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## Multivariable calculus

### Course: Multivariable calculus > Unit 4

Lesson 5: Double integrals# Double integrals 3

Let's integrate dy first! Created by Sal Khan.

## Want to join the conversation?

- Hi, could you help me please?

How do you invert the integration order? I cant find it in your videos.

Thanks!(5 votes)- In the video "Double Integrals 2", Sal integrate first the x (considering "y" like a constant), and then substitute the values of the definite integral. Now, in this video, he integrate first the y (considering "x" like a constant). In the7:20he mention that the result was the same, either you integrate first x or y.

I hope this helps.(16 votes)

- Why do you say Y to the third, instead of Y cubed or Y to the three, is it an 'American thing'? Im only wondering because it sounds very similar to Y to the 1/3 and could catch someone out.(5 votes)
- This is an interesting observation. As an American, I was always taught with the 'to the third' and never thought anything of it. I never realized the confusion that this could lead to!(12 votes)

- Does this work all the time? I mean can you always switch the way you integrate between dxdy and dydx and get the same answer? Also does this work for triple integrals as well?(6 votes)
- It will ALWAYS work (if f is well behaved), and is a great way to check if you have the right answer!

This result was proven by Fubini (and Tonelli):

http://mathworld.wolfram.com/FubiniTheorem.html

https://en.wikipedia.org/wiki/Fubini%27s_theorem

The latter link talks about the situation where Fubini fails.(6 votes)

- So does that make integration distributive?(2 votes)
- No, it just means that the order of integration does not matter, although sometimes, one order is much easier than another. Look up Fubini's Theorem for more on this.(5 votes)

- Hint: at6:14it looks like sal write dy instead of dx. If you compare it with the other writings of the "y" you find out that he means dx -> to avoid confusion(4 votes)
- When a 3-D graph changes colors like in the pictures, does this denote anything, or is it just designed to better show depth?(2 votes)
- In this case, just to help visualize depth. You can think of it like 'shadows' or 'ray tracing light' on the plane in the graph.

In other cases that you may have seen (*may see*) with Engineering software that graphically depicts the application of forces on an object, you may see color changes used to graphically display the magnitude of the force (*or heat being generated*) being exerted across an area of the object.

As a real world example that you can try: Get or think of bending a piece of plastic back and forth (*such as the tough plastic shells they package stuff in*). As you bend it back and forth, the area around the bend will start to discolor and warp as it breaks down and eventually snaps or tears so you can get the object out of the packaging. That 'discoloration' would be the coloring of the graphic display along the bend to show the heat you are generating by exerting the force to bend the plastic.(4 votes)

- Is there any difference between the double integrals in this video and the double integrals that appear to be one single sign?

I have noticed on the program MathType that there is a symbol that looks like two integrals glued together, and they share the same boundary condition, and the program sems to treat it like one sign that just "looks" like two integrals.

But in this video Khan is essentially just stacking up several separate integrals instead.(1 vote)- They are equivalent.

A double integral is what you have seen on MathType, the two integral symbols together with a region definition below. By Fubini's theorem, we can break the double (or triple) integral up into an iterated or repeated integral, which is what you are seeing here on Khan and which is the norm.

Doing the integration is usually easy to do; more often it is the calculating of the boundaries of the region of integration in order to apply Fubini that is the most difficult part of solving a double or triple integral (well, that and possibly the ensuing algebra post integration) - but that is just my opinion.(6 votes)

- what do you do when you have to evaluate a double integral when no limits are given?..how do you chose which "slice" to take first and then give it a depth?ie with whose respect do we integrate first dx or dy? i hope i made sense while asking.. i couldn't explain my confusion in a concise manner(2 votes)
- Choose which ever makes evaluating the integral easier since by Fubini's Theorem, the order of integration does not matter.(4 votes)

- Is it always okay to swap the order when you trying to find the volume under a surface?(3 votes)
- At 3 minutes in, why does Sal draw the green line for dx out of the axis labeled y? He did a similar thing in the previous video, Double Integrals 2, and I was wondering why dx of some point, a, wasn't the same as putting a line down at x=a, for instance.(2 votes)
- When we draw something that represents dx, whether it is a line, a surface, or a solid, we draw it so that every point on that line (or surface, or solid) has the same or nearly the same x value. So when he drew that green line parallel to the y axis, he was actually drawing that line because every point on that line has the same x value. Then he continued to expand that line in the z direction so that he drew a surface. And every point on that surface has the same exact x value. Then when he finally drew the depth by drawing a line parallel to the x axis and labeled it "dx", it became a solid where the x values of every point on that solid are not quite exact, but they are all nearly the same.

This is similar to any 2D example of integrals, where we typically draw rectangles with a height of y and a width of dx. For this 3D object, Sal drew a solid which had a height of x*y^2, width of y, and a depth of dx.(3 votes)

## Video transcript

In the last a video we figured
out the volume between this surface, which was xy squared
and the xy-plane when x went from 0 to 2 and y
went from 0 to 1. And the way we did it
is we integrated with respect to x first. We said, pick a y, and
let's just figure out the area under the curve. And so we integrated with
respect to x first, and then we integrated with respect to y. But we could have done it
the other way around. So let's do that and just make
sure we got the right answer. So let me erase a lot of this. So remember, our answer was 2/3
when we integrated with respect to x first, and then
with respect to y. But I will show you that we can
integrate the other way around. That's good when you can
get the same answer in two different ways. So let me redraw that graph
because I want to give you the intuition again. So that's my x-axis,
y-axis, z-axis. x, y, z. Then this is my xy-plane. down here. y goes from 0 to 1; x goes from
0 to 2, This is x equals 1, this is equals 2, this is y
equals 1, and then the graph-- I will do my best to draw it. Looks something-- let me get
some contrast going here. So the graph looks
something like this. Let me see if I can draw it. On this side it looks something
like that and then it comes down like that, straight. And then, the volume
we care about. It's actually this volume
underneath the graph. This is the top of the
surface on that side. We care about this volume
underneath the surface. And then when we draw the
bottom of the surface-- let me do it in a darker color--
looks something like this. This is the bottom
underneath the surface. I can even shade it a little
bit just to show you that it's the underneath. Hopefully that's a
decent rendering of it. Let's look back at
what we had before. It's like a page that I just
flipped up at this point, and we care about this volume,
kind of the colored area under there. So let's figure
out how to do it. Last time we integrated
with respect to x first. Let's integrate with
respect to y first. So let's hold x constant. So if we hold x constant what
we could do is for a given x-- let's pick an x. So if we pick a given x,
let's pick the x here. Then what we can do, for a
given x, you can view that function of x and y. If x is a constant, let's say
if x is 1 then z is just equal to y squared. That's easy to figure out the
area under because we can see that x isn't the constant, but
we can treat it as a constant. So for example, for any
given x, we would have a curve like this. What we could do is we could
try to figure out the area of this curve first. So how do we do that? Well, we just said, we could
kind of view this function up here as z is equal to xy
squared because that's exactly what it is. But we're holding x constant. We're treating it
like a constant. To figure out that area we
could take a dy, a change in y, multiply it by the
height, which is xy squared. So we take xy squared, multiply
it by dy, and then if we want this entire area we integrate
it from y is equal to 0 to y is equal to 1. Fair enough. Now once we have that area, if
you want the volume underneath this entire surface what we
could do is we can multiply this area times dx and
get some depth going. Let me pick a nice
color, that's green. So that's our dx. So if we multiply that times
dx we would get some depth. Let me do a darker color,
get some contrast. Sometimes I feel like that
guy who paints on PBS. So now we have the volume of
this, you kind of view it-- the area under the curve times a
dx, so we have some depth here. So it's time dx. And if we want to figure out
the entire volume under this surface-- between the surface
and the xy-plane given this constraint to our domain-- we
just integrate from x is equal to 0 to 2. All right, so let's
think about it. This area in green here
that we started with, that should be a function of x. We held x constant, but
depending on which x you pick this area is going to change. So when we evaluate this
magenta inner integral with respect to y we should
get a function of x. And then when you evaluate
the whole thing we'll get our volumes. So let's do it. Let's evaluate this
inner integral. Hold x constant. What's the antiderivative
of y squared? It's y of the third over 3. The x is a constant, right? We're going to evaluate
that at 1 and at 0. The outer integral is still
with respect to x dx. This is equal to-- let's see. When you evaluate y is equal
to 1 you get 1 to the third. That's 1. So it's x/3 minus when
y is 0 then that whole thing just becomes 0. This purple expression
is just x/3. And then we still have
the outside integral from 0 to 2 dx. So given what x we have, the
area of this green surface-- that was where we started. Given any given x, that
area-- I wanted something with some contrast. This area is x/3 depending
on which x you pick. If x is 1, this area
right here is 1/3. But now we're going to
integrate underneath the entire surface and get our volume. And like I said, when
you integrate it, it's a function of x. So let's do that. And this is just plain old
vanilla, standard integral. So what's the
antiderivative of x? It's x squared over 2. We have a 1/3 there so
it equals x squared over 2 times 3. So x squared over 6. And we're going to evaluate
it at 2 and at 0. 2 squared over 6 is 4/6. Minus 0/6, which is equal to 0. Equals 4/6. What is 4/6? Well, that's just the
same thing as 2/3. So the volume under the surface
is 2/3, and if you watched the previous video you will
appreciate the fact that when we integrated the other way
around, when we did it with respect to x first and then y,
we got the exact same answer. So the universe is in
proper working order. And I've surprisingly,
actually finished this video with extra time. So for fun, we can just spin
this graph and just appreciate the fact that we have figured
out the volume between this surface, xy squared
and the xy-plane. Pretty neat. Anyway, I'll see you
in the next video.