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Multivariable Calculus, Parameterization Question

  1. Jun 9, 2013 #1
    Hello everybody!

    My question has to do with the parameterization of 3D surfaces from 3 variables to 2. Specifically, I'm trying to figure out an aspect of the cross product of the directional derivatives of the parameterization to solve flux integrals. Trying to convert:
    ∫∫SF dS = ∫∫DF(r(u,v)) (ru x rv) du dv

    I hope that this is clear.

    1. The problem statement, all variables and given/known data
    What is the difference between r(R, theta) and r(theta, R), and its effects on the cross product:
    rR X rθ

    Suppose you have a circle on the x-y plane with:
    X = R * cos(θ)
    Y = R * sin(θ)
    Z = 0

    Where 0 ≤ R ≤ 1, 0 ≤ θ ≤ 2π

    Now, this can be written as r(R, theta) or r(theta, R), but then does r still end up being the same?:
    X = R * cos(θ)
    Y = R * sin(θ)
    Z = 0

    I know that:
    rR X rθ = - rθ X rR

    This cross product determines the orientation of the surface, so this is important to understand.


    2. Relevant questions

    How do you determine which variable gets to be the radius and which gets to be the angle?
    How do you determine which directional derivative gets to go first in the cross product?
    This is especially relevant when the variables that I'm converting to are U and V instead of something polar.


    3. The attempt at an explanation

    My guess would be that it would be something to do with the right-hand rule, where angles always go CCW. But then this is not easy to visualize, nor is it a really good explanation.

    EDITED: Not magnitude of cross product.
     
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2

    LCKurtz

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    Homework Helper
    Gold Member

    Good question, and a lot of texts aren't clear on that. You may not know whether to take ##r_u \times r_v## or the other way around. Just do one or the other. Then check to see whether it agrees with the given orientation, using the right hand rule or some other way. Most texts don't write Stokes' theorem as$$
    \oint_C \vec F\cdot d\vec R = \iint_{S} \nabla \times \vec F \cdot d\vec S
    =\pm \iint_{u,v} \nabla \times \vec F \cdot \vec R_u\times \vec R_v\, dudv$$where the sign is chosen to agree with the orientation, but they should. After you do the cross product you must check which sign to use. Sometimes you can tell by looking at one of the components of the cross product. For example if the surface is oriented upward, you could just check the z component. Or if it comes from a Stokes theorem problem, make sure it agrees with the orientation of the curve as given by the right hand rule.
     
  4. Jun 9, 2013 #3
    Ohhhhh, thank you! It seems obvious now, but the order I do the cross product is just another way to manipulate the orientation of the surface.

    To sum up, the order of r(R,θ) and r(θ,R) do not matter. But the cross product will give me an orientation that will point one direction and I have to compensate using negative signs to get the orientation I want.
     
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