Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Iterated integrals

  1. Aug 8, 2014 #1
    Ok for the purpose of this question let's stick to the flux integral:

    The general formula is ∫∫s (E-vector)*(dS-vector)=Flux where * stands for the dot-product.

    Now, I like it when my integrals make sense, and to do that I usually think of the Riemann Sum which might represent my integral. The problem I have though is this one:

    I can think of cases where this expression for the flux integral holds true, but I'm mainly thinking of very simple surfaces and vector fields. I just don't see how you can know this would be the expression of the flux integral considering the most complicated vector field you can think of on the most irregular surface you can imagine. Namely, I'm looking for the general argument that makes it necessary that the flux is "the double integral over some surface of E*dS".

    What is the rationale behind "Oh, sure it's simply the double integral of the dot product of the field and some differential of area". I'm specifically troubled by the double integral, what made people think the double integral was the operation needed? I can see how it is needed when I can think of a Riemann sum which represents the double integral where I have ΔxΔy or something on all terms, but I can't see this in a general way.

    I hope my question is clear, and thanks for helping me out.

    TL;DR: why is the flux the double integral of the dot product of the field and a differential of surface. I specifically want to know where the double integral came from.
    Last edited: Aug 8, 2014
  2. jcsd
  3. Aug 9, 2014 #2


    User Avatar
    Homework Helper

    Let me give this example, imagine a chessboard with a number in each cell. The sum of those numbers would be written like this: $$\sum_{i=0}^8 \sum_{j=0}^8 x_{ij}$$.

    Why write it in this way? Because it is scalable and is a lot neater than something like this: $$\sum_{i,j}^{0 ≤ i,j ≤ 8} x_{ij}$$.

    Is it two sums or one sum? I like to think of it as being one big sum. Can you relate this to the case of double integrals?
  4. Aug 9, 2014 #3
    Yeah that's kind of what I think of when I see some integral, but I don't see how you could think of it this way when you're integrating a dot product which might be varying (because it's a complicated vector field and a complicated surface, or whatever) all the time! I've found that there are ways to write a normal vector as a function, and that way you could simplify your integrand and think of the Riemann Sum representing it. Any ideas?
  5. Aug 9, 2014 #4


    User Avatar
    Homework Helper

    Is your question "why is flux of E defined to be the integral [itex]\int_S E \cdot dS[/itex]?", to which the answer is "that's the result of summing the component of E normal to a small surface element times the area of that element, and then making the areas arbitrarily small whilst still covering the whole of S", or is your question "why are surface integrals evaluated as double integrals?", to which the answer is "the surface integral notation is a shorthand for an integral over some region of the plane, and integrals over regions of the plane are defined as double integrals".

    The "surface integral" [itex]\int_S \mathbf{E} \cdot d\mathbf{S}[/itex] is defined to be the double integral [tex]\int_S \mathbf{E} \cdot d\mathbf{S} = \iint_A \mathbf{E}(\mathbf{x}(u,v)) \cdot \left(\frac{\partial \mathbf{x}}{\partial u} \times \frac{\partial \mathbf{x}}{\partial v}\right)\,du\,dv,[/tex] where [itex]S = \{\mathbf{x}(u,v) : (u,v) \in A \subset \mathbb{R}^2\}[/itex]. The point here is that [itex]d\mathbf{S} = \frac{\partial \mathbf{x}}{\partial u} \times \frac{\partial \mathbf{x}}{\partial v}\,du\,dv[/itex] is normal to [itex]S[/itex] and its magnitude is approximately the area of the image of a small rectangle in the [itex](u,v)[/itex] plane. There are of course two possible directions for the normal of the surface (we could equally have chosen [itex]d\mathbf{S} = \frac{\partial \mathbf{x}}{\partial v} \times \frac{\partial \mathbf{x}}{\partial u}\,du\,dv[/itex]), and the choice of which results in positive flux is largely a matter of convention.

    If [itex]A = [a,b] \times [c,d][/itex] is a rectangle then it should be easy to see how you get a Riemann sum for the double integral of [itex]f(u,v) = \mathbf{E}(\mathbf{x}(u,v)) \cdot \left(\frac{\partial \mathbf{x}}{\partial u} \times \frac{\partial \mathbf{x}}{\partial v}\right)[/itex],
    \sum_{i} \sum_{j} f(\xi_i,\eta_j)(u_{i+1} - u_i)(v_{j+1} - v_j)\qquad u_i \leq \xi_i \leq u_{i+1}, v_j \leq \eta_j \leq v_{j+1}[/tex] but if [itex]A[/itex] is not a rectangle then you have to extend the domain of integration to the smallest rectangle [itex]R[/itex] containing [itex]A[/itex], and define the integrand to be zero on [itex]R \setminus A[/itex].
  6. Aug 9, 2014 #5
    Yeah I can see how you get a Riemann sum for a simple surface like a rectangle. But I don't see how you could set up a Riemann sum for an abnormal surface and vector field; what makes people so confident it's still gonna be a double integral if there's no Riemann sum in sight?

    BTW, I'm reading about how you can construct the normal vector to a surface, in which case I think I'm beginning to see that you can establish a Riemann sum representing that surface integral. Is this the correct approach? The only problem is that I'm having trouble understanding and visualising what a level surface would be. Thanks.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Iterated integrals
  1. Iterated integral (Replies: 8)