Main content

### Course: Multivariable calculus > Unit 4

Lesson 7: Triple integrals# Triple integrals 3

Figuring out the boundaries of integration. Created by Sal Khan.

## Want to join the conversation?

- Towards the very end, Why isn't upper y boundary y= -2x + 6? I'm not sure why its just 6..(10 votes)
- Every time you integrate, you are essentially removing a dimension from your problem as you approach the answer. The first integral removes the z dimension so z=0 the rest of the way through because you integrated with respect to z, which is why you use the 2 dimensional projection in the xy plane to get the upper bound of x. Once you integrate with respect to x, you have finished with that dimension and x=0 from then on. Finally, now that z=0 and x=0, you can plug those into the original surface equation to get your upper y bound of 6. Then, you do your final one-dimensional integral to solve the problem.(46 votes)

- Can your x integral be from 0 to 3?(8 votes)
- Only if you integrate that last and you use y=6-2x as the upper bound for y. That boundary must be used at least once and your last integration must only have constants.(18 votes)

- sal you left me hangin in the end where is the next video??(15 votes)
- Why z=2 is that green triangle surface?(4 votes)
- SAL,

PLEASE do more on tripple integrals! I love your videos, but its a bummer you stop here with this trivial example. As someone else menioned in a comment years ago, can you continue into clindrical and sphereical coordinat systems?? Paraboloids, and frustrums, and more complex surfaces?(8 votes)

- Why do we put the variable bounds in the innermost integral?(4 votes)
- because if we use it in the ¿outermost integral? we will have in the final result a function, and we can´t have this, we need to have a fixed number...(6 votes)

- At11:11, if we want the area between z = 2 and 2x + 3z + y = 6, why do we integrate from x = 3 - y/2 to x= 0? Wouldn't we integrate from x = 3 to x = 3 - y/2 since x = 3 - y/2 is the lower bound?(4 votes)
- Very good observation. I followed in the footsteps of those who solved this hideous problem, without realizing this, and now I know my solution is meaningless since the limits are mixed up. I would recommend mentioning this as an error in the video so that others may be able to avoid similar meaningless work.

Thank you for your attention to detail and precision!(4 votes)

- At9:21, i don't understand why you used the x-y plane instead of the x-z plane. I am not very good at visualizing 3D diagrams, so could you please show me why?(3 votes)
- Because if your integration order takes care of Z first, i.e. dzdxdy, then once you find your Z limits in the first integral, then you are done with Z altogether, the next step to solve the triple integral is to project into the remaining variables' plane, in other words, project into the x-y plane. You would only project your solid into the x-z plane if you integrated with the order dydxdz or dydzdx, in other words, integrating y first.(6 votes)

- I Don't Get the boundaries you put for the z, x and Y ? i mean why don't we take boundaries for z to be from 0 to 2 and in x from 0 to 3 and y from 0 to 6 ? Im really confused(2 votes)
- Because that assumes that at every given point in your object, the lower Z height is 0, but as you can see, the lower Z heights vary, directly based on where you are in x and y, i.e. different values of x and y produce different values of Z. This relation between x, y, and z, is described by that Plane: 2x+3z+y=6.

Integrating with x from 0 to 3, y from 0 to 6, and z from 0 to 2 would give you the mass of a rectangular box, but as you can see, the object Sal drew is not a rectangular box. Instead, if we want the correct limits of integration, we need to find the height of that plane at every given x, and y value, in other words, we need to solve for Z in the plane equation 2X+3Z+Y=6, well using algebra, Z= (-2X -y)/3 +2. That equation is still the plane, but it's in terms of X and Y, so providing X and Y values yields a Z value. Essentially, that Plane is your lower bound in your integration limit (note that this would only be your integration limit if you were integrating Z first: dZdYdX or dZdXdY). The upper bound is the flat plane Z=2, because it's height does not depend on the X and Y values.

Hope this helps, I understand your struggle, this stuff used to confuse me as well.(5 votes)

- Why should the elemental volume dV be a cube here?

Since the shape under consideration is more like a wedge than a cuboid, shouldn't the elemental volume also take same shape?(2 votes)- Since the cube's volume will be infinitesimal, it doesn't matter that it's a cube. Also, cubes tesselate (they can be packed together without any empty space), wedges don't.(5 votes)

- Although I got the hang of doing a triple integral, something doesn't quite add up with the density example. If you say that dM = x^2*y*z (dxdydz) , wouldn't calculating M require just one integral? I mean if u integrate dM u get just M. By that reasoning , if we integrate both parts with just 1 integral we would get something like M = integral from x^2*y*z (dxdydz), and I realize that looks wrong, with a integral sign in regard with the 3 variables dx,dy,dz, but then what is the flaw in my reasoning.(3 votes)
- If the object in question was a uniform density, then yes, it would require only one integral. However, in the example the density is a function and not a constant, which means we need that triple integral.(2 votes)

## Video transcript

Let's now do another triple
integral, and in this one I won't actually evaluate
the triple integral. But what we'll do is we'll
define the triple integral. We're going to something
similar that we did in the second video where we
figured out the mass using a density function. But what I want to do in this
video is show you how to set the boundaries when the
figure is a little bit more complicated. And if we have time we'll try
to do it where we change the order of integration. So let's say I have the surface
-- let me just make something up -- 2x plus 3z plus
y is equal to 6. Let's draw that surface. It looks something like this. This will be my x-axis. This is going to be my z-axis. This is going to be my y-axis. Draw them out. x, y and z. And I care about the surface in
the kind of positive octant, right, because when you're
dealing with three-dimensionals we have, instead of four
quadrants we have eight octants. But we want the octant where
all x, y and z is positive, which is the one I drew here. So let's see, let me draw some
-- what is the x-intercept? When y and z are 0,
so we'll write here, that's the x-intercept. 2x is equal to 6, so
x is equal to 3. So 1, 2, 3. So that's the x-intercept. The y-intercept when x and
z are 0 are on the y-axis, so y will be equal to 6. so we have 1, 2, 3, 4, 5,
6 is the y-intercept. Then finally the z-intercept
when x and y are 0 we're on the z-axis. 3z will be equal to 6. So z is equal to 1, 2. So the figure that I care about
will look something like this -- it's be this
inclined surface. It will look
something like that. In this positive octant. So this is the surface
defined by this function. Let's say that I care about
the volume, and I'm going to make it a little
bit more complicated. We could say oh, well let's find the volume between the surface and the xy plane. But I'm going to make it a
little bit more complicated. Let's say the volume above
this surface, and the surface z is equal to 2. So the volume we care
about is going to look something like this. Let me see if I can
pull off drawing this. If we go up 2 here -- let me
draw the top in a different color, let me draw
the top in green. So this is along the zy plane. And then the other
edge is going to look something like this. Let me make sure I can draw it
-- this is the hardest part. We'll go up 2 here, and this
is along the zx plane. And we'd have another line
connecting these two. So this green triangle,
this is part of the plane z is equal to 2. The volume we care about is the
volume between this top green plane and this tilted plane
defined by 2x plus 3z plus y is equal to 6. So this area in between. Let me see if I can make
it a little bit clearer. Because the visualization, as I
say, is often the hardest part. So we'd have kind of a front
wall here, and then the back wall would be this wall back
here, and then there'd be another wall here. And then the base of it, the
base I'll do in magenta will be this plane. So the base is that plane
-- that's the bottom part. Anyway, I don't know if I
should have made it that messy because we're going to have to
draw dv's and d volumes on it. But anyway, let's try our best. So, if we're going to figure
out the volume -- and actually, since we're doing a triple
integral and we want to show that we have to do the triple
integral, instead of doing a volume, let's do the mass of
something of variable density. So let's say the density in
this volume that we care about, the density function is a
function of x, y and z. It can be anything. That's not the point of what
I'm trying to teach here. But I'll just define something. Let's say it's x squared yz. Our focus is really just
to set up the integrals. So the first thing I like to do
is I visualize -- what we're going to do is we're going to
set up a little cube in the volume under consideration. So if I had a -- let me do it
in a bold color so that you can see it -- so if I have a cube
-- maybe I'll do it in brown, it's not as bold but it's
different enough from the other colors. So if I had a little cube
here in the volume under consideration, that's a little
cube -- you consider that dv. The volume of that cube is kind
of a volume differential. And that is equal to dx --
no, sorry, this is dy. Let me do this in yellow,
or green even better. So dy, which is this. dy times dx, dx times dz. That's the volume of
that little cube. And if we wanted to know the
mass of that cube, we would multiply the density function
at that point times this dv. So the mass, you could call
it d -- I don't know, dm. The mass differential is going
to be equal to that times that. So x squared y z times this. dy, dx, and dz. And we normally switch this
order around, depending on what we're going to integrate with
respect to first so we don't get confused. So let's try to do this. Let's try to set
up this integral. So let's do it traditionally. The last couple of triple
integrals we did we integrated with respect to z first. So let's do that. So we're going to integrate
with respect to z first. so we're going to take this cube
and we're going to sum up all of the cubes in the z-axis. So going up and
down first, right? So if we do that, what
is the bottom boundary? So when you sum up up and down,
these cubes are going to turn to columns, right? So what is the bottom of the
column, the bottom bound? What's the surface? It's the surface
defined right here. So, if we want that bottom
bound defined in terms of z, we just have to solve
this in terms of z. So let's subtract. So what do we get. If we want this defined in
terms of z, we get 3z is equal to 6 minus 2x minus y. Or z is equal 2 minus
2/3x minus y over 3. This is the same thing as that. But when we're talking about
z, explicitly defining a z, this is how we get it,
algebraically manipulated. So the bottom boundary -- and
you can visualize it, right? The bottom of these columns
are going to go up and down. We're going to add up all
the columns in up and down direction, right? You can imagine summing them. The bottom boundary is
going to be this surface. z is equal to 2 minus
2/3x minus y over 3. And then what's
the upper bound? Well, the top of the column
is going to be this green plane, and what did we
say the green plane was? It was z is equal to 2. That's this plane, this
surface right here. Z is equal to 2. And, of course, what is the
volume of that column? Well, it's going to be the
density function, x squared yz times the volume differential,
but we're integrating with respect to z first. Let me write dz there. I don't know, let's say we want
to integrate with respect to -- I don't know, we want to
integrate with respect to x for next. In the last couple of
videos, I integrated with respect to y next. So let's do x just to show you
it really doesn't matter. So we're going to integrate
with respect to x. So, now we have these
columns, right? When we integrate with respect
to z, we get the volume of each of these columns wher the
top boundary is that plane. Let's see if I can
draw it decently. The top boundary is that plane. The bottom boundary
is this surface. Now we want to integrate
with respect to x. So we're going to add
up all of the dx's. So what is the bottom
boundary for the x's? Well, this surface is defined
all the way to -- the volume under question is defined all
the way until x is equal to 0. And if you get confused, and
it's not that difficult to get confused when you're imagining
these three-dimensional things, say you know what, we already
integrated with respect to z. The two variables I
have left are x and y. Let me draw the projection of
our volume onto the xy plane, and what does that look like? So I will do that. Because that actually does
help simplify things. So if we twist it, if we take
this y and flip it out like that, and x like that we'll get
in kind of the traditional way that we learned when we
first learned algebra. The xy-axis. So this is x, this is y. And this point is what? Or this point? What is that? That's x is equal to 3. So it's 1, 2, 3. That's x is equal to 3. And this point right here
is y is equal to 6. So 1, 2, 3, 4, 5, 6. So on the xy-axis, kind of the
domain -- you can view it that -- looks something like that. So one way to think about it is
we've figured out if these columns -- we've integrated
up/down or along the z-axis. But when you view it looking
straight down onto it, you're looking on the xy plane, each
of our columns are going to look like this where the
column's going to pop out out of your screen in
the z direction. But the base of each column is
going to dx like that, and then dy up and down, right? So we decided to integrate
with respect to x next. So we're going to add up each
of those columns in the x direction, in the
horizontal direction. So the question was what
is the bottom boundary? What is the lower bound
in the x direction? Well, it's x is equal to 0. If there was a line here, then
it would be that line probably as a function of y, or
definitely as a function of y. So our bottom bound here
is x is equal to 0. What's our top bound? I realize I'm already pushing. Well, our top bound is this
relation, but it has to be in terms of x, right? So what is this relation. So, you could view it as kind
of saying well, if z is equal to 0, what is this line? What is this line right here? So z is equal to 0. We have 2x plus y
is equal to 6. We want the relationship
in terms of x. So we get 2x is equal to 6
minus y where x is equal to 3 minus y over 2. And then finally we're going to
integrate with respect to y. And this is the easy part. So we've integrated up and
down to get a column. These are the bases of the
column, so we've integrated in the x direction. Now we just have to go up and
down with respect to y, or in the xy plane with respect to y. So what is the y
bottom boundary? Well, it's 0. y is equal to 0. And the top boundary
is y is equal to 6. And there you have it. We have set up the integral
and now it's just a matter of chugging through
it mechanically. But I've run out of time
and I don't want this video to get rejected. So I'll leave you there. See you in the next video.