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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
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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.
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So when you want to figure out
the volume-- when you want to
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figure out the mass-- which I
will do in the next video, we
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essentially will have to
integrate this function.
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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.
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And you'll see that it's really
just a lot of basic taking
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antiderivatives and avoiding
careless mistakes.
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I will see you in
the next video.
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