).

However we must be careful on how we define the open set that this is defined over

Here {x,yx2+y2(0,π)}.\{x,y|\sqrt {x^2+y^2} \in (0,\pi)\}.

Also, .r=+x2+y2r=+\sqrt{x^2+y^2}.

Clearly, the map is smooth since r0.r\neq 0. Also r<π.r<\pi.

Hence tan(rπ/2)tan(r-\pi/2) tan is well defined.

For surjection



Let (x,y,z)(x,y,z) be a given point on the cylinder. Since tan is continuous and injective on the range (π/2,π/2)(-\pi/2,\pi/2) and unbounded its surjective, and hence we get rr such that tan(rπ/2)=z.tan(r-\pi/2)=z. So take a=yr,b=zr.a=yr, b=zr. Now y2+z2=1y,z1\sqrt{y^2+z^2}=1\Rightarrow y,z\leq 1y2+z2=1y,z1\sqrt{y^2+z^2}=1\Rightarrow y,z\leq 1 . Hence a,br<πa,b\leq r<\pi. Hence the pre-image lies in our given domain.



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