∫Czds, where C is the helix of radius 1 which rises counterclockwise from (1,0,0) to (-1,0,2π).

To solve the line integral ∫Czds, where C is the helix of radius 1 which rises counterclockwise from (1,0,0) to (-1,0,2π), we first need to parameterize the helix.

A helix can be parameterized as follows:

r(t) = (cos(t), sin(t), bt)

where b is the vertical distance the helix travels in one complete revolution. In this case, the helix travels a distance of 2π in one revolution, so b = 2π.

So, our parameterization is:

r(t) = (cos(t), sin(t), 2πt)

Next, we need to find the limits of integration. The helix starts at (1,0,0) and ends at (-1,0,2π). This corresponds to t = 0 and t = 2π, respectively.

Now, we can compute the line integral. The line integral of a scalar field f over a curve C parameterized by r(t) from t=a to t=b is given by:

∫Cf ds = ∫ from a to b f(r(t)) ||r'(t)|| dt

In this case, our scalar field f is just z, which in terms of our parameterization is 2πt.

So, we need to compute r'(t) and ||r'(t)||.

r'(t) = (-sin(t), cos(t), 2π)

||r'(t)|| = sqrt((-sin(t))^2 + (cos(t))^2 + (2π)^2) = sqrt(1 + 4π^2)

So, our line integral becomes:

∫ from 0 to 2π (2πt) sqrt(1 + 4π^2) dt

This is an elementary integral, which can be computed to be:

[πt^2 sqrt(1 + 4π^2)] from 0 to 2π

= 4π^3 sqrt(1 + 4π^2)

So, ∫Czds = 4π^3 sqrt(1 + 4π^2).