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Multivariable calculus
Course: Multivariable calculus > Unit 4
Lesson 7: Triple integralsTriple integrals 1
Introduction to the triple integral. Created by Sal Khan.
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So what would the use be of a quadruple integral?(17 votes)
Quadruple integrals would be used if you are dealing with 4 dimensional space, which is something that has only ever been theorized. Theoretically, there is no limit to how many integrals you can have since mathematically, there's no limit to the number of spatial dimensions you can work with.
However, in practice, I've never seen quadruple integrals. There is very little, if any, practical application for them.(41 votes)
Sal do you have any videos on evaluating double or triple integrals using plane polar,cylindrical and spherical coordinates.I can't find them anywhere(22 votes)
There are articles on double integrals in Polar Coordinates in the previous section articles.(5 votes)
- Is there any plans to remake these videos? The entire multivariable calculus course is quite outdated and its a very fundamental topic for practically all engineering paths.(17 votes)
what is a triple integral good for other then solving a basic quadratic prism?(1 vote)
You can calculate the volume of any 3d object.(17 votes)
- Is there a difference between this and volume integrals?(6 votes)
Double and triple integrals are volume integrals--they are measuring the total volume of a 3-D object in the xyz-coordinate space. Analogously, "single" integrals measure the total area of a 2-D figure in the xy-coordinate plane.
Use: Function:
int -- y = f(x)
iint -- z = f(x, y)
iiint -- w = f(x, y, z)
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Note that "int" is the regular single integral, "iint" is a double integral, and "iiint" is a triple integral. While both double and triple deal with three dimensional space, the integrals are different. The triple integral measures 3-D objects while they are changing position, which brings it into the fourth dimension. Depending on the scenario, a triple integral may be a volume integral, but a double integral is, by default, a volume integral.(1 vote)
- Is it possible to derive the formula for the volume of a sphere using triple integrals?(2 votes)
I suppose you could, though it is really just make a simple problem more complicated.
http://www.wolframalpha.com/input/?i=triple+integral+calculator&f1=1&f=TripleIntegral.integrand_1&f2=z&f=TripleIntegral.intvariable1i_z&f3=-sqrt(r%5E2-x%5E2-y%5E2)&f=TripleIntegral.rangestart1i_-sqrt(r%5E2-x%5E2-y%5E2)&f4=sqrt(r%5E2-x%5E2-y%5E2)&f=TripleIntegral.rangeend1i_sqrt(r%5E2-x%5E2-y%5E2)&f5=y&f=TripleIntegral.intvariable2m_y&f6=-sqrt(r%5E2-x%5E2)&f=TripleIntegral.rangestart2m_-sqrt(r%5E2-x%5E2)&f7=sqrt(r%5E2-x%5E2)&f=TripleIntegral.rangeend2m_sqrt(r%5E2-x%5E2)&f8=x&f=TripleIntegral.intvariable3_x&f9=-r&f=TripleIntegral.rangestart3_-r&f10=r&f=TripleIntegral.rangeend3_r(10 votes)
- I don't quite understand the intuition behind taking the integral in the z axis. Sal describes it in the context of a volume, but it's a one dimensional quantity. Is it something akin to the Dirac Delta function? I understand in the context of the Dirac Delta how there could be some area under the curve equal to some quantity when one axis goes to infinity, but I'm not understanding the intuition behind how there could be AUC for a one dimensional quantity that doesn't go to infinity. Can anyone help me to better understand this?(1 vote)
The element dxdydz in not one-dimensional. 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 reason you can't see an area under the curve for the first integral is that, in a certain sense, there isn't one.
An 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.
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.
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.
A more useful intuition in this case is the one of adding up all the values.
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.
Integrating these across some area (integrating zdA) adds up all the z's and gives us the volume we wanted.
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.
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.
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.
The density p is a good start because it's the only thing which has a unit containing a mass term. 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 obvious choice is dV=dxdydz.
This gives us that dM=pdV=pdxdydz
If we then integrate this over the entire shape R, we obtain the mass M.
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.(8 votes)
- Does the order of the dx,dy, and dz matter in a general scope of taking the integral. I understand that different combinations can prove to be easier but must they be in a specific/alphabetical order? Ex. If dx is first does it need to be followed by dy, then by dz or same case but we start with dy then dz then dx.(3 votes)
All orders of integration will produce the same result - though as you said, some orders can be much more difficult.
The theorem: http://mathworld.wolfram.com/FubiniTheorem.html
It is a great way to check your work, eg if you did dxdydz try dxdzdy.(4 votes)
- I understand the entire concept as presented in the video. However I have a lingering doubt. The triple integral was solved by imagining the 3 dimensional space, and adding up tiny cubes all over the space. In the previous section of double integration, we solved again by imaging 3 D space and adding columns of area (dxdy), height f(x,y) over the 3 D space. Is it not possible to explain double integrals with only 2 D space?(4 votes)
- Shouldn't it be the Commutative Property rather than the Associative Property at3:17?(3 votes)
- Yeah, certainly seems so by context. Got the point across anyhow, I s'pose.(2 votes)
3:17Video 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 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.