Double integrals 2 | Double and triple integrals | Multivariable Calculus | Khan Academy

You hopefully have a little intuition now on what a double
integral is or how we go about figuring out the volume
under a surface.
So let's actually compute it and I think it'll all become
a lot more concrete.
So let's say I have the surface, z, and it's a
function of x and y.
And it equals xy squared.
It's a surface in three-dimensional space.
And I want to know the volume between this
surface and the xy-plane.
And the domain in the xy-plane that I care about is x is
greater than or equal to 0, and less than or equal to 2.
And y is greater than or equal to 0, and less
than or equal to 1.
Let's see what that looks like just so we have a
good visualization of it.
So I graphed it here and we can rotate it around.
This is z equals xy squared.
This is the bounding box, right? x goes from 0 to
2; y goes from 0 to 1.
We literally want this-- you could almost view it the
volume-- well, not almost.
Exactly view it as the volume under this surface.
Between this surface, the top surface, and the xy-plane.
And I'll rotate it around so you can get a little bit better
sense of the actual volume.
Let me rotate.
Now I should use the mouse for this.
So it's this space, underneath here.
It's like a makeshift shelter or something.
I could rotate it a little bit.
Whatever's under this, between the two surfaces--
that's the volume.
Whoops, I've flipped it.
There you go.
So that's the volume that we care about.
Let's figure out how to do and we'll try to gather a little
bit of the intuition as we go along.
So I'm going to draw a not as impressive version of that
graph, but I think it'll do the job for now.
Let me draw my axes.
That's my x-axis, that's my y-axis, and that's my z-axis.
x, y, z.
x is going from 0 to 2.
Let's say that's 2.
y is going from 0 to 1.
So we're taking the volume above this rectangle
in the xy-plane.
And then the surface, I'm going to try my best to draw it.
I'll draw it in a different color.
I'm looking at the picture.
At this end it looks something like this.
And then it has a straight line.
Let's see if I can draw this surface going down like that.
And then if I was really good I could shade it.
It looks something like this.
If I were to shade it, the surface looks
something like that.
And this right here is above this.
This is the bottom left corner, you can almost view it.
So let me just say the yellow is the top of the surface.
That's the top of the surface.
And then this is under the surface.
Let me show you the actual surface.
So this volume underneath here.
I think you get the idea.
So how do we do that?
Well, in the last example we said, well, let's pick an
arbitrary y and for that y, let's figure out the
area under the curve.
So if we fix some y-- when you actually do the problem, you
don't have to think about it in this much detail, but I want
to give you the intuition.
So if we pick just an arbitrary y here.
So on that y, we could think of it-- if we have a fixed y, then
the function of x and y you can almost view it as a function
of just x for that given y.
And so, we're kind of figuring out the value of this, of
the area under this curve.
You should view this as kind of an up down curve for a given y.
So if we know a y we can figure out then-- for example, if y
was 5, this function would become z equals 25x.
And then that's easy to figure out the value
of the curve under.
So we'll make the value under the curve as a function of y.
We'll pretend like it's just a constant.
So let's do that.
So if we have a dx that's our change in x.
And then our height of each of our rectangles is going to be a
function-- it's going to be z.
The height is z, which is a function of x and y.
So we can take the integral.
So the area of each of these is going to be our function, xy
squared-- I'll do it here because I'll run out of space.
xy squared times the width, which is dx.
And if we want the area of this slice for a given y, we just
integrate along the x-axis.
We're going to integrate from x is equal to 0
to x is equal to 2.
From x is equal to 0 to 2.
Fair enough.
Now, but we just don't want to figure out the area under the
curve at one slice, for one particular y, we want to
figure out the entire area of the curve.
So what we do is we say, OK, fine.
The area under the curve, not the surface-- under this curve
for a particular y, is this expression.
Well, what if I gave it a little bit of depth?
If I multiplied this area times dy then it would give me a
little bit of depth, right?
We'd kind of have a three-dimensional slice of the
I know it's hard to imagine.
Let me bring that here.
So if I had a slice here, we just figured out the area of a
slice and then I'm multiplying it by dy to give it a
little bit of depth.
So you multiply it by dy to give it a little bit of depth,
and then if we want the entire volume under the curve we add
all the dy's together, take the infinite sum of these
infinitely small volumes really now.
And so we will integrate from y is equal to 0
to y is equal to 1.
I know this graph is a little hard to understand, but you
might want to re-watch the first video.
I had a slightly easier to understand surface.
So now, how do we evaluate this?
Well, like we said, you evaluate from the
inside and go outward.
It's taking a partial derivative in reverse.
So we're integrating here with respect to x, so we can treat
y just like a constant.
Like it's like the number 5 or something like that.
So it really doesn't change the integral.
So what's the antiderivative of xy squared?
Well, the antiderivative of xy squared-- I want to make
sure I'm color consistent.
Well, the antiderivative of x is x to the 1/2--
sorry. x squared over 2.
And then y squared is just a constant, right?
And then we don't have to worry about plus c since
this is a definite integral.
And we're going to evaluate that at 2 and 0.
And then we still have the outside integral
with respect to y.
So once we figure that out we're going to integrate it
from 0 to 1 with respect to dy.
Now what does this evaluate?
We put a 2 in here.
If you put a 2 in there you get 2 squared over 2.
That's just 4 over 2.
So it's 2 y squared.
Minus 0 squared over 2 times y squared.
Well, that's just going to be 0.
So it's minus 0.
I won't write that down because hopefully that's a little
bit of second nature to you.
We just evaluated this at the 2 endpoints and
I'm short for space.
So this evaluated at 2y squared and now we evaluate
the outside integral.
0, 1 dy.
And this is an important thing to realize.
When we evaluated this inside integral, remember
what we were doing?
We were trying to figure out for a given y, what the
area of this surface was.
Well, not this surface, the area under the surface
for a given y.
For a given y that surface kind of turns into a curve.
And we tried to figure out the area under that curve
This ended up being a function of y.
And that makes sense because depending on which y we pick
we're going to get a different area here.
Obviously, depending on which y we pick, the area-- kind of a
wall dropped straight down-- that area's going to change.
So we got a function of y when we evaluated this and now we
just integrate with respect to y and this is just plain old
vanilla definite integration.
What's the antiderivative of 2y squared?
Well, that equals 2 times y to the third over 3,
or 2/3 y to the third.
We're going to evaluate that at 1 and 0, which
is equal to-- let's see.
1 to the third times 2/3.
That's 2/3.
Minus 0 to the third times 2/3.
Well, that's just 0.
So it equals 2/3.
If our units were meters these would be 2/3 meters
cubed or cubic meters.
But there you go.
That's how you evaluate a double integral.
There really isn't a new skill here.
You just have to make sure to keep track of the variables.
Treat them constant.
They need to be treated constant, and then treat them
as a variable of integration when it's appropriate.
Anyway, I will see you in the next video.