Why can't we let z = 2 in this parametric surface?

[flyingpig]

Homework Statement

[PLAIN]http://img191.imageshack.us/img191/5128/unledymj.jpg

My book says

r = <3cos\theta, 3sin\theta, z>

I understand what they are doing, but why don't they set z = 2 for the parametrization instead?

r = <3cos\theta, 3sin\theta, 2>

Like the radius, don't we just know how far it goes?

The Attempt at a Solution

 

[JThompson]

If you set z=2 you have the parametrization of a circle, not a cylinder. You do know that the cylinder is bounded, but you have to let z vary to obtain a cylinder.
Here's the thing. When you do these sorts of problems, the position vector is only part of the parametrization. The position vector by itself often isn't the surface in question. For instance in this problem r is not the cylinder between z=0 and z=2, it is just an unbounded cylinder. The intervals for the parametrization provide the rest of the information. So the whole parametrization is
\vec{r}=3cos\theta\hat{i}+3sin\theta\hat{j}+z\hat{k} for 0\leq\theta\leq 2\pi and 0\leq z\leq 2.

 

[rock.freak667]

You are parameterizing in θ and z, so when you are integrating wrt θ and z. If z=2, then you'd just be doing one plane and not from z=0 to z=2.