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Cross product integral?

  1. Feb 10, 2014 #1
    Hi everyone,

    What does [itex]\int \vec{f} × \vec{dl}[/itex] mean? does it mean "The Line Integral of Vector Function on a positive curve L"?

    And are the following named correctly?

    [itex]\int \vec{f} . dl [/itex] : Scalar Line Integral of a Vector Function

    [itex]\int udl [/itex] : Line Integral of a Scalar Function
  2. jcsd
  3. Feb 10, 2014 #2

    Simon Bridge

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    The integral sine should be thought of as a summation - a continuous summation.
    If you translate the integrals into their equivalent discrete sums, the expressions should make more sense to you.

    - f.dl : product of scalars. That would be an "area" or similar.
    Think how this would come about if f and dl were vectors ...

    - f.dl : I'm not sure about what you mean by this.
    Perhaps if f is a polynomial in x and dl=dx?
    A polynomial is a vector isn't it?
    How is this different from the first case?

    But perhaps you are thinking more of something like f=fxi+fyj+fzk
    so f.dl=fxdli+fydlj+fzdlk
    so everything depends on how dl is defined?

    Compare with the example of the polynomial.

    - fxdl
    ... these are clearer: consider what the dot and cross products do and how dl is defined.
    How do these definitions relate to the previous two examples?
    Last edited: Feb 10, 2014
  4. Feb 11, 2014 #3
    for f × dl: basically it will give an perpendicular vector, so it's much like fdl? (without dot)
  5. Feb 11, 2014 #4

    Simon Bridge

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    If f and dl are both vectors, then fdl (no dot) is ambiguous.

    f.dl would give, basically, the component of f in the direction of dl ... so dW=F.dl would be the work done by force F moving distance dl along a curve. We'd write: $$W=\int_C \vec F\cdot \text{d}\vec l$$ or something right?
    i.e. the fact that there is a parameterized curve is involved is represented by the "C" attached to the integral sign.

    If F and dl were always in the same direction, then the result is the same as just scalar multiplying the amplitudes and you get just ##W=\int F\text{d}l## ... which is the form you first learn for an integral. It's a special case.

    fxdl would give a vector perpendicular to both f and dl whose magnitude is the component of f perpendicular to dl.

    Now where, physically, would a cross product come up - where you have to add up lots of them incrementally?
    Last edited: Feb 11, 2014
  6. Feb 11, 2014 #5
    So.. if [itex]\vec{f}[/itex] is [itex](u-u^{2})\vec{i}+2u^{3}\vec{j}-3\vec{k}[/itex] and I were asked to evaluate [itex]\int fdu[/itex], I will integrate the i, j, and k components individually with respect to u, because du is not vector in this case, and my result would be a vector, right?

    And for your question, the only thing that came to my mind that is area of parallelogram, I don't know whether this is right or wrong.
  7. Feb 11, 2014 #6

    Simon Bridge

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    1. that would be correct ...

    2. In this case one of the sides of the parallelogram is dl ... you could presumably use it to evaluate surface areas - it's just that there are easier approaches.

    Can you come up with an example of such an integral?
  8. Feb 11, 2014 #7
    Actually I don't have a "physical" example of this integral, I was asked in a previous exam the following:

    "Let [itex]\vec{f}=x\vec{i}-2y\vec{j}+z\vec{k}[/itex] , L is a piece of regular circular spiral on the oy axis between the two planes [itex] 0 \leq y \leq \frac{3}{4\pi} [/itex] [itex] \phi[/itex] [itex]sin(2\phi) [/itex]

    Where [itex]x=3cos(2\phi)[/itex], [itex]y=3sin(2\phi)[/itex], [itex]z=6\phi[/itex]

    Evaluate the scalar line integral and the vector line integral of the vector [itex]\vec{f}[/itex] on L"

    I was told that the "scalar line integral" is dot product of f and dl, because the dot product result is a "scalar", and the vector line integral is the cross product of f and dl, because the cross product result is a "vector".

    We had an example in class includes cross product integral, which just says "Evaluate [itex]\int f × dl[/itex] where [itex]\vec{f}=xy\vec{i}-z\vec{j}+x\vec{k}[/itex], where [itex]x=t^{2}[/itex], [itex]y=2t[/itex], [itex]z=t^{3}[/itex], where t from 0 to 1".

    I just want to know if this "stuff" is even correct, because I couldn't find any book includes cross product integral, and (I'm sorry to say that) I don't trust my lecturer.

    Added the red line.
    Last edited: Feb 11, 2014
  9. Feb 11, 2014 #8

    Simon Bridge

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    That would appear to be an unusual use of the term.


    However - also see:
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