Triple Integrals 1 Introduction to the triple integral
Triple Integrals 1
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- 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 0, is less than or equal to,
- I don't know, 3.
- Let's say y is greater than or equal to 0, and is
- less than or equal to 4.
- And then let's say that z is greater than or equal to 0 and
- is less than or equal to 2.
- And I know, using basic geometry you could figure out--
- you know, just multiply the width times the height times
- the depth and 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, this is my z-axis, this is the y.
- x, y, z.
- So x is between 0 and 3.
- So that's x is equal to 0.
- This is x is equal to-- let's see, 1, 2, 3.
- y is between 0 and 4.
- 1, 2, 3, 4.
- So the x-y plane will look something like this.
- The kind of base of our cube will look something like this.
- And then z is between 0 and 2.
- So 0 is the x-y plane, and then 1, 2.
- So this would be the top part.
- And 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.
- You have a boundary here, come in like that.
- A boundary there.
- So we want to figure out the volume of this cube.
- And you could do it.
- You could say, well, 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 24 cubic units, whatever
- units we're doing.
- But let's do it as a triple integral.
- So what does a triple integral mean?
- Well, what we could do is we could take 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.
- Some place in this-- in 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 some place 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 its width is dy.
- So that length right there is dy.
- It's height is dx.
- Sorry, no, it's height is dz, right?
- The way I drew it, z is up and down.
- And it's depth is dx.
- This is dx.
- This is dz.
- This is dy.
- So you can say that a small volume within this larger
- volume-- you could 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.
- 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 that.
- But anyway, what can you do with it in here?
- Well, we can take the integral.
- All integrals help us do is help us take infinite sums of
- infinitely small distances, like a dz or a dx or
- a dy, et cetera.
- So, what we could do is we could take this cube and
- first, add it up in, let's say, the z direction.
- So we could take that cube and then add it along the up and
- down axis-- the z-axis-- so that 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 the lowest z value?
- Well, it's z is equal to 0.
- And what's the upper bound?
- Like if you were to just take-- keep adding these cubes, and
- keep going up, you'd run into the upper bound.
- And what's the upper bound?
- It's z is equal to 2.
- And of course, you would take the sum of these dv's.
- And 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 we'll do y next.
- And then we'll do x.
- So this integral, this value, as I've written it, will
- 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 constants here, it's actually going to be
- a constant value.
- It'll be the constant value of the volume of one
- of these columns.
- So essentially, it'll be 2 times dy dx.
- Because the height of one of these columns is 2,
- and then its with and its depth is dy and dx.
- So then if we want to figure out the entire volume-- 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.
- So if we're summing in the y direction, we could just take
- another integral of this sum in the y direction.
- And y goes from 0 to what? y goes from 0 to 4.
- I wrote this integral a little bit too far to the
- left, it looks strange.
- But I think you get the idea.
- y is equal to 0, to y is equal to 4.
- And then that'll give us the volume of a sheet that is
- parallel to the zy plane.
- And then all we have left to do is add up a bunch of those
- sheets in the x direction, and 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 is actually fairly
- So, first we're taking the integral with respect to z.
- Well, we don't have anything written 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 1 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're 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 the 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 a surface, is 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 a second.
- But anyway, to evaluate this, you could 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 8x from 0 to 3.
- So that'll be equal to 24 four units cubed.
- So I know the obvious question is, what is this good for?
- Well, 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 to tell 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 volume-- this area of space or
- whatever-- its mass is not uniform.
- If its mass was uniform, you could just multiply its uniform
- density times its volume, and you'd get its mass.
- But let's say the density changes.
- It could be a volume of some gas or it could be even some
- material with different compounds in it.
- So let's say that its density is a variable function
- of x, y, and z.
- So let's say that the density-- this row, this thing that looks
- like a p is what you normally use in physics for density-- so
- its density is a function of x, y, and z.
- Let's-- just to make it simple-- let's make
- it x times y times z.
- 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 kilograms
- per meter cubed.
- So if you multiply it times meter cubed, you get kilograms.
- So we could say that the mass-- well, I'll make up notation, d
- mass-- this isn't a function.
- Well, I don't want to write it in parentheses, because it
- makes it look like a function.
- So, a very differential 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 wanted to figure out the mass, since we're using
- rectangular coordinates, it would be the density function
- at that point times our differential volume.
- So times dx dy dz.
- And of course, we can change the order here.
- So when you want to figure out the volume-- when you want 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
- antiderivatives and avoiding careless mistakes.
- I will see you in the next video.
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