# Evaluate length of the spiral (Line Integral)

1. May 11, 2016

### says

1. The problem statement, all variables and given/known data
Evaluate the length of the spiral with parametric equation ψ(t) =< 2 cost, 2 sin t, π/t >, with t ∈ [0, 2π].

2. Relevant equations
Line integral ∫C f(x,y) dS

3. The attempt at a solution
f(x,y) = z = π/t
C π/t dS
[0, 2π] are the lower and upper bounds of integration
dS= √(dx/dt)2+(dy/dt)2

∫ π/t √(dx/dt)2+(dy/dt)2 ; 0≤t≤2π

Just wanted to check I've set this problem up correctly. At first I thought dS = √(dx/dt)2+(dy/dt)2 + (dz/dt)2 and that I would just integrate that with respect to the boundaries. I don't think this is correct though. I don't really understand why though.

2. May 11, 2016

### Ray Vickson

Since you have a curve in 3D, you need to use a different formula (as in your first thought):
$$ds = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \, dt$$
However, since you have an improper integral, you need to look at the limit
$$\lim_{a \to 0+} \int_a^{2 \pi} ds$$

3. May 11, 2016

### Paul Colby

Actually, right hand side of your expression for $dS$ is not really a differential is it? How would you make the right hand side equal $dS$?

4. May 11, 2016

### LCKurtz

First, it is standard to use $ds$ for the differential of arc length and $dS$ for the differential of surface area. That being said, I think Ray mis-spoke when he wrote $$\lim_{a \to 0+} \int_a^{2 \pi} ds\text{.}$$ If you were really going to parameterize it by arc length it would be$$\lim_{a \to 0+} \int_a^{l(C)} ds$$where $l(C)$ is the length of the curve $C$. But you wouldn't normally actually do that. You would use your parameterization and calculate$$\lim_{a \to 0+} \int_a^{2 \pi} |\vec R'(t)|~dt\text{.}$$That integrand is what you would call $ds$ (in terms of $dt$).

5. May 11, 2016

### Ray Vickson

That is actually what was meant by what I wrote, since I defined $ds$ as $|\vec{R'}(t)| \, dt$ in #2, so $t$ is the dummy variable of integration. Bad notation, perhaps.

Last edited: May 11, 2016