Parametrization of a Corkscrew Curve on a Paraboloid

[DougUTPhy]

Homework Statement

I'm doing a line integral and can't seem to figure out the parametrization of this curve:
x^2+y^2+z=2\pi

Homework Equations

Looking to get it to the form:
\textbf{c}(r,t)=(x(r,t),y(r,t),z(r,t)) (I don't even know if this is right though).

The Attempt at a Solution

Trying to use x=r \cos t and y=r \sin t but I still can't get anywhere.

I have a feeling I'm totally in the wrong direction.
The 2\pi is killing me too!

 

[HallsofIvy]

Your basic problem is NOT the "2\pi". It is that x^2+ y^2+ z= 2\pi does NOT define a line (or curve or path) in three dimensions. It can be written as z= 2\pi- x^2- y^2 which is a surface (specifically, a paraboloid). Essentially, given any x and y you can solve for z so this is a two dimensional figure, not one dimension.

Please tell us what the entire problem really is.

 

[DougUTPhy]

I realized this after thinking about for a while, the real parametrizaion I can't figure out is a curve that is a corkscrew getting narrower as it goes up around the parabolioid, starting at (\sqrt{2\pi},0,0) and ending at the top of the paraboloid, (0,0,2\pi)