What is the meaning of line integrals?

[JasonHathaway]

Hi everyone,


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

And are the following named correctly?

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

\int udl : Line Integral of a Scalar Function

 

[Simon Bridge]

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.

Thus:
- 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=f_xi+f_yj+f_zk
so f.dl=f_xdli+f_ydlj+f_zdlk
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?

 

[JasonHathaway]

for f × dl: basically it will give an perpendicular vector, so it's much like fdl? (without dot)

 

[Simon Bridge]

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.

Similarly:
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?

 

[JasonHathaway]

So.. if \vec{f} is (u-u^{2})\vec{i}+2u^{3}\vec{j}-3\vec{k} and I were asked to evaluate \int fdu, 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.

 

[Simon Bridge]

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?

 

[JasonHathaway]

Actually I don't have a "physical" example of this integral, I was asked in a previous exam the following:

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

Where x=3cos(2\phi), y=3sin(2\phi), z=6\phi

Evaluate the scalar line integral and the vector line integral of the vector \vec{f} 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 \int f × dl where \vec{f}=xy\vec{i}-z\vec{j}+x\vec{k}, where x=t^{2}, y=2t, z=t^{3}, 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.

Edit:
Added the red line.

 

[Simon Bridge]

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".

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

Compare:
http://farside.ph.utexas.edu/teaching/em/lectures/node17.html

However - also see:
https://www.physicsforums.com/showthread.php?t=524990
http://math.stackexchange.com/questions/555268/integral-with-cross-product