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All right, so let us now come to multidimensional integrals, and before we start with that, let us



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review how integrals were introduced in one dimension.



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So, for example, consider this function here in one dimension.



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It's a function F of X and X is a scalar and F as a scalar.



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Then if we want to calculate the area under the curve with respect to this x axis here, then this area



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is a sum of rectangles, or at least roughly it is the sum of rectangles.



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So you can see there is some error.



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For example, we do not account for this wide region here and we overestimate this region here, but



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it's kind of an approximate value.



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And what we have to do is we have to add up these rectangles here where Delta X is the length of this



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red line and F of X one, for example, is the value of the function at this position.



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Now we can make our approximation better if we make these rectangles smaller, smaller or narrower.



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So this brings about more rectangles.



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So you can see this time we have 12 rectangles and they're smaller.



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And you can also see that the arrow decreases and still the area is a sum of rectangles.



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But this time it would be 12 terms instead of six.



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Now, of course, we can make this better and better if we take the limit of Delta X, that goes to



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zero so that we add up an infinite number of rectangles and what we end up with is our integral, where



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we take the function, we integrate it, and we use the boundaries from A to B, which in our case is



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from here to here.



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Now, what you also have to consider is that functions can have positive and negative values, and when



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you integrate, you do not simply add up these three areas, but you also have to account for the sign.



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For example, here the curve is below the x axis.



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So this area is accounted for with a negative sign.



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Now, one thing to also realize is what is dysfunction, capital F. So when we have our function F of



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X. So does is here a function F effects, then we integrates this function to get this function capital



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F of X, and also we get an integration constant and vice versa.



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We get when we differentiate this function, we go back to our function small F of X.



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So quite often when you want to solve integrals like this, you have to think about which function do



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I have to consider?



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So that it's integral is the initial function.



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So we can take an example here.



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We have this polynomial here X to the power of three to one zero and now we integrate.



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And what we get is we have to increase the power of X by one and we have to divide by this new number



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here, because then we take our one over eight X to the depart for if we differentiate it, we get four



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times four times one over eight times X to the power of four minus one, which is exactly this one here.



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So if you know about differentiation, quite often it's very easy to calculate integrals, but generally



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speaking, it's a bit more difficult to calculate integrals than derivatives.



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And in our case, when we want to calculate the area from some starting value to some end value, then



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we have to consider the boundaries and get this value here.



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All right, so this is integration in one dimension and in multiple dimensions, it's not really much



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more difficult.



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So, for example, here, let's consider an example in two dimensions.



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So we have a function F that is a function of two arguments.



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So we have X and Y in our function.



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And if we want to calculate the volume under this curve in a certain region, we have to integrate twice



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and we have to integrate with respect to both variables, X and Y.



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So in this example here, we have the function just polynomial.



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And what the integral means in this case is that we calculate the area or better to say, the volume



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of this whole gray shape here.



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So that is the volume under this two dimensional shape here in red.



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And the volume is then, of course, three dimensional.



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So if the function F would be constant, then it would just be a cube.



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For example, if we choose appropriate boundary boundary conditions and then it would be very easy to



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calculate.



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But even in this case, it's quite easy.



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So we just have to take the function of this of this blue curve here and then integrate it twice with



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respect to X and Y and then use the boundaries so that the boundary accounts for this red shape here.



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So in our case, this means we have to take our function, integrate twice and use these boundaries



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here.



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So what happens again, since we have a polynomial here?



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First we integrate with with respect to X, so we have to increase the power of all of the axis.



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So, for example, here we get extra power for here we get X, because here we did not have any X,



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he would get X squared instead of X, and here we get X instead of nothing.



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Then we have to use to boundaries, so we plug in to in all of these terms and then subtract all of



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these terms with X equal to minus two point five, which is the other boundary in the X direction.



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And this gives us this one here.



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And now we are left with a one dimensional problem and we have to integrate once more.



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So let's do it here.



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We get in y here we get Y Square and here we get Y to the power of three.



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And once again, we used to boundaries and get a value for our volume.



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So I think that's really straightforward and it's not at all more complicated than a one dimensional



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integral.



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So if you know, integration one dimension, you also note for these examples.



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Now, sometimes in multiple dimensions, um, things become a bit more tricky because here you can see



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we have chosen a rectangular shape over which we integrate.



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So here it's important to realize that our function is two dimensional.



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So in terms of its arguments, it has two arguments and the value itself provides the third dimension



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so that everything spends a volume.



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Now, the problem here is that in reality, most of our functions are three dimensional, so they depend



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on X, Y and Z.



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So here the problem arises.



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What happens if we integrate over a three dimensional function?



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How could we imagine that?



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Because here for two dimensional function, our integral in the end was a volume.



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If we integrate over a three dimensional function, we get a four dimensional result, so to say.



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So that is very hard to imagine.



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So from now on, I want you to think of integrals in a bit of a different way, which makes it a bit



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easier.



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So instead of thinking about this integral here as being the volume under this curve, you could also



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imagine it to be just this shape, just the area.



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And then the value of the function is not anymore this height here.



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But you could imagine it like some density, for example.



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So if the function has a large value, there is a high density and if the function has a low value,



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there is a low density.



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So what you're then calculating is the mass, for example.



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And in one dimension this works as well.



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So if we have a one dimensional curve, as you typically do have in school, then you calculate integrals.



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You could just imagine it as being just this line and then your you're well, you have to function is



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not the height of this line, but it is the density.



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For example, imagine a wire and then the density of this wire.



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So the mass of a very short element is then indicated by the value of the function.



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So then in one dimension, you can imagine it as a one dimensional problem in two dimensions.



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You can imagine it as a two dimensional problem instead of having here this three dimensional object.



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And this allows us finally to calculate also three dimensional problems on the next slide.



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I will show you exactly what I mean with that.



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But he has one more comment in one dimensions.



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We can just integrate over a line from A to B, but already in two dimensions we have more possibilities.



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So here we consider the most simple case.



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We consider a rectangle from X1 to X to and from Y one to Y two.



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But in principle, you could also ask the question, what is the the integral if you integrate not over



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this rectangular shape here, but for example, over this triangle here, or you could even make it



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more difficult, you could take a circle or just some random shape.



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So what would happen then is you would not integrate from constant values, from numbers number one



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to number B and number three to number four.



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But instead, your boundaries of these integrations would then depend on X and Y, so you would have



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to parameterize your boundaries or better to say you would have to parametrized your shape and then



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put this into the boundaries.



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So here on the next slide, I want to show you how this works in three dimensions.



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So first of all, our space is now three dimensional.



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So we have X, Y and Z, X, Y and then the Z components here.



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So now we cannot anymore think of integrals as being in the middle of the area or the volume under some



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curve, because this will lead us here into the fourth dimension.



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So instead, I want you to think that the value of our function is something like the density or like



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the mass of a very small volume element of this shape here.



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And also you can see here is a pyramid.



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So this is not any more something rectangular.



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This is not a cube, for example.



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It's what happens here is that our integration boundary in the X and in the Y direction changes when



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we go to different values of Z, for example, here for the equals zero, which is the square here,



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we get quite a large square.



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But then the other extreme case for the for the highest point here are integration and X and Y, it



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just condenses to a single point.



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So before we make things too difficult, let's take the most easy example of this problem in three dimensions



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where our function itself is just one, or it could be any other constant.



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So this would mean we would have to just calculate the mass of this object here with a constant density,



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or in other words, we pull this value off the function out of the integral and just calculate the volume.



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So what we integrate here is the volume where we integrate over the X divides and we have these boundaries



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that you can see here and here.



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I have already introduced what I have explained earlier.



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So the boundaries for the X and for the Y integration are not just some numbers anymore, but they depend



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on the on the other coordinate Z, because you can see that the area over which we integrate shrinks



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down as we go to a larger Z values.



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So what I have done here, I have parametrized, the pyramid.



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So quite often this is the most difficult step of the whole integration to find the correct parameterization



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of your shape.



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So here I want to show you how I have done it.



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So if we take a cut of this pyramid along this triangle here, so you can see, then it looks like this,



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then we have here here I put my zero coordinate and then we have here into the negative X direction



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the value of minus eight over to here we have over two.



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So that this length here's a and so that this area here at the bottom is A squared and the height of



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our pyramid is H.



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And now what we want to have is we want to parametrized this function here, which is a linear function.



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And if you post the video and do it yourself and determine the function for this linear function here,



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you will get this one.



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So what is important here to check is that if you use Z equal to zero, which is this one here, this



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line, then our value for the boundary must reach from minus Aoba two to over two.



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So if we put Z to zero, we get minus Aoba two and here we get plus A over two.



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And if we put Z equal to H just as one here, then both of the boundaries have to collapse to zero.



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So here we put a Z equal to aged and we get one.



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So we get to zero and here we get also zero.



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And overall it's a linear function in Z.



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So it's correct.



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All right.



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So this was already the most difficult step.



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We have parametrized of a pyramid and now we just go straight forward and solve this integral here.



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And this is now particularly easy because what we have to integrate over is the function one.



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So once the integral of one when reintegrates goes back to X, it's of course X because the derivative



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of X is one.



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So we have X with the boundaries.



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X is equal to this one and this one.



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Now, of course, we just take the distance or the difference of these two values, which is two times



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this value, which is eight times in brackets, one minus the of H.



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Now, you can see that this value here is indeed c dependent, but it is not my dependent.



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So this means we can pull this out of the integral right at here.



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And so, once again, we have to integrate with respect to y over the function one.



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So we do the whole thing once again, but instead we have no idea why, why, why.



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But it doesn't matter.



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It will once again give us to value eight times in brackets, one minus C of H.



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So this means that when we calculate this whole integral here, we get this expression here squared.



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So this means we have a square.



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And then here the binomial Formula One square minus two times, one times Z of H.



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Plus, the square of a square just as one.



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OK, now we have the final step.



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We have to integrate this function here with respect to Z.



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So that's yeah, that's right.



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I did like this and let's integrate over Z.



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So we just have to increase the power of all of these these here.



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So we get at term proportional to Z to the power of one to three, and we have to account for the three



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factors.



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So please check that this is really correct.



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And then we have to use the boundaries from zero to age.



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So we integrate, set Z from zero to H, and what we get is, yeah, for Z equal to zero we get zero.



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And for the equal H we get a square times H minus H plus one third a square H.



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So you see these two cancel out and we have our result that the volume of a pyramid is one of a three



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a square times H.



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And this is a formula that you learned very, very early in school.



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But it is not at all clear where this formula comes from at this point.



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So now, you know, now you know about multidimensional integrals and now you can calculate the volume



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of a pyramid.



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That's quite good.



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So earlier I told you that I I want you to think of this function as being something like a density.



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So what would happen if we would have a more complicated function where F would not be equal to one,



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but we have some X, Y and Z dependent's then in terms of physics, you know that the mass is equal



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to the density times, the volume.



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And if the density is a constant throughout your whole whole shape, then you can write it down like



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this.



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But if your density is X, Y, Z dependent, then you can see it becomes exactly what we had before



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where this one is now a function.



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So you can think of calculating the mass of this object where you have some some changing density inside



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of your object.



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So that's a way to think about it.