Evaluate length of the spiral (Line Integral)

[says]

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 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&#039;(t)^2 + y&#039;(t)^2 + z&#039;(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]

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]

Since you have a curve in 3D, you need to use a different formula (as in your first thought):
ds = \sqrt{x&#039;(t)^2 + y&#039;(t)^2 + z&#039;(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 $dt$).

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.