# Double integral 1 | Double and triple integrals | Multivariable Calculus | Khan Academy

So far, we’ve used integrals
to figure out the area under a curve. And let’s just review a little
bit of the intuition, although this should hopefully be second
nature to you at this point. If it’s not, you might want
to review the definite integration videos. But if I have some function–
this is the xy plane, that’s the x-axis, that’s the y-axis–
and I have some function. Let’s call that, you know,
this is y is equal to some function of x. Give me an x and
I’ll give you a y. If I wanted to figure out the
area under this curve, between, let’s say, x is equal to
a and x is equal to b. So this is the area I
want to figure out. What I do is, I split it up
into a bunch of columns or a bunch of rectangles. Where– let me draw one of
those rectangles– where you could view– and there’s
different ways to do this, but this is just a review. Where you could review– that’s
maybe 1 of the rectangles. Well, the area of the
rectangle is just base times height, right? Well, we’re going to make these
rectangles really skinny and just sum up an infinite
number of them. So we want to make them
infinitely small. But let’s just call the
base of this rectangle dx. And then the height of this
rectangle is going to be f of x, at that point. It’s going to be f of– if this
is x0, or whatever, you can just call it f of x, right? That’s the height
of that rectangle. And if we wanted to take
the sum of all of these rectangles– right? There’s just going to
be a bunch of them. One there, one there. Then we’ll get the area, and if
we have infinite number of these rectangles, and they’re
infinitely skinny, we have exactly the area
under that curve. That’s the intuition behind
the definite integral. And the way we write that–
it’s the definite integral. We’re going to take the sums of
these rectangles, from x is equal to a, to x is equal to b. And the sum, or the areas that
we’re summing up, are going to be– the height is f of x,
and the width is d of x. It’s going to be f
of x times d of x. This is equal to the
area under the curve. f of x, y is equal to f
of x, from x is equal to a to x is equal to b. And that’s just a
little bit of review. But hopefully, you’ll now
see the parallel of how we extend this to taking the
volume under a surface. So first of all,
what is a surface? Well, if we’re thinking
in three dimensions, a surface is going to be
a function of x and y. So we can write a surface as,
instead of y is a function of f and x– I’m sorry. Instead of saying that y is a
function of x, we can write a surface as z is equal to
a function of x and y. So you can kind of view
it as the domain. Right? The domain is all of the set
of valid things that you can input into a function. So now, before, our domain was
just– at least, you know, for most of what we dealt with–
was just the x-axis, or kind of the real number line
in the x direction. Now our domain is the xy plane. We can give any x and any y–
and we’ll just deal with the reals right now, I don’t
want to get too technical. And then it’ll pop out another
number, and if we wanted to graph it, it’ll be our height. And so that could be the
height of a surface. So let me just show you what
a surface looks like, in case you don’t remember. And we’ll actually figure out
the volume under this surface. So this is a surface. I’ll tell you its function
in a second, but it’s pretty neat to look at. But as you can see,
it’s a server. It’s like a piece of
paper that’s bent. Let’s see, let me rotate it
to its traditional form. So this is the x direction,
this is the y direction. And the height is a function of
where we are in the xy plane. So how do we figure out
the volume under a surface like this? How do we figure
out the volume? It seems like a bit of a
stretch, given what we’ve learned from this. But what if– and I’m just
going to draw an abstract surface here– let
me draw some axis. Let’s see, that’s my x-axis. That’s my y-axis. That’s my z-axis. I don’t practice these videos
ahead of time, so I’m often wondering what I’m
about to draw. OK. So that’s x, that’s
y, and that’s z. And let’s say I
have some surface. I’ll just draw something. I don’t know what it is. Some surface. This is our surface. z is a function of x and y. So, for example, you give me a
coordinate in the xy plane. Say, here, I’ll put it into
the function and it’ll give us a z value now. And I’ll plot it there
and it’ll be a point on the surface. So what we want to figure
out is the volume under the surface. And we have to specify
bounds, right? From here, we said x is equal
to a, to x is equal to b. So let’s make a square
bound first, because this keeps it a lot simpler. So let’s say that the domain or
the region– not the domain– the region of– the x and y
region of this part of the surface under which we want
to calculate the volume. Let’s say, the shadow– if
the sun was right above the surface, the shadow
would be right there. Let me try my best to
draw this neatly. So this is what we’re
going to try to figure out the volume of. And let’s say– so, if we
wanted to draw it in the xy plane, like you can kind of
view the projection of the surface of the xy plane,
or the shadow of the surface of the xy plane. What are the bounds? You can almost view– what are
the bounds of the domain? Well, let’s say that this
point– let’s say that this right here, that’s 0,
0 in the xy plane. Let’s say that this is y is
equal to– I don’t know, that’s y is equal to a. That’s this line right here. Y is equal to a. And let’s say that this line
right here is x is equal to b. Hope you get that, right? This is the xy plane. If we have a constant x, it
would be a line like that. A constant y, a line like that. And then we have the
area in between it. So how do we figure out
the volume under this? Well, if I just wanted to
figure out the area of– let’s just say, this sliver. Let’s say we had a–
well, actually let me go the other way. Let’s say we had a constant y. So let’s say I had some sliver. I don’t want to confuse you. Let’s say that I had
some constant y. I just want to give
you the intuition. You know, let’s say. I don’t know what that is. It’s an arbitrary y. But for some constant y, what
if I could just figure out the area under the curve there? How would I figure out just
the area under that curve? It’ll be a function of
which y I pick, right? Because if I pick a y here,
it’ll be a different area. If I pick a y there, it’ll
be a different area. But I could view this now
as a very similar problem to this one up here. I could have my dx’s– let
me pick a vibrant color so you can see it. Let’s say that’s dx, right? That’s a change in x. And then the height is going
to be a function of the x I have and the y I picked. Although I’m assuming,
to some degree, that that’s a constant y. So what would be the area
of this sheet of paper? It’s kind of a constant y. It’s part of– it’s a sheet
of paper within this volume, you can kind of view it. Well, it would be– we said
the height of each of these rectangles is f of xy, right? That’s the height. It depends which x and
y we pick down here. And then its width is
going to be d of x. Not d of x, dx. And then if we integrated it,
from x is equal to 0, which was back here, all the way to x
is equal to b, what would it look like? It would look like that.
x is going from 0 to b. Fair enough. And this would actually
give us a function of y. This would give us an
expression so that I would know the area of this kind of sliver
of the volume, for any given value of y. If you give me a y, I can tell
you the area of the sliver that corresponds to that y. Now what could I do? If I know the area of any given
sliver, what if I multiply the area of that sliver times dy? This is a dy. Let me do it in a
vibrant color. So dy, a very small
change in y. If I multiplied this area
times a small dy, then all of a sudden I have
a sliver of volume. Hopefully that
makes some sense. I’m making that– that little
cut that I took the area of– by making it three dimensional. So what would be the
volume of that sliver? The volume of that sliver will
be this function of y times dy, or this whole thing times dy. So it would be the integral
from 0 to b of f of xy dx. That gives us the area
of this blue sheet. Now if I multiply this
whole thing times dy, I get this volume. It gets some depth. This little area that I’m
shading right here gets depth of that sheet. Now if I added all of those
sheets that now have depth, if I took the infinite sum– so if
I took the integral of this from my lower y bound– from 0
to my upper y bound, a, then– at least based on our intuition
here– maybe I will have figured out the volume
under this surface. But anyway, I didn’t
want to confuse you. But that’s the intuition of
what we’re going to do. And I think you’re going to
find out that actually calculating the volumes are
pretty straightforward, especially when you have
fixed x and y bounds. And that’s what we’re going
to do in the next video. See you soon.

## 48 thoughts on “Double integral 1 | Double and triple integrals | Multivariable Calculus | Khan Academy”

1. cooking shooking says:

you dont explain well because you yourself is confused

2. itachi Theonlylegend says:

thank you. I understood double integral now

I almost cried when i saw how logically and beautifully everything is explained in this video.

4. Subarna Subedi says:

Why do we use ds in line integral and dx here for area of one sheet ds=small arc lenght of curve dx=small change in x

5. Gabriel Mello says:

Excelent explanation.

6. Hanh Vo says:

very understandable compared to the larson textbook
Basically volume of a solid rectangular = LHW=f(x,y)dxdy

7. srayes1001 says:

Is the first integral(inside integral) like a ghost integral cos it can "vary"???

just ultimate

9. Sanjeev Kumar says:

wow thanks ..!! it was really getting itchy with my brain uptil now.!! thanks again

10. anonymous traveller says:

Water always finds a way to flow , khan you are that water , bringing free education.

11. JIA ABHIRAAJ says:

the explanation was soooo good . thank you sir

12. Darry Andrews says:

not now.

13. Allan Kálnay says:

8:04 why do you say that width is gonna be dx? I dont really understand what you mean by that at this moment. Please explain it to me

14. Boody says:

really you are the best

15. Sanjay Sahu says:

Please sir this videos also available in hindi

16. Nikhil More says:

Why answer is negative sometimes though it's a volume?

17. Nitin Mishra says:

Wow it was awesome video my whole confusion got cleared thanks Khan academy

18. Brandon Ignacio says:

Could a double integral also be described as the "mass" or something related to the mass of an object in 2-dimensions? Does this hold true for triple integrals as well? If so would the integrand be a density function?

19. G0tBlackOps says:

6:50

20. Harold H says:

Beast

21. PHOTON says:

that was an amazing explanation!!

amazing explanation

23. Rockstars Azam says:

Best teaching i have ever seen

24. Punith V H Puni says:

After I seeing this video I remember the thought of Einstein, that is "if you can't explain it simply ,then you don't understand it well enough ",sir u explained in simple manner, thanks sir🙏👍

25. Zubair Khan says:

Too beautifully explained…
Always rocking.

26. Josh Parkin says:

Thanks so much!

27. ANIL KUMAR says:

Nice sir

أكتب الطرجمة بالعربية من فضلك

29. Joan Conejos Jalencas says:

Lol it's only in 240 p xD

30. 男鲸Whaley says:

The comment section make me nt concentrating hahahaha

31. Para Bola says:

I love your work so much. Can someone do me a favour please? Between 3:19 and 3:52 is shown a 3D image of a surface. Could you or someone else tell me please what is the function of this surface.

32. vivek rana says:

God the effort you put in. Thanks a lot for that. You are changing the lives of millions of people by providing quality education for free. I scored well by watching your videos. I could not donate at this moment as I am a student. But I will definitely denote to your organization when I will start earning. Again thanks a lot for your effort.

33. 城下君 says:

Help a lot for me. Thanks!

34. Trey Gilliland says:

240p we meet again…

35. Shourya Arya says:

Thanks for the video Khan sir

36. Keita Tahara says:

I just started middle and I finally learned all the way up to multivariable calculus. But it would’ve been way faster if I just learned from this guy all the way up for sure.

37. n00bcake 547 says:

this could use an update

38. William Moura says:

Great! Where do we start seeing double integrals in an engineering course?

39. Dumberiable Squinchy says:

I love you

40. Shanker Lal Jat says:

thank you very much sir..

41. Justin Gwon says:

isn't that beautiful we still could learn from this video 11 years ago….

42. Aneesh Srinivas says:

Why is the double integral denoted ∫_{a}^b (∫_{α}^βf(x,y)dx)dy,i mean If in the first integral ∫_{α}^βf(x,y)dx what that one is saying is to integrate f(x,y) while treating y as a constant in that case shouldn’t the double integral notation be ∫_{a}^b(∫_{α}^βf(x,y)∂x)∂y

43. Alex H says:

Why do I even pay tuition, I can just watch Khan academy

44. minxxdia says:

45. Calvin Lai says:

where is d(z)?

46. Nick Nejat Can says:

PRAISE SAL KHAN

47. Aili Emory says:

Thank you so much for the work you do! It has helped me love math and see its beauty:)

48. David Rodriguez says:

That was an amazing introduction to double integration.

49. Gourab Chakraborty says:

Excellently explained……