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The area vector and its magnitude

  1. Sep 28, 2008 #1
    I have a slight conceptual problem with the use of a vector to represent a differential area (as used in calculating flux). It makes sense in a vetor field such as velocity, where each point has a speed and direction, but in the case of area where its magnitude is a measure of space it confuses me as to how that measurement of space can be made in an infinitessimally small area. Wouldn't area vectors only contain a direction and not a magnitude?
  2. jcsd
  3. Sep 29, 2008 #2


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    You can indicate the direction of the plane by any vector perpendicular to it (for example, if [itex]\vec v, \vec w[/itex] span the plane then anything along [itex]\vec v \times \vec w[/itex] will do. The area of the plane can be given by setting the magnitude of this vector equal to the area of the plane.
  4. Sep 29, 2008 #3
    That makes sense if the surface is a flat plane, but what of the general case of any surface? How do you talk about the magnitude of the area vector on a curved surface?
  5. Sep 29, 2008 #4
    In boring calculus, we can approximate curves with tangent lines.

    In hard calculus, we can approximate surfaces with tangent planes.

    Just as we get better and better approximations of a curve as the number of tangent lines goes to infinity and their size shrinks to zero, we can get better and better approximations of a surface as the number of planes goes to infinity and their area shrinks to zero.
  6. Sep 29, 2008 #5
    I guess I'm not articulating my question well. In understand what you are telling me, but when an area is represented as a vector, is it just a unit vector normal to the surface at that point (i.e., direction without magnitude) or can it be said to have some magnitude? I dont think it can have a magnitude since you are talking about a point.
  7. Sep 29, 2008 #6
    Could you be more specific with what you mean exactly? "Represent" can mean many different things. An "an area" could mean a rectangular segment of a plane or it could mean the geometric area of such a segment.

    It might be helpful to give an example of a problem you're working on, too.
  8. Sep 29, 2008 #7
    You're trying to do a flux problem in E&M I take it.
    Look at it this way. You can take an infinitesimal change in area (that concept isn't too hard to grasp), and conveniently define the vector sticking out normal to it as having the magnitude OF that area.

    When you let that area 'go to zero', you end up with a surface integral. Usually for what I think you're doing, problems are usually chosen such that the surface integral turns out to just be the surface area of a sphere (4 pi r^2) or something of that sort. If not, say you've got a plane at some angle to an electric field...

    We 'conveniently define' the vector whose direction is normal to the plane to have the magnitude of the area of that plane, and the integral E (dot) da turns into a simple dot product of the electric field dotted with the 'area vector', or |E| |A| Cos(angle between the two vectors).

    Hope this helps.
    Hope it's right too, have a test in like 4 hours on it.
  9. Sep 29, 2008 #8
    Its not one particular problem, but what spurred my question was learning about calculating the flux of a vector field [tex]\int_S\vec{E}\cdot d\vec{a}[/tex] where the differential area of surface S is a vector with both magnitude (geometric area) and direction. How can the area at a point on a surface have any magnitude?

    Edit: Cvan, you are right, it is an E&M problem. I am just having a hard time understanding how a differential area can have any magnitude. It makes sense to me for other vector quantities like velocity, etc. in which their magnitude is not composed of space. Since the magnitude of area is composed of space, how can an infinitessimal area have any area?

    I seem to be getting further from understanding the more i elaborate my question, haha.
    Last edited: Sep 29, 2008
  10. Sep 29, 2008 #9
    Ah. I think I see what you're getting at.

    The "da" in the equation you gave is still a vector. It has a definite direction, and that direction is always normal to the surface.

    Suppose we have a sphere of radius 1 we want to approximate. One way to do this would be to approximate our sphere with 6 tangent planes... imagine sticking it snugly inside a cube. Each surface on the cube has an area of 2. The total area is 6 x 2 = 12, which is of course a very rough approximation of there sphere's actual surface area: 4*pi*r^2 = 4*pi.

    What are the "da"'s in our cube approximation? They are the (unit) normals to each surface scaled by the area of the surface. That means the top face is a vector pointing up with a length of 2. The one on the opposite side is a vector pointing down with a length of 2.

    Let's put this sphere in a uniform field F(x, y) = (0, -1). What is the flux?

    Let's use a Riemann style approach. Take our cube approximation of the sphere. There are four "da" sections (those vectors we talked about above).

    For the top side, the normal is (0, 2). (Note how the total length is 2 and the direction is up along the straight up along the y-axis). The dot product between the field and this "da" is (0, 1) * (0, 2) = 2.

    Similarly, for the bottom side, the normal is (0, -2). The dot product between this side and the field is (0, 1) * (0, -2) = -2.

    For the remaining four sides, the normal will be at a right angle to the field F. The dot product between orthogonal vectors (vectors at right angles) is zero, so the remaining terms are all 0.

    Our (approximate) flux in this field is thus 2 + -2 + 0 + 0 + 0 + 0 = 0.

    Now, 0 is the actual answer for a sphere, because our situation was so simple. But in general, if you wanted a better approximation, you'd have to use a more complicated shape. You might use an octrahedron, a dodecahedron, or a icosidodecahedron to approximate our sphere. There will be more surface normals to consider (more "da"s) but at the same time, their individual areas will be much smaller. Also, the direction of each normal will not be as easy to calculate, and you probably need to start plugging things into a calculator or a computer program.

    lol =-D
  11. Sep 29, 2008 #10
    Thanks for your response (it also clarified what flux means to me). I guess its no more a logical jump to understand differential area than it is any other differential quantity, but my problem was that the units of area are dimensions of space, which would be zero in the differential amount.
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