Surface Integral: Finding Distance Along Sphere

In summary, the conversation discusses finding the distance between two points on a sphere, not in a linear manner but along the surface. The equations for calculating distance in two and three dimensions are mentioned, with a focus on the 3D equation involving integration. It is noted that the unit vectors in this equation are not constant. An example is provided to illustrate how to integrate the equation to find the distance.
  • #1
Gear.0
66
0

Homework Statement


This isn't actually homework, but just related to something I am working on.

Basically I need to find the distance between two points along a sphere. Not the linear distance, but the distance along the surface.
This would be akin to "arc length" except in three dimensions.

Homework Equations



2D arc length: [itex]s = r \alpha[/itex]
3D:
[tex]ds = dr \hat{r} + r d\theta \hat{\theta} + r sin(\theta) d\phi \hat{\phi}[/tex]

The Attempt at a Solution



I'm sure I already have an equation and a method of solving this by changing the coordinates to Cartesian, and then taking the dot product to find the angle and then using the 2D arc length equation to get the distance along the surface of the sphere between the two points.

However, I am writing this because I am interested in using the second equation I gave above to find the distance. I'm not sure if I'm doing it right but it looks like I could just integrate so I have "s" instead of "ds" and the unit vectors are constant so they come outside of the integrals:
[tex]s = \hat{r}\int_{r}\int_{\theta}\int_{\phi}dr + \hat{\theta}\int_{r}\int_{\theta}\int_{\phi}r d\theta + \hat{\phi}\int_{r}\int_{\theta}\int_{\phi}r sin(\theta) d\phi[/tex]

The first term would become {r-hat}*(r2-r1) right?
But what I'm not sure of are the other terms because just looking at the second term by itself you can split it up in this way:
[tex]\hat{\theta}\int_{r}r \int_{\theta}d\theta[/tex]

The thing that concerns me is that there is no "dr", so how do I go about integrating this? Similarly with the third term.

Can I not find the distance in this way? I don't see why I shouldn't be able to...
perhaps the "r" is a constant in this case...

I hope this question isn't too dumb, and if it is please just bare with me and try to help me understand. Any advice is appreciated.

Thanks.
 
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  • #2
You're totally on the right path. r = constant, therefore
[tex]
dr = 0
[/tex]
therefore
[tex]
ds = r d\theta \hat{\theta} + r sin(\theta) d\phi \hat{\phi}
[/tex]
but you also need to parametrize your theta a phi; i.e. a path is one dimensional, so you need to have only one free parameter (e.g. 't' for theta(t), phi(t) )
plug that parametrization into 'ds' above, and integrate over 't' (e.g.).

Because you're looking for the shortest distance (presumably), your solution will be a geodesic (or great circle in this case).
 
  • #3
Gear.0 said:
2D arc length: [itex]s = r \alpha[/itex]
3D:
[tex]ds = dr \hat{r} + r d\theta \hat{\theta} + r sin(\theta) d\phi \hat{\phi}[/tex]
Your 3D equation is the vector equation

[tex]d\mathbf{s} = dr\,\mathbf{\hat{r}}+r\,d\theta\,\boldsymbol{\hat{\theta}}+r\sin \theta\,d\phi\,\boldsymbol{\hat{\phi}}[/tex]

for the line element. If you integrate ds, you'll get the vector s, not a distance. What you want is its length ds where

[tex]ds^2=d\mathbf{s} \cdot d\mathbf{s} = dr^2 + r^2 (d\theta^2+\sin^2\theta\, d\phi^2)[/tex]

(On a side note, in spherical coordinates, the unit vectors are not constant. They're functions of r, θ, and ϕ, so you can't just yank them out of integrals like you did.)
 
  • #4
Thanks.

But um, how would I do:
[tex]\int_{r} dr^{2}[/tex]

is that even possible?
 
  • #5
You wouldn't. It's probably easiest to see how you do this type of calculation by example.

Say the path is on the surface of a unit sphere and satisfies ϕ=θ where the angles run from ϕ=θ=0 to ϕ=θ=π/2. (See the attached plot.) On the surface of a sphere, r=1 is constant, so dr=0. Since ϕ=θ, we get dϕ=dθ. Let's use θ as the integration variable, so we'll get

[tex]\begin{align*}
ds &=\sqrt{ds^2}=\sqrt{dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)} \\
& = \sqrt{0^2+1^2(d\theta^2+\sin^2\theta\,d\theta^2)} \\
& = \sqrt{1+\sin^2\theta}\,d\theta
\end{align*}
[/tex]

The length would therefore be given by

[tex]s=\int ds = \int_0^{\pi/2} \sqrt{1+\sin^2\theta}\,d\theta[/tex]
 

Attachments

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What is a surface integral?

A surface integral is a mathematical concept used in calculus to calculate the total value of a function over a three-dimensional surface. It involves finding the area of infinitesimal pieces of the surface and summing them up to find the total value.

What is the purpose of finding distance along a sphere using a surface integral?

The purpose of finding distance along a sphere using a surface integral is to calculate the length of a curve on the surface of a sphere. This can be useful in various applications, such as calculating the distance between two points on a globe or finding the shortest path between two points on a curved surface.

What is the formula for calculating distance along a sphere using a surface integral?

The formula for calculating distance along a sphere using a surface integral is ∫∫dS, where dS represents the infinitesimal area element and the integral is taken over the entire surface of the sphere. This formula can be modified to suit different situations, such as finding the distance along a specific curve on the sphere.

How is the surface of a sphere divided for the purpose of a surface integral?

The surface of a sphere can be divided into infinitesimal pieces using different methods, such as spherical coordinates, parametric equations, or using the unit normal vector. These methods involve breaking the surface into smaller parts and summing them up to calculate the total value using the surface integral formula.

What are some applications of finding distance along a sphere using a surface integral?

The applications of finding distance along a sphere using a surface integral include calculating the shortest path between two points on a globe, finding the distance traveled by an object moving along a curved path on the surface of a sphere, and determining the surface area of a sphere.

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