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Surface Integral, flux. Boundary and orientation

  1. Mar 10, 2012 #1
    In solving a flux integral over a flat surface, inclined above the xy-plane, does the boundary of the surface influence the flux only through the integral limits? (and not through its normal vector)

    Let's say that there is an elliptic surface inclined above the xy-plane. The orientation is given by the plane: z=3-y. Now, when I am supposed to solve this surface integral, it seems I can easily use the normal vector of the plane z=3-y in the dS conversion:

    dS = [-(∂f/∂x)[itex]^{2}[/itex], -(∂f/∂y)[itex]^{2}[/itex], 1]dA, where f(x,y)=3-y

    This means that I do not need to think about the parametrization of the elliptic surface...
    This would ofcourse only be true for flat surfaces. So, for clarification, am I doing something wrong?
     
  2. jcsd
  3. Mar 10, 2012 #2
    To clarify, because this might seem like a strange question... Does the magnitude of the Normal vector play a role, or just its direction?
     
  4. Mar 10, 2012 #3

    chiro

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    As a disclaimer you should probably get some further input since I haven't taken vector calculus in a long time, but as far as I understand it, the normal vector should be a unit normal and the magnitude should be contained in the information for the actual flux component itself.

    If the magnitude was not unit (i.e. 1) then you would have to just normalize the integral in the appropriate way. You have to remember that you are calculating a projection of the flux with respect to the surface normal and if you review the definition of such a projection, you should find that the normal needs to be unit length.

    This implies that only the direction matters and not the magnitude.
     
  5. Mar 10, 2012 #4
    Thanks, so the reason why one projects the surface onto a plane like the xy-plane is to find how much of the different vector field components go trough the surface.
     
  6. Mar 10, 2012 #5

    chiro

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    In flux integrals its best to think of each point on your surface having its own normal vector at a point. On a plane, this doesn't change but on a curved surface it does.

    What you are doing is just doing a projection of the 'flux component' with respect to the normal.

    If the flux is parallel to the normal then the magnitude of the flux is at a maximum since it is going through the surface with maximum intensity. If the normal is perpendicular, then the flux will not even touch the surface resulting in a zero contribution since the flux is parallel to the surface itself.
     
  7. Mar 10, 2012 #6

    HallsofIvy

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    The answer could be either "yes" or "no" depending upon exactly how you define the "normal vector". If you are given some vector function, [itex]\vec{v}(s,t)[/itex], defined over a surface F(s, t)= constant, to integrate the vector function over the surface by finding the gradient, [itex]\nabla F[/itex] which is, of course, normal to the surface.

    Many texts form the unit normal, [itex]\vec{v}[/itex], by dividing that by the length [itex]\left|\nabla F\right|[/itex] but then form the "differential of surface area" [itex]dS= \left|\nabla F\right|dsdt[/itex]. But I have never seen much sense in dividing by [itex]\left|\nabla\right|[/itex] to get the unit normal, then multiplying by it to get the differential of surface area! Instead, I prefer to define the "vector differential of surface area", as done in many other texts, [itex]\vec{dS}= \nabla F dsdt[/itex].

    On the one hand you can calculate the flux as
    [tex]\int \vec{v}\cdot\vec{n}dS= \int \vec{v}\cdot\vec{n} \left|\nabla F\right| dsdt[/tex]
    or as
    [tex]\int \vec{v}\cdot\vec{dS}= \int \vec{v}\cdot\nabla F dsdt[/tex]

    I prefer the second but they are equivalent.
     
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