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Homework Help: Vector Field Describing Fluid Flow in a Torus

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Write a vector field equation which describes fluid flowing around a pipe of radius r whose axis is a circle of radius R in the (x,y)-plane.

    2. Relevant equations
    Equation of a torus?

    3. The attempt at a solution
    What I've gathered from the question: the pipe is in the shape of a torus of radius r and the circle of radius R runs through the center of the inside of the pipe.

    I know that two things describe this flow:

    1. The magnitude of the flow decreases the farther away from the axis line on the inside of the torus that the point (x,y,z) is.
    2. The flow goes either clockwise or counterclockwise around the origin in the (x,y)-plane. So the vector field equation for that piece is F(x,y)=<y,x> or F(x,y)=<y,-x>

    Otherwise, I have no idea where to start.
  2. jcsd
  3. Sep 7, 2008 #2


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    No, that's not the equation of a torus. That's the equation of a circle in the xy-plane or of a cylinder in three dimensions. Assume that the z-axis runs through the center of the torus. We can get a particular cross section of the pipe by drawing a plane containing the z-axis that cuts the torus at angle [itex]\theta[/itex]. The center of the torus will be [itex](R cos(\theta), R sin(\theta), 0)[/itex]. From that point, let [itex]\phi[/itex] be that angle made with the xy-plane. The point will be an additional [itex](r cos(\phi)cos(\theta), r sin(\phi)cos(\theta), r sin(\phi))[/itex]. That means that a point on the torus is given by [itex]x= R cos(\theta)+ r cos(\phi) cos(\theta)[/itex], [itex]y= Rsin(\theta)+ r sin(\theta)cos(\phi)[/itex], [itex]z= r cos(\phi)[/itex].

  4. Sep 7, 2008 #3
    I know x2+y2=r2 is the equation of the circle that acts as the axis of the pipe (torus). I wrote "Equation of a torus?" below it because I figured it would be of use, but since I didn't know what it was, I wrote that.
  5. Sep 8, 2008 #4
    Update on my progress:

    The distance from any point (x,y,z) contained within the torus to its axis (x2+y2=1) is given by nr, where n and r (the radius of the pipe itself) are any number between 0 and 1, inclusive (the upper restriction on r accounts for the fact that the pipe cannot intersect itself in the center). Using this, along with the parameterization of the torus, the position vector of any point in the torus with the unit circle in the (x,y)-plane as its axis is:


    I've placed this equation into Cartesian coordinates as well: [(x2+y2)1/2-1]2+z2=(nr)2

    The magnitude of the vector in the field at the point (x,y,z) in the torus is then r2-(nr)2 = r2-[(x2+y2)1/2-1]2-z2 (after distributing the - through).

    Now, I have no clue what this vector's direction is.
  6. Sep 9, 2008 #5
    Nevermind, I figured it out.
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