Triple integrals 3 | Double and triple integrals | Multivariable Calculus | Khan Academy
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Triple integrals 3 | Double and triple integrals | Multivariable Calculus | Khan Academy

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 this
was a 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.

About James Carlton

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91 thoughts on “Triple integrals 3 | Double and triple integrals | Multivariable Calculus | Khan Academy

  1. You waste too much time yapping. I cant help but to fast forward through your video. Consequently I end up missing bits of important information. You are mind numbing. You are redundant. You keep saying the same things over and over. To say nothing of you being repetitive. And redundant too. And in a ten minute video you might only have two minutes of content. Quit treating us like were retarded fourth graders. If we cant keep up, we shouldnt even be watching this video yet.

  2. Its really frustrating to be intellectually capable of learning this stuff, but to be turned off from doing so because you repeat yourself like we didnt catch it the first time. It would take a tenth the time… or contain ten times the material… if you just get to the damn point. If its beyond some of us to understand, we need to stick to basic algebra until we are ready.

  3. Some of your competitor math video contributors are a hell of a lot better. The only reason why I come to you is that you do frequently get to valuable information. Plus you have more videos and are better organized. But in the end I still cant tolerate hearing you yap on for ten minutes about nothing or about something conceptually trivial or about something you already stated.

  4. Dont get me wrong. I will still watch. I have subscribed and dont mean to insult. I just wish you would clean up your videos a bit. Respect your viewers a bit more by actually pretending we dont need out hand held.

  5. Clearly the guy below me shares a minority opinion.

    Anyways, at the very end, when integrating with respect to y, how could the upper bound be y=6? It's only y=6 at the far left when x=0? It should be a variable bound.

    I know for almost certain that I'm wrong because (a) Sal knows what he's doing and (b) a variable bound would not have been placed on the outermost integral!

  6. well, thanks to you. I managed to understand what you are talking about. A lot of abstract thinking, but I see what you mean. Good thing I know how to set up the integral and my professor is going to teach that next month. So thanks to you I have a headstart. 🙂

  7. I can't understand what the problem is. You make it sound as if you are forced to watch this man. He is giving you his time and knowledge and yet you are complaining.

  8. You are amazing person ! My son is just wondering from topic to topic….he is only 9, but love to do calculus , I really am amazed … you are the driving engine of his journey.

    Michael, Vancouver BC

  9. Cool! Except for that last part…why are the last boundaries 0 to 6 and not 0 to 6 – 2x?
    That's kind of what you did for your x boundaries, why change up technique for y boundaries? I understand you want a constant for the last integral, but does that mean you can you just add a constant when convenient? The y does change… a lil confused.

  10. if the volume we're going to find is the region between the green plane and the plane, wouldn't the x boundaries be from the line x=3-1/2y to 3 since the base is the plane, wouldnt x boundaries from 0 to x=3-1/2y be the region under the slanted plane and not above it?

  11. @missjust4chill Usually when setting up ur 3 integrals u want to make it so that ur last integration is from one constant to another. Because ur doing a projection, u can think of it as though you were figuring out the aread of a triangle. Because he wrote x as a function of y, well then you just plug in your values of y which only range from 0 to 6.
    Kind of like when you're graphing a line you usually have y as a function of x. here u plug in y, where it's only from 0 to 6.

  12. @MarvelsofaLifetime Congratulations on proving how big of a douchebag you are, going out of your way to antagonize a third party when you obviously failed to understand their point. Not only that, but you have also proven you lack of faith in people, assumed the worst in me, and shown your own prejudices and bigotries. Congrats, again.

  13. @CogitoErgoCogitoSum
    The problem is this…
    You could consolidate all 5000 hours of play time of Sals videos into one five minute clip and youd probably learn more. The problem is Sal baby-walks retards through math and science… people who are predestined to failure, predisposed to ignorance, and should… SHOULD… fail. But, Sal pushes them through anyway, lowering the standard of American academia… because retards have diplomas and degrees when they shouldnt.

  14. @CogitoErgoCogitoSum Sal reiterates EVERYTHING a dozen times until youre bored of listening, close your ears, and then suddenly you wonder why you missed some vital tidbit – the only new piece of information he spat out in the last ten videos of a playlist. Its annoying.

    His examples are great. His teaching is great. BUT… he needs to stream line it.

  15. @CogitoErgoCogitoSum LITERALLY. No joke. No one WANTS to listen and watch if they arent learning anything. Half the people who attend his site have no desire to learn anyway, or else they'd be paying attention in class and wouldnt need to be baby-walked now. Add to that the fact rarely anything new gets taught, and everything is reiterated – memorization via "beating it into the skull" approach is the last thing our society needs.

  16. @CogitoErgoCogitoSum Its painful to sit here and listen. And I ONLY come to him when there is NOTHING else available. Im more inclined to buy an expensive university text book or sit through a dry lecture.

    Simple fact. These videos are five minutes a piece and they are all archived in order. If you dont understand a given lecture, watch the previous video. If you dont understand that one, watch the one previous. So on so forth. Regulate yourself, watch what is appropriate to your intellect.

  17. @CogitoErgoCogitoSum Nobody is willing to dumb down, admit ignorance, and watch earlier videos. I watch his most advanced, most recent topic, and he is still explaining what he already did a hundred videos ago. Hello? If you dont get it already, what the fudge are you watching THIS ONE for?

  18. whats the title of the next video? I wanna watch more… I'm watching while checking if my assignment is correct, but nope,so i am thankful that I watch this video first before I pass my assignment this afternoon..:)

  19. @Incognito6543 IThe mass of the 'object' between the plane z=2 and the plane 2x+3z+y=6; for which you need the object's volume and density (p=M/V); where density p=yzx^2.

  20. @elementoxygen 9 years old? Well, I guess I should stop thinking I'm intelligent for being a freshman in Cal III… I mean Christ!

  21. i dont know if someone already asked but… why y=6 at the ending, i thought x=3-y/2 because it was limitated by that purple line but what happened with "y"?

  22. Well, from 11:16 and onward you are taking the sum of the rectangles in the xy plane. Obviously, you start with the rectangle at y=0 (so 0 is the first bound of the graph). Then, you add more rectangles, from y=0 to y=1 to y=5 until you stop at y=6. In that last case (the rectangle at y=6), x=0, right? Okay, so lets substitute 0 for x in x=3-y/2 –> 0=3-y/2 –> -3 = -y/2 –> 3 = y/2 –> 6=y

  23. I have a question. So drawing a sketch is pretty much mandatory for doing these kinds of problems, right? Cause otherwise I don't see how you even get started on finding the bounds unless you can do everything in your head.

  24. So when you integrate Z, the value of it goes to zero for the next step of integration. Similarly, once you integrate X, it's value goes to zero for the last step of integration. If you then draw the same triangle where X always equal to 0, you would get a line going from Y=0 to Y=6, which are your bounds. If it makes it easier, you can imagine you lose a dimension every time you integrate a variable. When you have 3 variables left, 3 dimensions. 2 var., 2 dim. 1 var., 1 dim.

  25. The boundaries really make sense if you think of them in terms of the minimum amount of information necessary to describe the border of each plane, considering of course, the information provided in the previous (inner) integrals. That's definitely the most clear way to think about it. I suppose it could easily be related to sculpting. 🙂

  26. Bit late in the day but I cant understand zero being the lower bound when setting the lower limits for x and y.
    x = y = 0 are NOT points on  the inclined surface.
    Looked at another way x=3 is on the surface but x=2.9 is not
    Likewise y=6 is on the surface y=5.9 is not.

    x=y=z=0 are the limiting values of the corner of the solid.

    Am I wrong ?

  27. i have no problem setting these up, it's the amount of calculations that are made, so hard not to make a mistake somewhere 🙁

  28. Hi, Mr. Sal. Is correct that there is six different ways to calculate triple integrals:
    dx dy dz.      dx dz dy.
    dy dx dz.      dy dz dx.
    dz dx dy.      dz dy dx.
    How do we calculate the triple integral using all these different ways.

  29. I need explanation why the bounds of integration in this video for each dx, dy, dz 
    change when we change the order of integration.
    for example dx dy dz or dz dy dx.
    Shouldn't the bounds stay the same any way we do the integration?
    is the bounds for dz (z=2 – 2/3x – y/3  to  z=2)?
    is the bounds for dy (y=0  to  y=6 – 2x)?
    is the bounds for dx (x=0  to  x=3 – y/2)?  Thanks, Khan Academy.

  30. shouldnt the lower bound of z be 0 during the integration of dz ? since the bottom most part of the shaded area is touching the x-y plane ?

  31. thank you very much but i think there was a mistake in the end the last integral of the "y"s the interval should be from 0 to y=6-2x

  32. y's upper bound has to equal 6. It is also thought of in 2D much like with x, but if there is a 2x in the upper bound of y, the integral isn't solvable. You get a variable you are not integrating with respect to in a definite integral, so you have to set both x and z = 0 to get y=6 and an actual number.

  33. A continuation of the video lecture would be great, along with triple integrals in cylindrical and spherical coordinates.

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