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## Multivariable calculus

### Course: Multivariable calculus > Unit 4

Lesson 5: Double integrals# Double integrals 4

Another way to conceptualize the double integral. Created by Sal Khan.

## Want to join the conversation?

- Where do I get these almonds for power?(56 votes)
- Does Sal have any videos that integrate using polar coordinates?(13 votes)
- I agree, he should definitely make one using polar coordinates. I'm having trouble figuring out the limits of integration for theta.(8 votes)

- At7:57, Sal mentions that dA is used as a shorthand. Isn't it also used to generalize dxdy so that the order of integration (and the coordinate system) is not specified? I feel like this would make it more applicable in proofs and theorems and such.(8 votes)
- dA is often used to indicate integration over an area without specifying how the integration will be performed. dA can even indicate integration over a curved surface in 3-D.(4 votes)

- In-spite of the approach to the problem being different from the first video. The way Sal goes about solving it is exactly the same. Is it only a way to visualise the problem differently? (since the volume of a column is different from a sheet etc.)(3 votes)
- Pretty much. In the first video it was cross sections of sheets intersecting, but in this video it's going strait to the matter of taking infinitely many columns of the space under the graph. They both get you the end, but take a slightly different point of view.(5 votes)

- Question: Do I always need to draw the two dimensional graph of the bounds? Or can I jump into integrating? Do I NEED to do this for the limits of integration? I'm not opposed to it, I Just prefer a mathematical way so that I don't have to think about how to draw it or take the time to do it on my calculator. Do I just solve for things in terms of x when I do dydx, and vice versa? And my teacher said you can't always just choose which order you want to do it. How do I know when I can and when I can't? Oh, and can you do one on polar coordinates? I'm having trouble finding the limits of integration for theta =/(3 votes)
- i can`t understand what is the difference between simple integration and double integration,,in simple one we used to find area under the curve and in this(double integration) volume under the curve,,, so please explain me the difference b/w them,,maybe i am missing something(3 votes)
- I don't think you are missing anything at this stage. The only comment I have is that, yes, single integration is the area under a curve, whereas the double is volume under a surface (not curve as you wrote) if you are integrating a function such that z=f(x, y) (Remember - there is no volume under a curve - why?).

Is there something in particular*you think*should be different?

When you get to triple integrals, there is also a volume relationship, but there is also much more depending on the function being integrated - so the interpretation of the result of triple integrals can be a bit more complicated.(3 votes)

- what is double integration of cos(x+y)dx.dy first integrating from pi/2 to x and second integration from 0 to pi/2 ?(2 votes)
- Hint: start by using the trig identity
`cos(x+y) = cos(x)cos(y) - sin(x)sin(y)`

... I think you will find that this problem is relatively straightforward after that. HTH(3 votes)

- I didn't knew in America, Europe education is this clear concept. I wish I would have done bachelor's degree in USA.

CAn you make video on double integral in polar just like the first method ? So we could visualize ??(2 votes) - what is the key point I seem to be missing with calculating the volume under a surface defined as Z=f(x,y), and above an area in the xy plane using double integrals or triple integrals. I've seen examples where I can't see the difference between the two, yet one uses a double and the other a triple?(2 votes)
- (any chance you could send a reference to the video/problems you mention?)

It is easy to set up a double integral of the form z=f(xy) into a triple integral where the bounds of z are 0 from below and the function f(x,y) above: ∫∫f(x,y)dydx = ∫∫∫dzdydx <from z=0 to z=f(xy)>

Notice that the volume is between an arbitrary surface z=f(x,y) and the plane z=0.

With the triple integral we can find the volume between two arbitrary surfaces, and moments and centroids etc. of the volume if given the appropriate function, eg ∫∫∫f(x,y,z).(1 vote)

- At8:52Sal says we can do this in polar coordinates. Can someone explain to me how this would work?(4 votes)
- 3D polar could be either with an extra angle coordinate or another radius coordinate , spherical or cylindrical respectively. You might try 2D polar first, I would go back and watch integration videos in 2D and do a parallel situation in polar.

But basically, the idea is to split the area into sectors of a circle where dTHETA becomes infinitely small. The area of a sector is 1/2*pi*r^2. Be mindful of your endpoints - integrating from pi/4 to 3pi/2 is much different than -pi/4 to pi/4. Hope this helps.(2 votes)

## Video transcript

I think it's very important to
have as many ways as possible to view a certain type of
problem, so I want to introduce you to a different way. Some people might have taught
this first, but the way I taught it in the first integral
video is kind of the way that I always think about when
I do the problems. But sometimes, it's more useful
to think about it the way I'm about to show you, and maybe
you won't see the difference, or maybe you'll say, oh, Sal,
those are just the exact same thing. Someone actually emailed me and
told me that I should make it so I can scroll things, and I
said, oh, that's not too hard to do. So I just did that, and
I scrolled my drawing. But anyway, let's say we have
a surface in 3 dimensions. It's a function of x and y. You give me a coordinate down
here, and I'll tell you how high the surface
is at that point. And we want to figure out the
volume under that surface. So. We can very easily figure out
the volume of a very small column underneath the surface. So this whole volume is what
we're trying to figure out, right, between the
dotted lines. I think you can see it. You have some experience
visualizing this right now. So let's say that I have
a little area here. We could call that da. Let me see if I can draw this. Let's say we have a little
area down here, a little square in the x-y plane. And it's, depending on how you
view it, this side of it is dx, its length is dx, and the
height, you could say, on that side, is dy. Right? Because it's a little small
change in y there, and it's a little small change in x here. And its area, the area of
this little square, is going to be dx times dy. And if we wanted to figure out
the volume of the solid between this little area and the
surface, we could just multiply this area times the function. Right? Because the height at this
point is going to be the value of the function,
roughly, at this point. Right? This is going to be an
approximation, and then we're going to take an infinite sum. I think you know
where this is going. But let me do that. Let me at least draw the little
column that I want to show you. So that's one end of it, that's
another end of it, that's the front end of it, that's
the other end of it. So we have a little figure that
looks something like that. A little column, right? It intersects the
top of the surface. And the volume of this
column, not too difficult. It's going to be this little
area down here, which is, we could call that da. Sometimes written
like that. da. It's a little area. And we're going to multiply
that area times the height of this column, and that's the
function at that point. So it's f of x and y. And of course, we could have
also written it as, this da is just dx times
dy, or dy times dx. I'm going to write it in
every different way. So we could also have
written this as f of xy times dx times dy. And of course, since
multiplication is associative, I could have also written it as
f of xy times dy dx. These are all equivalent, and
these all represent the volume of this column, that's the
between this little area here and the surface. So now, if we wanted to figure
out the volume of the entire surface, we have a couple
of things we could do. We could add up all of the
volumes in the x-direction, between the lower x-bound and
the upper x-bound, and then we'd have kind of a thin sheet,
although it will already have some depth, because we're
not adding up just the x's. There's also a dy back there. So we would have a volume of a
figure that would extend from the lower x all the way to the
upper x, go back dy, and come back here. If we wanted to sum
up all the dx's. And if we wanted to do that,
which expression would we use? Well, we would be summing with
respect to x first, so we could use this expression, right? And actually, we could
write it here, but it just becomes confusing. If we're summing with respect
to x, but we have the dy written here first. It's really not incorrect, but
it just becomes a little ambiguous, are we summing
with respect to x or y. But here, we could say, OK. If we want to sum up all the
dx's first, let's do that. We're taking the sum with
respect to x, and let me, I'm going to write down the actual,
normally I just write numbers here, but I'm going to say,
well, the lower bound here is x is equal to a, and the upper
bound here is x is equal to b. And that'll give us the volume
of, you could imagine a sheet with depth, right? The sheet is going to be
parallel to the x-axis, right? And then once we have that
sheet, in my video, I think that's the newspaper people
trying to sell me something. Anyway. So once we have the sheet, I'll
try to draw it here, too, I don't want to get this picture
too muddied up, but once we have that sheet, then we can
integrate those, we can add up the dy's, right? Because this width right
here is still dy. We could add up of all the
different dy's, and we would have the volume
of the whole figure. So once we take this sum,
then we could take this sum. Where y is going from it's
bottom, which we said with c, from y is equal to c to y's
upper bound, to y is equal to d. Fair enough. And then, once we evaluate
this whole thing, we have the volume of this solid, or the
volume under the surface. Now we could have
gone the other way. I know this gets a little
bit messy, but I think you get what I'm saying. Let's start with that little
da we had originally. Instead of going this way,
instead of summing up the dx's and getting this sheet, let's
sum up the dy's first, right? So we could take, we're summing
in the y-direction first. We would get a sheet that's
parallel to the y-axis, now. So the top of the sheet would
look something like that. So if we're coming the dy's
first, we would take the sum, we would take the integral with
respect to y, and it would be, the lower bound would be y is
equal to c, and the upper bound is y is equal to d. And then we would have that
sheet with a little depth, the depth is dx, and then we could
take the sum of all of those, sorry, my throat is dry. I just had a bunch of almonds
to get power to be able to record these videos. But once I have one of these
sheets, and if I want to sum up all of the x's, then I could
take the infinite sum of infinitely small columns, or in
this view, sheets, infinitely small depths, and the lower
bound is x is equal to a, and the upper bound is
x is equal to b. And once again, I would have
the volume of the figure. And all I showed you here is
that there's two ways of doing the order of integration. Now, another way of saying
this, if this little original square was da, and this is a
shorthand that you'll see all the time, especially in
physics textbooks, is that we are integrating along
the domain, right? Because the x-y plane
here is our domain. So we're going to do a double
integral, a two-dimensional integral, we're saying that the
domain here is two-dimensional, and we're going to take that
over f of x and y times da. And the reason why I want to
show you this, is you see this in physics books all the time. I don't think it's a
great thing to do. Because it is a shorthand, and
maybe it looks simpler, but for me, whenever I see something
that I don't know how to compute or that's not obvious
for me to know how to compute, it actually is more confusing. So I wanted to just show you
that what you see in this physics book, when someone
writes this, it's the exact same thing as this or this. The da could either be dx times
dy, or it could either be dy times dx, and when they do this
double integral over domain, that's the same thing is just
adding up all of these squares. Where we do it here, we're
very ordered about it, right? We go in the x-direction, and
then we add all of those up in the y-direction, and we
get the entire volume. Or we could go the
other way around. When we say that we're just
taking the double integral, first of all, that tells us
we're doing it in two dimensions, over a domain,
that leaves it a little bit ambiguous in terms of
how we're going to sum up all of the da's. And they do it intentionally in
physics books, because you don't have to do it using
Cartesian coordinates, using x's and y's. You can do it in polar
coordinates, you could do it a ton of different ways. But I just wanted to show
you, this is another way to having an intuition of the
volume under a surface. And these are the exact same
things as this type of notation that you might
see in a physics book. Sometimes they won't write
a domain, sometimes they'd write over a surface. And we'll later do
those integrals. Here the surface is easy, it's
a flat plane, but sometimes it'll end up being a curve
or something like that. But anyway, I'm
almost out of time. I will see you in
the next video.