WEBVTT

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Let's say I wanted to find the
volume of a cube, where the

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values of the cube-- let's say
x is between-- x is greater

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than or equal to 0, is less
than or equal to,

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I don't know, 3.

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Let's say y is greater than
or equal to 0, and is

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less than or equal to 4.

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And then let's say that z is
greater than or equal to 0 and

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is less than or equal to 2.

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And I know, using basic
geometry you could figure out--

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you know, just multiply the
width times the height times

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the depth and you'd
have the volume.

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But I want to do this example,
just so that you get used to

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what a triple integral looks
like, how it relates to a

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double integral, and then later
in the next video we could do

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something slightly
more complicated.

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So let's just draw
that, this volume.

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So this is my x-axis, this is
my z-axis, this is the y.

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x, y, z.

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OK.

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So x is between 0 and 3.

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So that's x is equal to 0.

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This is x is equal to--
let's see, 1, 2, 3.

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y is between 0 and 4.

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1, 2, 3, 4.

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So the x-y plane will look
something like this.

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The kind of base of our cube
will look something like this.

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And then z is between 0 and 2.

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So 0 is the x-y plane,
and then 1, 2.

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So this would be the top part.

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And maybe I'll do that in a
slightly different color.

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So this is along the x-z axis.

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You'd have a boundary
here, and then it would

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come in like this.

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You have a boundary here,
come in like that.

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A boundary there.

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So we want to figure out
the volume of this cube.

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And you could do it.

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You could say, well, the depth
is 3, the base, the width is 4,

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so this area is 12
times the height.

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12 times 2 is 24.

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You could say it's 24
cubic units, whatever

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units we're doing.

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But let's do it as
a triple integral.

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So what does a triple
integral mean?

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Well, what we could do is we
could take the volume of a very

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small-- I don't want to say
area-- of a very small volume.

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So let's say I wanted to take
the volume of a small cube.

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Some place in this-- in the
volume under question.

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And it'll start to make more
sense, or it starts to become a

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lot more useful, when we have
variable boundaries and

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surfaces and curves
as boundaries.

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But let's say we want to
figure out the volume of this

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little, small cube here.

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That's my cube.

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It's some place in this larger
cube, this larger rectangle,

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cubic rectangle, whatever
you want to call it.

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So what's the volume
of that cube?

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Let's say that its width is dy.

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So that length
right there is dy.

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It's height is dx.

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Sorry, no, it's
height is dz, right?

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The way I drew it,
z is up and down.

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And it's depth is dx.

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This is dx.

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This is dz.

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This is dy.

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So you can say that a small
volume within this larger

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volume-- you could call that
dv, which is kind of the

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volume differential.

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And that would be equal to,
you could say, it's just

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the width times the
length times the height.

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dx times dy times dz.

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And you could switch the
orders of these, right?

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Because multiplication is
associative, and order

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doesn't matter and all that.

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But anyway, what can you
do with it in here?

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Well, we can take the integral.

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All integrals help us do is
help us take infinite sums of

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infinitely small distances,
like a dz or a dx or

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a dy, et cetera.

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So, what we could do is we
could take this cube and

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first, add it up in, let's
say, the z direction.

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So we could take that cube and
then add it along the up and

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down axis-- the z-axis--
so that we get the

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volume of a column.

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So what would that look like?

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Well, since we're going up and
down, we're adding-- we're

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taking the sum in
the z direction.

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We'd have an integral.

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And then what's the
lowest z value?

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Well, it's z is equal to 0.

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And what's the upper bound?

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Like if you were to just take--
keep adding these cubes, and

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keep going up, you'd run
into the upper bound.

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And what's the upper bound?

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It's z is equal to 2.

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And of course, you would
take the sum of these dv's.

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And I'll write dz first.

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Just so it reminds us
that we're going to

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take the integral with
respect to z first.

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And let's say we'll do y next.

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And then we'll do x.

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So this integral, this value,
as I've written it, will

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figure out the volume of a
column given any x and y.

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It'll be a function of x and y,
but since we're dealing with

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all constants here, it's
actually going to be

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a constant value.

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It'll be the constant value
of the volume of one

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of these columns.

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So essentially, it'll
be 2 times dy dx.

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Because the height of one
of these columns is 2,

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and then its with and
its depth is dy and dx.

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So then if we want to figure
out the entire volume-- what

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we did just now is we figured
out the height of a column.

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So then we could take those
columns and sum them

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in the y direction.

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So if we're summing in the y
direction, we could just take

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another integral of this
sum in the y direction.

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And y goes from 0 to what?
y goes from 0 to 4.

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I wrote this integral a
little bit too far to the

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left, it looks strange.

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But I think you get the idea.

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y is equal to 0, to
y is equal to 4.

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And then that'll give us the
volume of a sheet that is

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parallel to the zy plane.

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And then all we have left to do
is add up a bunch of those

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sheets in the x direction, and
we'll have the volume

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of our entire figure.

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So to add up those sheets,
we would have to sum

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in the x direction.

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And we'd go from x is equal
to 0, to x is equal to 3.

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And to evaluate this
is actually fairly

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straightforward.

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So, first we're taking the
integral with respect to z.

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Well, we don't have anything
written under here, but we

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can just assume that
there's a 1, right?

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Because dz times dy times
dx is the same thing as

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1 times dz times dy dx.

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So what's the value
of this integral?

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Well, the antiderivative
of 1 with respect to

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z is just z, right?

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Because the derivative
of z is 1.

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And you evaluate
that from 2 to 0.

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So then you're left with--
so it's 2 minus 0.

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So you're just left with 2.

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So you're left with 2, and you
take the integral of that from

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y is equal to 0, to y is equal
to 4 dy, and then

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you have the x.

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From x is equal to 0,
to x is equal to 3 dx.

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And notice, when we just took
the integral with respect to

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z, we ended up with
a double integral.

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And this double integral is the
exact integral we would have

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done in the previous videos on
the double integral, where you

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would have just said, well,
z is a function of x and y.

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So you could have written, you
know, z, is a function of x

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and y, is always equal to 2.

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It's a constant function.

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It's independent of x and y.

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But if you had defined z in
this way, and you wanted to

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figure out the volume under
this surface, where the surface

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is z is equal to 2-- you
know, this is a surface, is z

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is equal to 2-- we would
have ended up with this.

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So you see that what we're
doing with the triple

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integral, it's really,
really nothing different.

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And you might be wondering,
well, why are we

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doing it at all?

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And I'll show you
that in a second.

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But anyway, to evaluate
this, you could take the

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antiderivative of this with
respect to y, you get 2y-- let

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me scroll down a little bit.

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You get 2y evaluating
that at 4 and 0.

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And then, so you get 2 times 4.

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So it's 8 minus 0.

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And then you integrate
that from, with respect

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to x from 0 to 3.

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So that's 8x from 0 to 3.

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So that'll be equal to
24 four units cubed.

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So I know the obvious question
is, what is this good for?

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Well, when you have a kind
of a constant value within

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the volume, you're right.

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You could have just done
a double integral.

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But what if I were to tell you,
our goal is not to figure out

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the volume of this figure.

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Our goal is to figure out
the mass of this figure.

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And even more, this volume--
this area of space or

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whatever-- its mass
is not uniform.

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If its mass was uniform, you
could just multiply its uniform

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density times its volume,
and you'd get its mass.

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But let's say the
density changes.

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It could be a volume of some
gas or it could be even some

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material with different
compounds in it.

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So let's say that its density
is a variable function

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of x, y, and z.

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So let's say that the density--
this row, this thing that looks

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like a p is what you normally
use in physics for density-- so

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its density is a function
of x, y, and z.

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Let's-- just to make it
simple-- let's make

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it x times y times z.

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If we wanted to figure out the
mass of any small volume, it

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would be that volume times
the density, right?

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Because density-- the units of
density are like kilograms

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per meter cubed.

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So if you multiply it times
meter cubed, you get kilograms.

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So we could say that the mass--
well, I'll make up notation, d

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mass-- this isn't a function.

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Well, I don't want to write it
in parentheses, because it

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makes it look like a function.

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So, a very differential mass,
or a very small mass, is going

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to equal the density at that
point, which would be xyz,

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times the volume of that
of that small mass.

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And that volume of that small
mass we could write as dv.

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And we know that dv is the
same thing as the width times

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the height times the depth.

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dv doesn't always have to
be dx times dy times dz.

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If we're doing other
coordinates, if we're doing

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polar coordinates, it could be
something slightly different.

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And we'll do that eventually.

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But if we wanted to figure out
the mass, since we're using

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rectangular coordinates, it
would be the density function

00:10:03.550 --> 00:10:07.030
at that point times our
differential volume.

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So times dx dy dz.

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And of course, we can
change the order here.

00:10:13.870 --> 00:10:16.386
So when you want to figure out
the volume-- when you want to

00:10:16.386 --> 00:10:19.000
figure out the mass-- which I
will do in the next video, we

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essentially will have to
integrate this function.

00:10:21.290 --> 00:10:27.400
As opposed to just
1 over z, y and x.

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And I'm going to do that
in the next video.

00:10:28.690 --> 00:10:32.050
And you'll see that it's really
just a lot of basic taking

00:10:32.050 --> 00:10:34.700
antiderivatives and avoiding
careless mistakes.

00:10:34.700 --> 00:10:37.280
I will see you in
the next video.

00:10:37.280 --> 00:10:37.900