Parameterizing z-value of Cylinder in Line Integral Projection (Using Stokes Theorem)

The Head

I am a little confused about how to generally go about applying Stokes's Theorem to cylinders, in order to calculate a line integral. If, for example you have a cylinder whose height is about the z axis, I get perfectly well how to parameterize the x and y components, using polar coordinates, for example, but what about z (and thus dz)? If you have a vector field that is dependent on z, such as F= (z,x,yz) (it doesn't matter which, I am just listing one for clarity) and you want to replace with polar coordinates, what approach do you take with z, when projecting your surface about a closed loop?

What confuses me is if you have a sphere, or something similar, you find the shadow of the surface, which is effectively the z-value where the circumference is widest-- but with the cylinder, all z-values are at locations with equal circumference about the x-y plane. Is it the bottom? Top? Somewhere in between?

Thanks!!!

Stephen Tashi

It isn't clear to me what sort of problem you are describing. Can you give a specific example or post a link to such a problem on the web?

When you talk about "the components" in your original post, you aren't making it clear whether you mean the components of a vector field F(x,y,z) or whether you mean the components of a vector that describes the surface involved or whether you mean the components of the vector that traces out the curve to used for the line integral.

The Head

For example, you have a surface defined by a cylinder whose circular faces are in the x-y plane with x^2 +y^2=1 and -1<z<3. Then let's say you have a vector field F=(zx, xy, zy) and you want to calculate the the double integral of the curl of F. You could do this directly, or by using Stokes's Theorem and taking the line integral of the projection, right?

Well, parameterizing x=cos(t), y=sin(t) seems logical, and dx and dy follow from these. But I don't know what to do with z. If this were a sphere, you could just rewrite z in terms of x and y, or by some other method. But I don't know what to do with it when it is with a cylinder.

Stephen Tashi

If you dealing with a curve in the xy plane, you set z = 0. Perhaps I don't understand what curve you want to use in the line integral.

The Head

Do you have a choice? If you have a cylinder whose upper and lower surfaces reside somewhere on the z-axis, say z1 and z2, can you project the surface of the cylinder to any value of z (where obviously z=0 is the simplest)?

This is where I am getting confused: If it were the same problem, but the surface was the part of z=9-x^-y^2 that lies above z=5, you would still parameterize C the same way with x and y, but for z, you would set it equal to 5, right? That is the only way I have seen that type of problem, so I assume it is either the only way to do it, or the best way to do it.

But with a cylinder, do you have more flexibility parameterizing C (and thus choose z=0)? What if the cylinder's base were at z=5 and went up from there (so that z=0 was not inside your surface)-- does that change anything, or can you still project your curve so that z=0?

I really appreciate your help!

Stephen Tashi

First, don't get the idea that Stokes theorm must involve a projection. All it needs is a surface and a curve that "bounds" the surface. The curve doesn't have to be a projection of the surface onto a plane. Yes, it often turn out in problems that the way to find a nice curve is to project something onto a plane. In the case of a cylinder about the z-axis, it is true that you can find a nice curve by projecting the sides of the cylinder onto a plane. But there is nothing in Stokes theorem that requires that the curve be such a projection.

If you have cylindrical surface about the z-axis (one that has a top) which is bounded below by the plane z = 5 then you can apply Stokes theorem to the curve where it intersects that plane. On that curve you set z = 5.

You would not us the curve in xy plane where z = 0 because that curve does not bound the surface. If you got the right answer by using that curve, it would only be a happy coincidence.