## anonymous 3 years ago 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.

1. anonymous

|dw:1352924458586:dw|

2. anonymous

I'm not sure what you mean by parametrise

3. anonymous

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

4. anonymous

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

5. anonymous

$0 <t<2 \pi$

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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.

12. anonymous

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

13. anonymous

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

14. anonymous

That is, a spiral rather than a full curve

15. anonymous

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