All right, so let us now come to multidimensional integrals, and before we start with that, let us
review how integrals were introduced in one dimension.
So, for example, consider this function here in one dimension.
It's a function F of X and X is a scalar and F as a scalar.
Then if we want to calculate the area under the curve with respect to this x axis here, then this area
is a sum of rectangles, or at least roughly it is the sum of rectangles.
So you can see there is some error.
For example, we do not account for this wide region here and we overestimate this region here, but
it's kind of an approximate value.
And what we have to do is we have to add up these rectangles here where Delta X is the length of this
red line and F of X one, for example, is the value of the function at this position.
Now we can make our approximation better if we make these rectangles smaller, smaller or narrower.
So this brings about more rectangles.
So you can see this time we have 12 rectangles and they're smaller.
And you can also see that the arrow decreases and still the area is a sum of rectangles.
But this time it would be 12 terms instead of six.
Now, of course, we can make this better and better if we take the limit of Delta X, that goes to
zero so that we add up an infinite number of rectangles and what we end up with is our integral, where
we take the function, we integrate it, and we use the boundaries from A to B, which in our case is
from here to here.
Now, what you also have to consider is that functions can have positive and negative values, and when
you integrate, you do not simply add up these three areas, but you also have to account for the sign.
For example, here the curve is below the x axis.
So this area is accounted for with a negative sign.
Now, one thing to also realize is what is dysfunction, capital F. So when we have our function F of
X. So does is here a function F effects, then we integrates this function to get this function capital
F of X, and also we get an integration constant and vice versa.
We get when we differentiate this function, we go back to our function small F of X.
So quite often when you want to solve integrals like this, you have to think about which function do
I have to consider?
So that it's integral is the initial function.
So we can take an example here.
We have this polynomial here X to the power of three to one zero and now we integrate.
And what we get is we have to increase the power of X by one and we have to divide by this new number
here, because then we take our one over eight X to the depart for if we differentiate it, we get four
times four times one over eight times X to the power of four minus one, which is exactly this one here.
So if you know about differentiation, quite often it's very easy to calculate integrals, but generally
speaking, it's a bit more difficult to calculate integrals than derivatives.
And in our case, when we want to calculate the area from some starting value to some end value, then
we have to consider the boundaries and get this value here.
All right, so this is integration in one dimension and in multiple dimensions, it's not really much
more difficult.
So, for example, here, let's consider an example in two dimensions.
So we have a function F that is a function of two arguments.
So we have X and Y in our function.
And if we want to calculate the volume under this curve in a certain region, we have to integrate twice
and we have to integrate with respect to both variables, X and Y.
So in this example here, we have the function just polynomial.
And what the integral means in this case is that we calculate the area or better to say, the volume
of this whole gray shape here.
So that is the volume under this two dimensional shape here in red.
And the volume is then, of course, three dimensional.
So if the function F would be constant, then it would just be a cube.
For example, if we choose appropriate boundary boundary conditions and then it would be very easy to
calculate.
But even in this case, it's quite easy.
So we just have to take the function of this of this blue curve here and then integrate it twice with
respect to X and Y and then use the boundaries so that the boundary accounts for this red shape here.
So in our case, this means we have to take our function, integrate twice and use these boundaries
here.
So what happens again, since we have a polynomial here?
First we integrate with with respect to X, so we have to increase the power of all of the axis.
So, for example, here we get extra power for here we get X, because here we did not have any X,
he would get X squared instead of X, and here we get X instead of nothing.
Then we have to use to boundaries, so we plug in to in all of these terms and then subtract all of
these terms with X equal to minus two point five, which is the other boundary in the X direction.
And this gives us this one here.
And now we are left with a one dimensional problem and we have to integrate once more.
So let's do it here.
We get in y here we get Y Square and here we get Y to the power of three.
And once again, we used to boundaries and get a value for our volume.
So I think that's really straightforward and it's not at all more complicated than a one dimensional
integral.
So if you know, integration one dimension, you also note for these examples.
Now, sometimes in multiple dimensions, um, things become a bit more tricky because here you can see
we have chosen a rectangular shape over which we integrate.
So here it's important to realize that our function is two dimensional.
So in terms of its arguments, it has two arguments and the value itself provides the third dimension
so that everything spends a volume.
Now, the problem here is that in reality, most of our functions are three dimensional, so they depend
on X, Y and Z.
So here the problem arises.
What happens if we integrate over a three dimensional function?
How could we imagine that?
Because here for two dimensional function, our integral in the end was a volume.
If we integrate over a three dimensional function, we get a four dimensional result, so to say.
So that is very hard to imagine.
So from now on, I want you to think of integrals in a bit of a different way, which makes it a bit
easier.
So instead of thinking about this integral here as being the volume under this curve, you could also
imagine it to be just this shape, just the area.
And then the value of the function is not anymore this height here.
But you could imagine it like some density, for example.
So if the function has a large value, there is a high density and if the function has a low value,
there is a low density.
So what you're then calculating is the mass, for example.
And in one dimension this works as well.
So if we have a one dimensional curve, as you typically do have in school, then you calculate integrals.
You could just imagine it as being just this line and then your you're well, you have to function is
not the height of this line, but it is the density.
For example, imagine a wire and then the density of this wire.
So the mass of a very short element is then indicated by the value of the function.
So then in one dimension, you can imagine it as a one dimensional problem in two dimensions.
You can imagine it as a two dimensional problem instead of having here this three dimensional object.
And this allows us finally to calculate also three dimensional problems on the next slide.
I will show you exactly what I mean with that.
But he has one more comment in one dimensions.
We can just integrate over a line from A to B, but already in two dimensions we have more possibilities.
So here we consider the most simple case.
We consider a rectangle from X1 to X to and from Y one to Y two.
But in principle, you could also ask the question, what is the the integral if you integrate not over
this rectangular shape here, but for example, over this triangle here, or you could even make it
more difficult, you could take a circle or just some random shape.
So what would happen then is you would not integrate from constant values, from numbers number one
to number B and number three to number four.
But instead, your boundaries of these integrations would then depend on X and Y, so you would have
to parameterize your boundaries or better to say you would have to parametrized your shape and then
put this into the boundaries.
So here on the next slide, I want to show you how this works in three dimensions.
So first of all, our space is now three dimensional.
So we have X, Y and Z, X, Y and then the Z components here.
So now we cannot anymore think of integrals as being in the middle of the area or the volume under some
curve, because this will lead us here into the fourth dimension.
So instead, I want you to think that the value of our function is something like the density or like
the mass of a very small volume element of this shape here.
And also you can see here is a pyramid.
So this is not any more something rectangular.
This is not a cube, for example.
It's what happens here is that our integration boundary in the X and in the Y direction changes when
we go to different values of Z, for example, here for the equals zero, which is the square here,
we get quite a large square.
But then the other extreme case for the for the highest point here are integration and X and Y, it
just condenses to a single point.
So before we make things too difficult, let's take the most easy example of this problem in three dimensions
where our function itself is just one, or it could be any other constant.
So this would mean we would have to just calculate the mass of this object here with a constant density,
or in other words, we pull this value off the function out of the integral and just calculate the volume.
So what we integrate here is the volume where we integrate over the X divides and we have these boundaries
that you can see here and here.
I have already introduced what I have explained earlier.
So the boundaries for the X and for the Y integration are not just some numbers anymore, but they depend
on the on the other coordinate Z, because you can see that the area over which we integrate shrinks
down as we go to a larger Z values.
So what I have done here, I have parametrized, the pyramid.
So quite often this is the most difficult step of the whole integration to find the correct parameterization
of your shape.
So here I want to show you how I have done it.
So if we take a cut of this pyramid along this triangle here, so you can see, then it looks like this,
then we have here here I put my zero coordinate and then we have here into the negative X direction
the value of minus eight over to here we have over two.
So that this length here's a and so that this area here at the bottom is A squared and the height of
our pyramid is H.
And now what we want to have is we want to parametrized this function here, which is a linear function.
And if you post the video and do it yourself and determine the function for this linear function here,
you will get this one.
So what is important here to check is that if you use Z equal to zero, which is this one here, this
line, then our value for the boundary must reach from minus Aoba two to over two.
So if we put Z to zero, we get minus Aoba two and here we get plus A over two.
And if we put Z equal to H just as one here, then both of the boundaries have to collapse to zero.
So here we put a Z equal to aged and we get one.
So we get to zero and here we get also zero.
And overall it's a linear function in Z.
So it's correct.
All right.
So this was already the most difficult step.
We have parametrized of a pyramid and now we just go straight forward and solve this integral here.
And this is now particularly easy because what we have to integrate over is the function one.
So once the integral of one when reintegrates goes back to X, it's of course X because the derivative
of X is one.
So we have X with the boundaries.
X is equal to this one and this one.
Now, of course, we just take the distance or the difference of these two values, which is two times
this value, which is eight times in brackets, one minus the of H.
Now, you can see that this value here is indeed c dependent, but it is not my dependent.
So this means we can pull this out of the integral right at here.
And so, once again, we have to integrate with respect to y over the function one.
So we do the whole thing once again, but instead we have no idea why, why, why.
But it doesn't matter.
It will once again give us to value eight times in brackets, one minus C of H.
So this means that when we calculate this whole integral here, we get this expression here squared.
So this means we have a square.
And then here the binomial Formula One square minus two times, one times Z of H.
Plus, the square of a square just as one.
OK, now we have the final step.
We have to integrate this function here with respect to Z.
So that's yeah, that's right.
I did like this and let's integrate over Z.
So we just have to increase the power of all of these these here.
So we get at term proportional to Z to the power of one to three, and we have to account for the three
factors.
So please check that this is really correct.
And then we have to use the boundaries from zero to age.
So we integrate, set Z from zero to H, and what we get is, yeah, for Z equal to zero we get zero.
And for the equal H we get a square times H minus H plus one third a square H.
So you see these two cancel out and we have our result that the volume of a pyramid is one of a three
a square times H.
And this is a formula that you learned very, very early in school.
But it is not at all clear where this formula comes from at this point.
So now, you know, now you know about multidimensional integrals and now you can calculate the volume
of a pyramid.
That's quite good.
So earlier I told you that I I want you to think of this function as being something like a density.
So what would happen if we would have a more complicated function where F would not be equal to one,
but we have some X, Y and Z dependent's then in terms of physics, you know that the mass is equal
to the density times, the volume.
And if the density is a constant throughout your whole whole shape, then you can write it down like
this.
But if your density is X, Y, Z dependent, then you can see it becomes exactly what we had before
where this one is now a function.
So you can think of calculating the mass of this object where you have some some changing density inside
of your object.
So that's a way to think about it.