# Evaluate length of the spiral (Line Integral)

## Homework Statement

Evaluate the length of the spiral with parametric equation ψ(t) =< 2 cost, 2 sin t, π/t >, with t ∈ [0, 2π].

## Homework Equations

Line integral ∫C f(x,y) dS

## 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.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Evaluate the length of the spiral with parametric equation ψ(t) =< 2 cost, 2 sin t, π/t >, with t ∈ [0, 2π].

## Homework Equations

Line integral ∫C f(x,y) dS

## 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.

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
$$\lim_{a \to 0+} \int_a^{2 \pi} ds$$

Paul Colby
Gold Member
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##?

LCKurtz
Homework Helper
Gold Member
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
$$\lim_{a \to 0+} \int_a^{2 \pi} ds$$

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##?

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##).

Ray Vickson
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##).