OK, so now you know how to calculate three dimensional integrals in the three dimensional space.
However, since we are now in multiple dimensions, we can also calculate different types of integrals.
For example, we can calculate one dimensional integrals along some path.
So this is a quasi one dimensional problem.
We are still in the three dimensional space, but we only consider a certain line and for simplicity
so that you understand what is happening here.
I want to show you an example of a two dimensional function where we consider a path.
And I found this nice animation on Wikipedia that I want to explain to you on the following slide.
So here you go.
This is a contour plots, as we had previously.
So here the value of the function F is colored by these different colors here.
And now we have a path see from point A to point B, so you can see that this path here is along this
red line and that the blue line indicates the value of the function f with respect to the zero line.
And I can see here the path is projected into one dimensions.
And so what you have to do is you have to integrate over this blue function with respect to the red
zero line.
So let's look at it one more time.
So here again, you can see the colors and the path and of course, the color is just an indication
for the value of the function.
So you could plot it in such a three dimensional plot where the value of the function is the third dimension.
Now here the important thing happens.
Now the path will be projected into one dimension and now it's just a one dimensional problem.
And when we integrate the integral here, we integrate over F, but we also have to account for this
additional term that comes from projecting the path into one dimension.
So this is what we get for the equation.
This was the same as in the animation that you have just seen.
So it's pretty, pretty basic.
We integrate over the function.
But here this is new.
We have the absolute value of the derivative of our path.
So this may look a bit, uh, yeah.
But, uh, theoretical.
Let's let's look at the detail so that you will understand better what's happening here.
OK.
So here in blue, we have our function.
This is a function that we had already considered previously and.
Now, you introduce a pass and here I've just chosen this Redpath here path, see, and our path is
parametrized by a parameter T. So it's a two dimensional quantity.
So we have X and Y component and you change T..
I have chosen to boundaries for this path to be minus pi over two and PI over two.
And the X component is just T and the Y component is a sign of T.
Now, of course, you may say that this path can be parametrized also in a different way, and that
is absolutely fine.
For example, you could choose a different parameter to not that goes from zero to one, and then you
would have to multiply here a constant and here a constant.
And that would be exactly the same path.
And it would also be correct.
But let's go with what I've chosen here.
So this is our path are of Tea Party is this parameter that has these two boundaries that will later
be A and B, and now we calculate the derivative of this path are with respect to tea.
So we get for the X component one and for the Y component, we get cosine of T.
I guess that's pretty, pretty easy.
And now the absolute value of this is of course the square root of X squared plus Y square.
So we get the absolute value is square root of one plus cosine square of tea.
And no, we are almost done.
The only thing we have to be careful about is that here we do not have any more of our I mean, we still
have half of our pots.
We have our of tea.
So we have to introduce our path here in terms of the argument of this function.
So this is because we later will integrate over the parameter T because it is a quasi one dimensional
problem.
This is also why you have here only one integral symbol and not to even though our function has two
arguments, X and Y.
OK, so here our function is one we had previously already.
It's this polynomial where we have this additional term that makes us X and Y.
And now I introduced a path hour of tea.
So that means our X component will be T and our Y component will be a sign of T.
So this means F of R of T is minus three T to the part three.
And then here we have four times sine square left and here we have four X Y, we get tea times sign
of T.
So this is our parametrized function and now we can calculate this line integral here with the boundaries
minus pi a two and plus pi over two of this function.
And now this is something you just have to calculate.
So to be honest, this is very, very difficult.
I think it's not not even possible to integrate analytically.
So here I want to show you a cool trick that I use quite often that you can use as well.
So if you have an integral, you can go to the website that is called Wolfram Alpha.
This is using the language of Mathematica, which is a program that I'm using quite often.
And there you you do not really have to know the correct syntax.
Even you can just type what you want to do.
And most of the time, Wolfram Alpha will be able to do what you want them to do.
So you just write integrate them in these brackets.
I wrote the function and then the boundaries.
I want to integrate from minus pi over to two pi over to.
I didn't even write.
What's the what's the variable here or if I just understood what I want to do.
That's good.
So here's what do we get as the output.
That's exactly what we want to calculate and the result is four hundred and six.
So yeah, it doesn't really matter what's the result.
The important thing is that you understand that it is possible to integrate over a path and that's why
we want to do this.
We have to introduce the path in the argument of the function and we have to account here for this additional
factor, which comes from projecting this path onto a straight line.
So that really matters if our path is straight or not.
So, of course, if you would have a straight path, then this one here would become one.
OK, so this was the line integral for a scalar function F and we can also calculate the line integrals
for vector functions in two dimensions.
And so here the calculation is quite similarly.
But what happens here?
I want to again show you at the example of this nice animation that I found on Wikipedia.
So here we have a path and our function is now a vector field.
It is indicated by these blue arrows.
So this means at each position we do not have one single value, but we have to.
And a way to visualize this is to use such a vector plot.
So now let's look at this animation from the beginning for restart soon.
So we have here our function capital F, which is just two dimensional vector, and we have once again
our path C from A to B, and now we calculate the DOT product of D-R and.
F, which is the product of these two arrows that you can also see here, so they are almost parallel.
So it's quite a large value, as you can see here.
And now if we go along the line, as was just done in this animation, you will see, for example,
at some point our our arrow for D.R will be pointing in the opposite direction of the value of the function.
This is where we get these negative values that you have just seen.
So let's look at it one more time.
This time, let's let's consider this red arrow here.
You can see it always follows along the line.
And for example, at this point here, it will be pointing along the opposite direction to this blue
arrow here just now.
And this is where we get this negative value.
So once again, we have a quasi one dimensional problem over which we have to integrate.
And, of course, we know how this is done.
And so we end up with this formula here.
So when you compare this scalar function of the vector function, it is quite similarly.
But here you have to take the absolute value and you have to dot product.
So in the end you will get a scalar result.
It would just get a number, of course.
So I think, yeah, since we had this example for the scalar function, it's OK to not have an example
for the vector function because in the end it's just an exercise in math.
What do you have?
The only thing that's a bit new from a conceptual side is that you have to introduce a path and then
you have to parameterize.
And this is exactly the same as in this case.
But here you just take the vector of these derivatives and calculate the DOT product.
OK, so this is basically all I want to tell you about these line integrals, the only final comment
that I have is that quite often, especially in physics, we will integrate over a closed path.
Where are the boundaries?
A and B will be the same.
So whenever this is the case, you typically indicate this is such an integral sign with a circle.
So this means that you integrate of a closed path.
But of course, even though the starting and the ending position will be the same, the integral will
most of the time still be different from zero.
And also the value of the integral will strongly depend on the path itself.
So this is quite different from a purely one dimensional problem where the value of the integral is
just determined by the boundaries and b in multiple dimensions.
When you consider such a line integral, which is a quasi one dimensional problem, the result strongly
depends on the path.