1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface integrals of vector fields

  1. May 9, 2012 #1
    The integral for calculating the flux of a vector field through a surface S with parametrization r(u,v) can be written as:

    [itex]\int\int_{D}F\bullet(r_{u}\times r_{v})dA[/itex]

    But what's to stop one from multiplying the normal vector [itex]r_{u}\times r_{v}[/itex] by a scalar, which would result in a different flux? When you're asked to find the flux of a vector field through a surface, how do you know which scalar to use?
     
  2. jcsd
  3. May 9, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    dA is du*dv, right? That's your formula. It doesn't say you can multiply the normal vector in the formula by an arbitrary number. Why would you do that?
     
  4. May 9, 2012 #3
    I got confused because I was trying to solve a problem where you had to use Stoke's theorem to find the flux of the vector field <2x,3y,z3> through the curve of intersection of the paraboloid z=x2+y2 and the plane 2x+4y+z=1, when C is traversed conterclockwise when viewed from the origin. So I found the curl to be <0,0,5>, the normal to the plane <2,4,1> and the region D traced out by the curve C to be a circle of radius sqrt(6). The flux I think is just -<0,0,5>*<2,4,1>π(sqrt(6))2=-30π. But then I started thinking, since any normal can be used to find the scalar equation of a plane, why can't the normal <2,4,1> be multiplied by, say, 2 to get <4,8,2>. It'll still be normal to the plane, right? Then the flux would be -60π. But now that I think of it more, that doesn't make sense, since the normal vector was derived from the scalar equation of the plane, which dictates that the normal vector be <2,4,1>.
     
  5. May 9, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Your formula uses a specific normal vector and a specific surface measure dA=du*dv. Sometimes you do it with a unit normal vector and a different notion of surface measure. The two are interrelated. You can't change one without changing the other.
     
  6. May 10, 2012 #5
    Thank you. So it can't be changed just by virtue of the definition of the differential dS=(ru x rv)dudv, which relies on a unique parameterization r(u,v) (or at least unique within the scope of a single coordinate system)? I really don't understand why this was troubling me; I must have been very tired last night.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Surface integrals of vector fields
Loading...