Find a parametrization of the surface Σ.

Σ is the part of a cylinder x^2+y^2=1 that lies between the planes z=-1 and z=1.

Question

Find a parametrization of the surface Σ.

Σ is  the part of a cylinder x^2+y^2=1 that lies between the planes z=-1 and z=1.

anonymous · Answer

|dw:1352924458586:dw|

anonymous · Answer

I'm not sure what you mean by parametrise

anonymous · Answer

Meaning put it into parametric form. <x,y,z>

anonymous · Answer

OK. Forget about the z for a minute $$f(t)=cost \hat{x}+sint \hat{y} $$ Is the circle

anonymous · Answer

$$0 <t<2 \pi$$

anonymous · Answer

Gotcha. Im mostly fine with finding. x=rcos(theta) y=rsin(theta). Im not sure how to find z though.

anonymous · Answer

Now, to include the z co-ordinate:$$cost \hat{x}+sint \hat{x}+(\frac{t}{\pi}-1)\hat{z}$$ $$ 0<t<2 \pi $$I basically twisted the z thing to get something that linearly went from z=-1 at t=0 to z=1 at t=2 pi

anonymous · Answer

Hmmm, not 100% sure how you go to that point. Could you break it down for me?

anonymous · Answer

Mainly, where did t/pi-1 come from? And what goes into t? The z coordinate says the answer is simply 'zk'

anonymous · Answer

Because $$t=0, (\frac{0}{\pi}-1))=-1$$
$$t=0, (\frac{2 \pi}{\pi}-1))=1$$

anonymous · Answer

Hmmm I think I get it a little more now. :/ Still really murky on it though. the way I learned it was to just plug your parametrized x and y back into an equation with z in it and you find your z that way. Ehhh.

anonymous · Answer

I really don't think I even understand the basic concept of this section to be honest.

anonymous · Answer

It certainty doesn't help that I was incorrect. What I actually coded for was|dw:1352927439044:dw|

anonymous · Answer

That is, a spiral rather than a full curve

anonymous · Answer

http://tutorial.math.lamar.edu/Classes/CalcIII/ParametricSurfaces.aspx There's cylinder in there. I'll explain if more help needed, sorry for mistake earlier