![]() Out the entire volume- what we did just now is we figured Of these columns is 2, and then its with and It'll be a function of x and y,īut since we're dealing with all constants here, it'sĪctually going to be a constant value. So this integral, this value,Īs I've written it, will figure out the volume of aĬolumn given any x and y. That we're going to take the integral with And what's the upper bound? It's z is equal to 2. Keep adding these cubes, and keep going up, you'd run And what's the upper bound? Like if you were to just take. Lowest z value? Well, it's z is equal to 0. So what would that look like? Well, since we're going up andĭown, we're adding- we're taking the sum in Then add it along the up and down axis- the z-axis. So, what we could do is weĬould take this cube and first, add it up in, let's Help us take infinite sums of infinitely small distances, But anyway, what can youĭo with it in here? Well, we can take the integral. Orders of these, right? Because multiplication isĪssociative, and order doesn't matter and all that. You could say, it's just the width times the Volume within this larger volume- you could call thatĭv, which is kind of the volume differential. Of that cube? Let's say that its width is dy. It's some place in this largerĬube, this larger rectangle, cubic rectangle, whatever Variable boundaries and surfaces and curvesįigure out the volume of this little, small cube here. Sense, or it starts to become a lot more useful, when we have Integral mean? Well, what we could do is weĬould take the volume of a very small- I don't want to sayĪrea- of a very small volume. You could say it's 24Ĭubic units, whatever units we're doing. Is 3, the base, the width is 4, so this area is 12 ![]() You have a boundary here,Ĭome in like that. Here, and then it would come in like this. In the next video we could do something slightly Like, how it relates to a double integral, and then later Just so that you get used to what a triple integral looks Width times the height times the depth and you'd Geometry you could figure out- you know, just multiply the Greater than or equal to 0 and is less than or equal to 2. Or equal to 0, and is less than or equal to 4. X is between- x is greater than or equal to 0, is less Volume of a cube, where the values of the cube- let's say Now, you'll notice, in this working, we weren't thinking about any areas under curves because that would mean we had to work in 4D in our heads and that's a nuisance so, instead we just thought of integrating as adding up a load of infinitesimals. If we then integrate this over the entire shape R, we obtain the mass M. Density is measured in kgm^-3 so, in order to get it into a mass (kg) we're going to need to multiply it by something with units of m^3. The density p is a good start because it's the only thing which has a unit containing a mass term. Now, the integral of dM over the entire shape should be M fairly intuitively so we need to find a way to express dM in terms of other things. If we have some arbitrary shape R (for a Region of space) with some arbitrary density function p and we want to find the mass M of the entire thing. With a triple integral, the answer is possibly a little harder to interpret so I'll work in terms of density because it's a good example. Integrating these across some area (integrating zdA) adds up all the z's and gives us the volume we wanted. If we go back to a double integral, you remember how we integrated zdA (or zdxdy) to get the volume? Instead of thinking about z as a height above the xy plane, let's instead work in the 2D xy world where we've labelled each point in our world with a value, z. This intuition doesn't help us much when we want to do triple integrals because in this case we're actually finding the 4D-hypervolume under the curve.Ī more useful intuition in this case is the one of adding up all the values. In 2D space, this comes out nicely as the area under the curve and, if we move to three dimensions, by integrating zdA we can find the volume under a curve. The reason you can't see an area under the curve for the first integral is that, in a certain sense, there isn't one.Īn integral (of ydx for the sake of argument) is not so much about finding the area under y, but rather is about adding up all of the values of y along the line x with some form of weighting to make sure it converges. It's an infinitesimal volume element (dV, just like we could use dA in double integrals) so, in order to get the total volume, we need to add up, by integrating, the volume elements dV in each direction in turn. The element dxdydz in not one-dimensional.
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