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# Triple integrals 1

## Video transcript

let's say I wanted to find the volume of a cube where the values of the cube let's say X is between X is greater than or equal to zero is less than or equal to I don't know three let's say Y is greater than or equal to zero and is less than or equal to four and then let's say that Z is greater than or equal to zero n is less than or equal to two now I know using basic geometry you could figure out you know just met just multiply the width times the height times the depth then you'd have the volume but I want to do this example just so that you get used to what a triple integral looks like how it relates to a double integral and then later in the next video we could do something slightly more complicated so let's just draw that this volume so this is my x-axis so my z axis this is the y X Y Z X Y Z okay so X is between zero and three so that's X is equal to 0 this is X is equal to say 1 2 3 y is between 0 & 4 1 2 3 4 so in the XY plane it looks something like this the kind of base of our cube will look something like this and then Z is between 0 to 2 so 0 is the xy-plane and then 1/2 so this would be the top part and I'll maybe I'll do that in a slightly different color so this is along the x z axis you'd have a boundary here and then it would come in like this if a boundary here come in like that boundary there then like that so we want to figure out the volume of this cube and you could do it you could say well this is the the depth is 3 the base the width is 4 so this area is 12 times the height 12 times 2 is 24 you could say it's 20 for the cubic units whatever units were doing but let's do it as as a triple integral so what is a triple integral mean what we can do is we could say we can take the the smart the volume of a very small I don't want to say area of a very small volume so let's say I wanted to take the volume of a small cube someplace in this the volume under question and it'll start to make more sense or it starts to become a lot more useful when we have variable boundaries and surfaces and curves as boundaries but let's say we want to figure out the volume of this little small cube here that's my cube it's someplace in this larger cube this larger rectangle cubic rectangle whatever you want to call it so what's the volume of that cube let's say that it's its width is dy its width is dy so that length right there is dy its height is DX all right now its height is DZ right the way I drew it Z is up up and down and then its depth is DX this is DX so this is DX this is DZ this is dy so you could say that the small volume when this larger volume you can call that DV which is kind of the volume differential and that would be equal to you could say it's just the width times the length times the height dy a DX times dy times DZ and you could switch the orders of these right because multiplication is associative and order doesn't matter and all of that but anyway what can you do with it in here well we can we can take the integral we get integrals all integrals help us do is it helps us take infinite sums of infinitely small distances like a DZ or DX or dy is let cetera so what we can do is we could take this cube and first add it up and let's say the Z direction so we can take that cube and then add it and then add it along up the up and down axis the z axis so we get the volume of a column so what would that look like well since we're going up and down we're adding we're taking the sum in the z direction we'd have an integral and then what's art what's the lowest Z value well it's Z is equal to zero Z equal to zero and let's say the upper bet like if you were to just take keep adding these cubes and keep going up you would run into the upper bound and what's the upper bound its Z is equal to two Z is equal to two and of course you would take the sum of these DVS I'll write DZ first just so it reminds us that we're going to take the integral with respect to Z first and let's say well do Y next and then we'll do X so this integral this value as I have written it we'll figure out the volume of a column given any X and y it'll be a function of x and y but since we're dealing with all constant here it's actually going to add be a constant value it'll be it'll be the VAT the constant value of the volume of one of these columns and it'll essentially be will be two times dy/dx because the height of one of these columns is 2 and then its width and its depth is dy and DX so then if we want to figure out the entire volume so what we did just now is we figured out the height of a column so then we could take those columns and sum them in the y-direction right so for summing in the y-direction we could just take another integral of this sum in the Y direction and Y goes from zero to what Y goes from zero to four I wrote this integral a little bit too far to the left looks strange but I think you get the idea Y is equal to 4 y y is equal to 0 to Y is equal 4 and then that'll give us the volume of kind of a sheet that is parallel to the zy plane and then all we have to left left to do is add up a bunch of those sheets in the X direction we'll have the volume of our entire figure so to add up those sheets we would have to sum in the X direction and we'd go from X is equal to 0 to X is equal to 3 and to evaluate this it's actually fairly straightforward so first we're taking the integral with respect to Z well we don't have anything written in under here but we can just assume that there's a 1 right because DZ times dy times DX is the same thing as 1 times DZ times dy DX so what's the value of this integral well the antiderivative of one it with respect to Z is just Z right because the derivative of Z is 1 and you evaluate that from 2 to 0 so then you're left with so it's 2 minus 0 so you just left with 2 so you're left with 2 and you take the integral of that from Y is equal to 0 to Y is equal to 4 dy and then you have X from X is equal to 0 to X is equal to 3 DX and notice when we just took the integral with respect to Z we ended up with a double integral and this double integral is the exact integral we would have done in the previous videos on the double integral where you would have just said well Z is a function of x and y so you could have written you know Z is a function of x and Y is always equal to 2 it's a constant function it's independent of x and y but if you had defined Z in this way and you wanted to figure out the volume under this surface where the surface is Z is equal to 2 you know this is the surface Z is equal to 2 we would have ended up with this so you see that what we're doing with the triple integral it's really really nothing different and you might be wondering well why are we doing it at all and I'll show you that in in a second anyway to evaluate this you could you know take the antiderivative of this with respect to Y you get 2y let me scroll down a little bit you get 2y evaluating that at 4 and 0 and then so you get 2 times 4 so it's 8 minus 0 and then you integrate that from with respect to X from 0 to 3 so that's 8 X from 0 to 3 so that'll be equal to 24 units cute so I know the obvious question is what is this good for well when you have constant when you have a kind of a constant value within the volume you're right you could have just done a double integral but what if I were told you our goal is not to figure out the volume of this figure our goal is to figure out the mass of this figure and even more this this this this volume this area of space or whatever its mass is not uniform if its mass was uniform you could just you know multiply its its uniform density times its volume and you would get its mass but let's say the density changes it could be a volume of some gas or it could be even you know some material with different compounds in it so let's say that its density is a variable function of XY and Z so let's say that the density this Rho this thing that looks like a P is what you normally use in physics for density so its density as a function of X Y & Z let's just to make it simple let's make it X times y times Z so if we wanted to figure out if we wanted to figure out the mass of any small volume it would be that volume times the density right because density the units of density are like you know kilograms per meter cubed so if you multiply it times meter cubed you get kilograms so we could say that the mass I'll make up notation D mass this isn't a function well I don't write a parenthesis it makes look a function so the a very a differential of mass or a very small mass is going to equal the density at that point which would be XYZ times the volume of that of that small mass and that volume of that small mass we could write as DV and we know that DV is the same thing as the width times the height times the depth DV doesn't always have to be DX times dy times DZ if we're doing other coordinates if we're doing polar coordinates it could be something slightly different and we'll do that eventually but if we want to figure out the mass since we're using rectangular coordinates it would be the density function at that point times that our differential volume so x DX dy DZ and of course we can change the order here so when you want to figure out the volume so we're going to figure out the mass which I will do in the next video we essentially will have to integrate this function as opposed to just 1 over Z Y and X and I'm going to do that in the next video and you'll see that it's really just a lot of basic taking anti derivatives and avoiding careless mistakes I will see you in the next video