line integral

But it is still a line integral so it is still going to turn into a single integral when you plug in the correct values.

Since the line of light radiation measurement involves a line integral of the radiating region, data handling and analysis are necessary.

One of them involves computing a line integral from the origin to a point in the plane by going first along the x axis, then vertically.

It tells you, if you take the line integral of the gradient of a function, what you get back is the function.

This one is a line integral. So, you use the method to explain here, namely, you express x and y in terms of a single variable.

But, you know, it gives you an example where you can turn are really hard line integral into an easier double integral.

It gives a relationship between a surface integral over an oriented surface and a line integral along a simple closed curve .

We said the line integral of F dot n ds is actually the line integral of - Q dx P dy.

If we want to compute the line integral along this guy then we have to break it into a sum of three terms.

Then, we just have to, well, the line integral is just the change in value of a potential.

It tells us, if you compute the line integral along C1, it's just f at this point minus f at this point.

OK, so it has this system of wheels and everything that compute for you the line integral along C of, well, it depends on the model.

OK, so the proof, ; so just going to prove that the line integral is path independent; the others work the same way.

In the line integral in the plane, you had two variables that you reduced to one by figuring out what the curve was.

That is correct. If you are trying to compute a line integral for flux, wait, where did I put it?

So, to say that a vector field with conservative means 0 that the line integral is zero along any closed curve.

This relates a line integral for one field to a surface integral from another field.

And then you know that the line integral inside will be equal to 0 the double integral of curl, which is the double integral of zero.

And, so the total line integral for this thing is equal to the integral along C prime, I guess the outer one.

So, that's a really strange statement if you think about it because the left-hand side is a line integral.

The results are of use value for simplifying the computation of line integral and surface integral.

How do I compute the line integral along the curve that goes all around here?

If we have to compute a line integral, we have to do it by finding a parameter and setting up everything.

Then the flux of F across a curve C is, by definition, a line integral but I will use notation F dot n ds.

Then I can actually -- --replace the line integral for flux by a double integral over R of some function.

We say that the line integral is path independent.

And so, when I take the line integral along a closed curve, 0 I don't have to compute it. I notes going to be zero.

If the field changes slowly, we can replace the time integral by a line integral over the unperturbed orbit.

Let's us replace a line integral along a closed curve by a double integral Well, here it is the same.

And, I still want to compute the line integral along a closed curve.

Well, we cannot really think of flux as a line integral.

And, that will give us the line integral, OK?

We formed the flux integral that was a line integral F dot nds.

It says that the line integral is zero along any closed curve.

Remember, you have to know how to set up and evaluate a line integral of this form.

If you cannot parameterize the curve then it is really, really hard to evaluate the line integral.

But some of them compute the line integral of x dy.

So, food for thought: let's say that I want to apply Stokes theorem to simplify a line integral along the curve here.

I mean, your line integral doesn't make sense so there's nothing to compute.

What is flux? Well, flux is actually another kind of line integral.

If you want to do it by direct calculation then you have to parametize this circle and figure out what the line integral will be.

Well, it's here, we have both x and y together, but we want to, because it's the line integral, there should be only one variable.

But, they compute some line integral, OK?

On the math it is a line integral of something dx plus something dy.

A line integral for flux just becomes this.

You will see it doesn't change things much when it comes to computing a line integral.

So, if I have a gradient field, and I do the line integral, then I will get f at the endpoint minus f at the starting point.

OK, so there's the line integral.

And so it will be a surface integral, not a line integral anymore.

If you are in the plane you have two components, and then this becomes the line integral of M dx plus N dy.

So, now I claim this is the same thing as a second property, which is that the line integral of F is path independent.

This side here is a usual line integral.

And this we can compute using the definition of the line integral.

If there are no other questions then I guess we will need to figure out how to compute this guy and how to actually do this line integral.

And so that's why you have this line integral that makes perfect sense, but you can't apply Green's theorem to it.

We have a line integral along a curve.

So that will be the line integral of Pdx plus Qdy.

And then I add these together. That is what the line integral means.

That is a similar way to do it because it is, again, a line integral so it should work the same way.

So, what it does actually is it computes a line integral.

And so I want to compute for the line integral along that curve.

Well, the line integral along C1 minus C2, well, C1-C2 let's just form a closed curve that is C1 minus C2.

Why is that easy? Well, that is because Mdx Ndy when we do the line integral it becomes M dx plus N dy.

And then, you'd still get, if a curl is zero 0 then this line integral is zero. That one is also zero.

So, you should remember, what is this line integral, and what's the divergence of a field?

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