I'll try and help you out, I've never really used cylindrical co-ordinates but I do know that they're defined by three properties; the radial distance from the z-axis, the azimuth (angle to) of the point and height z.
Your equation would result in a straight line on the z-axis since the distance from the z-axis isn't mentioned. Are you sure the equation is correct as it clearly doesn't correspond with the diagram.
Try it, let , and on this grapher, the result is just a straight line.
surf_graph_cylin
Why $\displaystyle r=0$ ?
Try the plotter with $\displaystyle r=t$, $\displaystyle t=0$ to $\displaystyle 5, $ and $\displaystyle \theta = z = s,$ $\displaystyle s = 0$ to$\displaystyle 15.$
Hmmm, I understand what you're saying, but if r isn't explicitly defined surely that means it's either irrelevant hence or it's true for all values of r: , why would the equation imply that ?
I'm not criticising you, just brainstorming. =)
I think an appropriate parametric representation would be:
where: and
That does seem a bit silly but it's all I've got at the moment.
I chose $\displaystyle r = 0\rightarrow 5,$ (and $\displaystyle \theta = z = 0\rightarrow 15),$ for no particular reason other than it gives a reasonable plot (and produces something like the surface shown in the original question) whereas $\displaystyle r = 0\rightarrow\infty$ does not. Just for example really.
If there are no limits specified then I think $\displaystyle \infty$ is correct, afterall if you were asked to plot, in [2D], $\displaystyle y=1$ for example, you would draw, or at least imply, a line of infinite length, not a single point at $\displaystyle (0,1).$
If the diagram weren't present I'd think the answer I gave was correct but I'm thinking that some information is missing. Given how the diagram looks
where: and
...doesn't seem possible.