# Double integrals 5 | Double and triple integrals | Multivariable Calculus | Khan Academy

In all of the double integrals we've done so
far, the boundaries on x and y were fixed.
Now we'll see what happens when the boundaries on
x and y are variables.
So let's say I have the same surface, and I'm not going to
draw it the way it looks, I'll just kind of draw
it figuratively.
But the problem we're actually going to do is z, and this is
the exact same one we've been doing all along.
The point of here isn't to show you how to integrate, the point
of here is to show you how to visualize and think
And frankly, in double integral problems the hardest part is
figuring out the boundaries.
Once you do that, the integration is pretty
straightforward.
It's really not any harder then single variable integration.
So let's say that's our surface: z is equal
to xy squared.
Let me draw the axes again.
So that's my x-axis.
That's my z-axis.
That's my y-axis.
x, y, and z.
And you saw what this graph looked like several videos ago.
I took out the whole grapher and we rotated and things.
I'm not going to draw the graph the way it looks; I'm just
going to brought fairly abstractly as just an
abstract surface.
Because the point here it's really to figure out the
boundaries of integration.
Before I actually even draw the surface, I'm going
to draw the boundary.
The first time we did this problem we said, OK, x goes
from 0 to 2, y goes from 0 to 1, and then we figured out the
volume above that bounded domain.
Now let's do something else.
Let's say that x goes from 0 to 1.
And let's say that the volume that we want to figure out
under the surface, it's not from a fixed y to
an upper-bound y.
I'll show you: it's actually a curve.
So this is all on the xy plane, everything I'm drawing here.
And this curve, we could view it two ways: we could say y is
a function of x, y is equal to x squared.
Or we could write is equal to square root of y.
We don't have to write plus or minus or anything like that
because we're in the first quadrant.
So this is the area above which we want to
figure out the volume.
Let me, yeah, it doesn't hurt to color it in just so we
can really hone in on what we care about.
So that's the area above which we want to
figure out the volume.
You could kind of say, that's our bounded domain.
And so x goes from 0 to 1, and then this point
is going to be what?
That point's going to be 1 comma 1, right?
1 is equal to 1 squared, 1 is equal to the square root of 1.
So this point is y is equal to 1.
And then I'm not going to draw this surface exactly.
I'm just trying to give you a sense of what the volume of
the figure we're trying to calculate is.
If this is just some arbitrary surface-- let me do it in a
different color --so this is the top.
This line is going vertical in the z-direction.
Actually, I could draw it like this, like it's a curve.
And then this curve back here is going to be like a wall.
And maybe I'll paint this side of the wall just so you can see
what it kind of looks like.
Trying my best.
Think you get an idea.
Let me make it a little darker; this is actually more of an
exercise in art than in math, in many ways.
You get the idea.
And then the boundary here is like this.
And this top isn't flat, you know, it could
be curved surface.
I do a little like that, but it's a curved surface.
And we know in the example we're about to do that the
surface right here is z is equal to x squared.
So we want to figure out the volume under this.
So how do we do it?
We could actually use the intuition that I just gave you.
We're essentially just going to take a da, which is a little
small square down here, and that little area, that's the
same thing as the dx-- let me use a darker color --as a dx
times a dy, and then we just have to multiply it times f of
xy, which is this, for each area, and then
some them all up.
And then we could take a sum in the x-direction first
or the y-direction first.
Now before doing that, just to make sure that you have
the intuition because the boundaries are the hard part,
let me just draw our xy plane.
So let me rotate it up like that.
I'm just going to draw our xy plane.
Because that's what matters.
Because the hard part here is just figuring out our
bounds of integration.
So the curve is just y is equal to x squared, look
something like that.
This is the point y is equal to 1.
This is y-axis, this is the x-axis, this is the
point x is equal to 1.
That's not an x, that's a 1.
This is the x.
Anyway, so we want to figure out, how do we sum up this dx
times dy, or this da, along this domain?
So let's draw it.
Let's visually draw it and it doesn't hurt to do this when
you actually have to do the problem because this
frankly is the hard part.
A lot of calculus teachers will just have you set up the
integral and then say, OK, well the rest is easy.
Or the rest is Calc 1.
OK, so this area, this area here is the same thing
as this area here.
So its base is dx and its height is dy.
And then you could imagine that we're looking at
this thing from above.
So the surface is up here some place and we're looking
straight down on it, and so this is just this area.
So let's say we wanted to take the integral with
respect to x first.
So we want to sum up, so if we want the volume above this
column, first of all, is this area times dx, dy, right?
So let's write the volume above that column.
It's going to be the value of the function, the height at
that point, which is xy squared times dx, dy.
This expression gives us the volume above this area, or
this column right here.
And let's say we want the sum in the x direction first.
So we want to sum that dx, sum one here, sum here,
et cetera, et cetera.
So we're going to sum in the x-direction.
So my question to you is, what is our lower
bound of integration?
Well, we're kind of holding our y constant, right?
And so if we go to the left, if we go lower and lower x's we
kind of bump into the curve here.
So the lower bound of integration is
actually the curve.
And what is this curve if we were to write x
is a function of y?
This curve is y is equal to x squared, or x is equal
to the square root of y.
So if we're integrating with respect to x for a fixed y
right here-- we're integrating in the horizontal direction
first --our lower bound is x is equal to the square root of y.
That's interesting.
I think it's the first time you've probably seen a
variable bound integral.
But it makes sense because for this row that we're adding up
right here, the upper bound is easy.
The upper bound is x is equal to 1.
The upper bound is x is equal to 1, but the lower bound is
x is equal to the square root of y.
Because you go back like, oh, I bump into the curve.
And what's the curve?
Well the curve is x is equal to the square root of y because we
don't know which y we picked.
Fair enough.
So once we've figured out the volume-- so that'll give us the
volume above this rectangle right here --and then we
want to add up the dy's.
And remember, there's a whole volume above what
I'm drawing right here.
I'm just drawing this part in the xy plane.
So what we've done just now, this expression, as it's
written right now, figures out the volume above
that rectangle.
Now if we want to figure out the entire volume of the solid,
we integrate along the y-axis.
Or we add up all the dy's.
This was a dy right here, not a dx.
My dx's and dy's look too similar.
So now what is the lower bound on the y-axis if I'm summing
up these rectangles?
Well, the lower bound is y is equal to 0.
So we're going to go from y is equal to 0 to what--
what is the upper bound?
--to y is equal to 1.
And there you have it.
Let me rewrite that integral.
So the double integral is going to be from x is equal to square
root of y to x is equal to 1, xy squared, dx, dy.
And then the y bound, y goes from 0 to y to 1.
I've just realized I've run out of time.
In the next video we'll evaluate this, and then we'll
do it in the other order.
See you soon.