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.

you dont explain well because you yourself is confused

thank you. I understood double integral now

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

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

Excelent explanation.

very understandable compared to the larson textbook

Basically volume of a solid rectangular = LHW=f(x,y)dxdy

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

just ultimate

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

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

the explanation was soooo good . thank you sir

not now.

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

really you are the best

Please sir this videos also available in hindi

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

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

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?

6:50

Beast

that was an amazing explanation!!

amazing explanation

Best teaching i have ever seen

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🙏👍

Too beautifully explained…

Always rocking.

Thanks so much!

Nice sir

أكتب الطرجمة بالعربية من فضلك

Lol it's only in 240 p xD

The comment section make me nt concentrating hahahaha

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.

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.

Help a lot for me. Thanks!

240p we meet again…

Thanks for the video Khan sir

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.

this could use an update

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

I love you

thank you very much sir..

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

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

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

was really helpful🤗

where is d(z)?

PRAISE SAL KHAN

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

That was an amazing introduction to double integration.

Excellently explained……