# Cross product integral?

1. Feb 10, 2014

### 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

2. Feb 10, 2014

### 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=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
3. Feb 11, 2014

### JasonHathaway

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

4. Feb 11, 2014

### 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?

Last edited: Feb 11, 2014
5. Feb 11, 2014

### 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.

6. Feb 11, 2014

### 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?

7. Feb 11, 2014

### 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:

Last edited: Feb 11, 2014
8. Feb 11, 2014

### Simon Bridge

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: