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Compute the flux from left to right across a curve

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Compute the flux of [itex]\overrightarrow{F}(x,y) = (-y,x)[/itex] from left to right across the curve that is the image of the path [itex]\overrightarrow{\gamma} : [0, \pi /2] \rightarrow \mathbb{R}^2[/itex], [itex]t \mapsto (t\cos(t), t\sin(t))[/itex].

    A (2-space) graph was actually given, and the problem referenced "the curve" as given in polar coordinates by [itex]r=\theta[/itex] and [itex]0\leq \pi/2[/itex], so the above parameterization was my doing.

    2. Relevant equations

    Flux is given by [itex]\int\int_{S}{\overrightarrow{F} \cdot d\overrightarrow{S}} = \int\int_{D}{\overrightarrow{F} \cdot \overrightarrow{n} dA} = \int\int_{D}{- F_1 g_x - F_2 g_y + F_3} dA[/itex], where [itex]\overrightarrow{F}[/itex] is a vector field defined on a surface [itex]S[/itex] given by [itex]z = g(x,y)[/itex], oriented by a unit normal [itex]\overrightarrow{n}[/itex].

    3. The attempt at a solution

    First, I am thrown off by the "from left to right" requirement--I don't know what that means. The idea I have is that it has something to do with the direction of the normal.

    Also, I have only really dealt with (3-space) surfaces.

    Second, I am used to computing the normal as the cross product of the parametrized surface of [itex]S[/itex] differentiated with respect to each of the two variables parameterizing it. Here, I only have one variable. Would I change my parameterization to have two variable but only use one?

    Thanks.
     
  2. jcsd
  3. Mar 24, 2013 #2

    Dick

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    Yes, it's the choice of which normal to use. If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve. So use the normal pointing away from the origin. And if you take the derivative with respect to your parameter you'll get a tangent vector. In 2d finding the normal to that is easy. Just rotate it 90 degrees.
     
  4. Mar 25, 2013 #3
    Thanks for the response.

    "If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve."

    I don't understand what you are saying here. Could you please explain more explicitly? I am thinking you are talking about drawing the vector field of [itex]\overrightarrow{F}[/itex] over the graph of the curve. If that is the case, I am having trouble seeing that the vector field flows in the direction of increase of the curve.
     
  5. Mar 25, 2013 #4

    Dick

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    No, I'm not paying any attention to F. Your curve goes out from the origin and then back to the y-axis in the first quadrant. You have two choices of normal, it can be rotated right or left from the tangent. If they say flux from left to right, I think you should pick the normal to the right.
     
    Last edited: Mar 25, 2013
  6. Mar 31, 2013 #5
    Thanks for the reply.

    Alright... I think I need to work on my intuition for this stuff to better understand your help.

    Anyway, I think I've got something decent, and I'm done with it. For anyone looking at this thread for help, I found this link helpful:

    http://web.mit.edu/jorloff/www/18.02a-esg-iap/topic40.pdf

    Thanks again Dick.
     
  7. Mar 31, 2013 #6

    Dick

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    That's great! But let me try and explain again. If they give you a closed curve and say compute the flux they usually mean outward going flux. So you should pick the outward normal. So if they give you a curve and describe the direction (here left to right) you should pick the normal pointed outward from that direction. So that's rightwards.
     
  8. Apr 1, 2013 #7
    Gotcha; that makes a bit more sense now. Thanks for the generalization!
     
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