If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:7:26

In the last video, we had this
rectangle, and we used a triple integral to figure
out it's volume. And I know you were probably
thinking, well, I could have just used my basic geometry
to multiply the height times the width times the depth. And that's true because this
was a constant-valued function. And then even once we
evaluated, once we integrated with respect to z, we ended up
with a double integral, which is exactly what you would have
done in the last several videos when we just learned the
volume under a surface. But then we added a twist
at the end of the video. We said, fine, you could have
figured out the volume within this rectangular domain, I
guess, very straightforward using things you already knew. But what if our goal is not
to figure out the volume? Our goal was to figure out the
mass of this volume, and even more, the material that we're
taking the volume of-- whether it's a volume of gas or a
volume of some solid-- that its density is not constant. So now the mass becomes
kind of-- I don't know-- interesting to calculate. And so, what we defined, we
defined a density function. And rho, this p looking thing
with a curvy bottom-- that gives us the density
at any given point. And at the end of the
last video we said, well, what is mass? Mass is just density
times volume. You could view it another way. Density is the same thing
as mass divided by volume. So the mass around a very, very
small point, and we called that d mass, the differential of the
mass, is going equal the density at that point, or the
rough density at exactly that point, times the volume
differential around that point, times the volume of this
little small cube. And then, as we saw it on the
last video, if you're using rectangular coordinates, this
volume differential could just be the x distance times the y
distance times the z distance. So, the density was that our
density function is defined to be x, y, and z, and we
wanted to figure out the mass of this volume. And let's say that our x, y,
and z coordinates-- their values, let's say they're in
meters and let's say this density is in kilograms
per meter cubed. So our answer is going to be in
kilograms if that was the case. And those are kind of the
traditional Si units. So let's figure out the mass of
this variably dense volume. So all we do is we have the
same integral up here. So the differential of mass
is going to be this value, so let's write that down. It is x-- I want to make sure
I don't run out of space. xyz times-- and I'm
going to integrate with respect to dz first. But you could actually
switch the order. Maybe we'll do that
in the next video. We'll do dz first, then we'll
do dy, then we'll do dx. Once again, this is just
the mass at any small differential of volume. And if we integrate with z
first we said z goes from what? The boundaries on
z were 0 to 2. The boundaries on
y were 0 to 4. And the boundaries on
x, x went from 0 to 3. And how do we evaluate this? Well, what is the
antiderivative-- we're integrating with
respect to z first. So what's the antiderivative
of xyz with respect to z? Well, let's see. This is just a constant so
it'll be xyz squared over 2. Right? Yeah, that's right. And then we'll evaluate
that from 2 to 0. And so you get-- I know I'm
going to run out of space. So you're going to get
2 squared, which is 4, divided by 2, which is 2. So it's 2xy minus 0. So when you evaluate just this
first we'll get 2xy, and now you have the other
two integrals left. So I didn't write the
other two integrals down. Maybe I'll write it down. So then you're left
with two integrals. You're left with dy and dx. And y goes from 0 to 4
and x goes from 0 to 3. I'm definitely going
to run out of space. And now you take the
antiderivative of this with respect to y. So what's the antiderivative
of this with respect to y? Let me erase some stuff just
so I don't get too messy. I was given the very good
suggestion of making it scroll, but, unfortunately,
I didn't make it scroll enough this time. So I can delete this
stuff, I think. Oops, I deleted some of that. But you know what I deleted. OK, so let's take
the antiderivative with respect to y. I'll start it up here
where I have space. OK, so the antiderivative of
2xy with respect to y is y squared over 2, 2's cancel out. So you get xy squared. And y goes from 0 to 4. And then we still have the
outer integral to do. x goes from 0 to 3 dx. And when y is equal
to 4 you get 16x. And then when y is 0
the whole thing is 0. So you have 16x integrated
from 0 to 3 dx. And that is equal to what? 8x squared. And you evaluate
it from 0 to 3. When it's 3, 8 times 9 is 72. And 0 times 8 is 0. So the mass of our figure-- the
volume we figured out last time was 24 meters cubed. I erased it, but if you
watched the last video that's what we learned. But it's mass is 72 kilograms. And we did that by integrating
this 3 variable density function-- this function
of 3 variables. Or in three-dimensions
you can view it as a scalar field, right? At any given point, there
is a value, but not really a direction. And that value is a density. But we integrated the scalar
field in this volume. So that's kind of the new
skill we learned with the triple integral. And in the next video I'll
show you how to set up more complicated triple integrals. But the real difficulty with
triple integrals is-- and I think you'll see that your
calculus teacher will often do this-- when you're doing triple
integrals, unless you have a very easy figure like this, the
evaluation-- if you actually wanted to analytically evaluate
a triple integral that has more complicated boundaries or more
complicated for example, a density function. The integral gets very
hairy, very fast. And it's often very difficult
or very time consuming to evaluate it analytically just
using your traditional calculus skills. So you'll see that on a lot of
calculus exams when they start doing the triple integral, they
just want you to set it up. They take your word for it that
you've done so many integrals so far that you could
take the antiderivative. And sometimes, if they really
want to give you something more difficult they'll just say,
well, switch the order. You know, this is the integral
when we're dealing with respect to z, then y, then x. We want you to rewrite
this integral when you switch the order. And we will do that
in the next video. See you soon.