Evaluate length of the spiral (Line Integral)

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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.
 

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  • #2
Ray Vickson
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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):
[tex]ds = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \, dt [/tex]
However, since you have an improper integral, you need to look at the limit
[tex] \lim_{a \to 0+} \int_a^{2 \pi} ds [/tex]
 
  • #3
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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
LCKurtz
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Since you have a curve in 3D, you need to use a different formula (as in your first thought):
[tex]ds = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \, dt [/tex]
However, since you have an improper integral, you need to look at the limit
[tex] \lim_{a \to 0+} \int_a^{2 \pi} ds [/tex]
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##).
 
  • #5
Ray Vickson
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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.
 
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