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. OK. 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 straightforward. 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.

cuboid, but thanks for the tutorials 😀

maybe hes just really good with the mouse.

Is very confuse the explication

maybe if you understood English you would understand easier?

I think you mean "this is a very confusing explanation"

he uses a tablet… he says so on his bio.

Yes, it's true, I try to say very confusing explanation, I just believe the problem to solve is very easier and seem me the explanation was complex. Excuse me if I offend someone, I just to say my opinion. Lienciada Lujan Tringaniello, Universidad de La Plata, Argentina

this is so hellpful! thsnks!

@lujan2608 licenciada en que? se puede saber?

porque en general en las matematicas se explican las cosas con ejemplos sencillos. Independientemente de eso, no le veo lo complicado a la explicacion, es en realidad lo que uno hace cuando integra, y bastante bien lo esta explicando.

mucho mejor que decir "suma 1 al exponente y dividi por el exponente resultante" sin mostrarte nada mas

saludos

Cuando llegues a ver esos temas veras que son muy faciles en verdad y no me dedico a subir videos en la web sino a explicarles a mis alumnos de la universidad. Saludos

fak yu bich

@lujan2608

You totally missed the point. This video is an introduction to tripple integrals. He is using a problem that is easy to solve without a tripple integral to make it easier to understand how a tripple integral works. Ofcourse this is not the kind of thing you would normally use a tripple integral for.

with your help im going to ace all my assignments thank you for the vids

@naseltzer its called a graphics tablet

ola un pekeño favor traducelo españolll…

no hay un video en youtube de integrales triples (en español y bueno), maldicion!

how the hell do you dislike this?

but why do you start with a "function" 1

and not 2 because the z boundary is 2 ?

Hmm this is similar to Double Integrals plus there is a z-axis.

@Obamanation154 dont be stupid. Its clearly Math10B lololol

@ATERIFA agreed, it clearly is 10B!

@Porongaz4 mejor aprende inglés amigo, te ayudará en tu carrera montones.

Thanks a lot dude, this has been very helpful. Keep making math videos like this please!

Truly useful and professionally explained.tyvm mate B-)

How about a quadruple integral? =P

My Hero..You are the reason why i survived engineering calculus

this is not a cube

Actually, I had a math teacher that could write as well as in pen with the mouse. When I went to his place for private paid lessons, he wrote In paint instead of a blackboard.

Could you do this as a double integral where you say f(x,y)=2 ?

Commutative*

it does exist and youll come up with a 4 spatial dimensional integral (4d figure)

"Private paid lessons" lol, suuuuuure

This video was really helpful. I'm finishing my PHD in Triple Integrals this year, and this vid helped me along with my thesis.

actually its a cuboid at first

may be mouse.

I believe you meant multiplication is commutative

hi thank you very much for that

Thanx man!!real helpfull

briliant!

Great tutor

Khan rules

Why z=2y

a parallelopiped

show

thanks for the clear explanation

this would extremely help in time, using a cheap stylus to cut time on your amazing videos without using a mouse…for reasons of time, to bring to the masses faster methods of explanation the details of math…or anything else that you would intend to promote to the masses….

I wonder at what level this math is taught in the US. In Sweden, this is something one would learn at the university level. Probably only in math-heavy majors like engineering.

I love triple integrals…!

Who da fook said tripple integral gives us volume😡😢😴it gives us mass😂

Integration is nothing but summation dont say its an area.never ever ever nevver.

what kind of people would press the dislike button for such a video. Thanks Sal (Ultimate Respect :))

salam to u

This what launch the space shuttle

Thank you Khan Academy! Your videos are always really help me visualise math. Triple integrals make sense now!

My hero

p =/= rho 🙂

why would you choose triple integral over double integral seems pointless to me……

well i know as a fact that this is not part of calculus 12 in Canada….. it is first year of university in Canada….

Very good

Thank you.

∭ are fun

AMAZING REPRENSENTATION