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Generalization of Surface Integral

  1. Jul 4, 2011 #1
    Given that a surface integral of a function, f(x,y,z), is written as [itex]\int\int f(x,y,z) dS[/itex] where dS= |df/dx x df/dy| dA, how can this be generalized into more dimensions? In other words, is it possible to find a way to convert dS into a differential piece of area for more than 3 dimensions? What type of cross product would be able to incorporate the cross product of something differentiated with respect to three parameters, or four, or so on?
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  3. Jul 4, 2011 #2

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    Hi schaefera! :smile:

    What you write is a bit ambiguous.

    It should be something like:
    [tex]\iint_S f(x,y,z) dS = \iint_S f(\phi(u,v)) ~ \left|\frac {\partial \phi} {\partial u} \times \frac {\partial \phi} {\partial v}\right| ~ du~dv[/tex]
    where [itex]\phi(u,v)[/itex] is a function that maps (u,v) to points (x,y,z) on the surface S.

    For n dimensions this is generalized using the absolute value of the Jacobian determinant.
    Your cross product is a special case for 3 dimensions that happens to be the same.

    See for instance: http://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables

    The formula shown in the article is:
    Last edited: Jul 4, 2011
  4. Jul 4, 2011 #3
    Ok, I believe I understand that (although I have to admit it's a bit beyond what I have studied)! My one remaining question is: why is the cross product in the 3D version a special case of the Jacobian determinant? Because the parameterized function relies on two variables, wouldn't the Jacobian be a 2x3 matrix (the top row being the partials with respect to u and the second row being the partials with respect to v)? I'm obviously missing something, but how do you get the determinant of that?
    Last edited: Jul 4, 2011
  5. Jul 4, 2011 #4

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    The cross product can be represented in many ways, as you can see on wikipedia.
    One of them is as the determinant of a matrix:
  6. Jul 5, 2011 #5
    But isn't that different from taking a Jacobian and then finding the determinant of that matrix?
  7. Jul 5, 2011 #6

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    Ah, I see your point.
    I guess I was hasty in answering your question with the Jacobian.

    Well, in n dimensions, the cross product is generalized to:

    I guess you can use this to calculate the integral.
    Last edited: Jul 5, 2011
  8. Jul 5, 2011 #7
    There is a beautiful generalization. We will define something called a k-dimensional parametrized-manifold in R^n, that is analogous to say, a 2-dimensional parametrized-surface in R^3.

    Here is the definition of a parametrized-manifold:

    Let k <= n. Let A be open in R^k, and let g: A ---> R^n be a map of class C^r. The set Y = g(A), together with the map g, constitute what is called a parametrized-manifold of dimension k. We denote this parametrized-manifold by Y_g; and we define the k-dimensional volume of Y_g by the equation [tex] v(Y_g) = \int_A V(Dg) [/tex], provided the integral exists.

    Here, Dg is the derivative of g, and [tex] V(Dg) = \sqrt{det[Dg^{tr}Dg]} [/tex].

    Note that if f is a continuous map from Y_g to R then the integral of f over Y_g, with respect to volume, is defined by [tex] \int_{Y_g} f dV = \int_A (f \circ g) V(Dg) [/tex]

    We use the notation dV in the integral to denote integral with respect to volume

    EDIT: Also note that [tex] Dg^{tr}Dg [/tex[ is always a square matrix, so taking the determinant of it is never a problem.
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