Consider the surface z=thaita (sorry i dont know how to spell it) in cylindrical coordinates shown in figure 3. Find aparametric representation of this surface

thanks

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 $\large z=\theta$ 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 $\large r = 0$, $\large \theta =t$ and $\large z = t$ 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 $r=0$ or it's true for all values of r: $\large r \in \mathbb{R}$, why would the equation $\large z=\theta$ imply that $\large r \in \left [ 0;5 \right ]$?

I'm not criticising you, just brainstorming. =)

I think an appropriate parametric representation would be:

$\large r = t$

$\large z = s$

$\large \theta = s$

where: $\large t \in \mathbb{R}$ and $\large s \in \mathbb{R}$

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

$\large r = t$

$\large z = s$

$\large \theta = s$

where: $\large t \in \mathbb{R}$ and $\large s \in \mathbb{R}$

...doesn't seem possible.