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Vector Calculus: Air flowing through loop of straight lines

  1. Feb 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Air is flowing with a speed v in the direction (-1, -1, 1,) calculate the volume of air flowing through the loop consisting of straight lines joining (in order i presume) (1,1,0) (1,0,0) (0,0,0) (0,1,1) (1,1,0)

    2. Relevant equations

    3. The attempt at a solution
    I assume you have to do a surface integral to calculate the area of the loop, calculate the component of air that is flowing in that direction and do a simple velocity * area for volume per second? Not sure how to get started with the first two parts though.
    Last edited: Feb 11, 2015
  2. jcsd
  3. Feb 11, 2015 #2

    Ray Vickson

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    Draw a rough sketch of the loop.

    There well be infinitely many surfaces having that loop as perimeter (why?). Can you tell if it matters which surface you pick?
  4. Feb 11, 2015 #3
    Fab, thanks for the response.

    There are infinite surfaces with that loop as the perimeter as its an open surface, and all we are given is the boundary. And no it shouldn't matter which surface we pick as we are only interested in the volume of air passing through the boundary. What now? Im still unsure of what integral im supposed to be doing, the loop.

    http://imgur.com/hHiq1o9 is my very rough sketch

    EDIT: changed my mind, im not sure if we can pick any surface
  5. Feb 11, 2015 #4


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    It isn't usually proven in most calculus courses, but if ##\nabla \cdot \vec V = 0## in a simply connected region, then there is a vector field ##\vec F## such that ##\vec V = \nabla \times \vec F##. That says ##\iint_{S} \vec V\cdot d\vec S =\iint_{S} \nabla \times F\cdot d\vec S = \int_C \vec F\cdot d\vec R## for any open surface ##S## bounded by the curve ##C##. Since the right side doesn't depend on ##S## but only on ##\vec F## and ##C##, you can use any surface bounded by the curve to calculate the left side. There are a couple of obvious surfaces you could use.
    Last edited: Feb 11, 2015
  6. Feb 11, 2015 #5
    Your figure doesn't look right to me. Consider subdividing the loop into two abutting triangles: (1,1,0) (1,0,0) (0,0,0) and (0,0,0) (0,1,1) (1,1,0). It's easy to find the unit normal to each of these triangles, or, better yet, the unit normal times the area, by using cross products.

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