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# Double integrals 2

Video transcript

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. So we care about this
volume underneath here. 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 volume that we care about. 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 in the traditional sense. 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.