Triple Integrals (Part - 1) Video Lecture - Engineering Mathematics

Let's say I wanted to find the
volume of a cube, where the
values of the cube-- let's say
x is between-- x is greater
than or equal to 0, is less
than or equal to,
I don't know, 3.
Let's say y is greater than
or equal to 0, and is
less than or equal to 4.
And then let's say that z is
greater than or equal to 0 and
is less than or equal to 2.
And I know, using basic
geometry you could figure out--
you know, just multiply the
width times the height times
the depth and you'd
have the volume.
But I want to do this example,
just so that you get used to
what a triple integral looks
like, how it relates to a
double integral, and then later
in the next video we could do
something slightly
more complicated.
So let's just draw
that, this volume.
So this is my x-axis, this is
my z-axis, this is the y.
x, y, z.
OK.
So x is between 0 and 3.
So that's x is equal to 0.
This is x is equal to--
let's see, 1, 2, 3.
y is between 0 and 4.
1, 2, 3, 4.
So the x-y plane will look
something like this.
The kind of base of our cube
will look something like this.
And then z is between 0 and 2.
So 0 is the x-y plane,
and then 1, 2.
So this would be the top part.
And maybe I'll do that in a
slightly different color.
So this is along the x-z axis.
You'd have a boundary
here, and then it would
come in like this.
You have a boundary here,
come in like that.
A boundary there.
So we want to figure out
the volume of this cube.
And you could do it.
You could say, well, the depth
is 3, the base, the width is 4,
so this area is 12
times the height.
12 times 2 is 24.
You could say it's 24
cubic units, whatever
units we're doing.
But let's do it as
a triple integral.
So what does a triple
integral mean?
Well, what we could do is we
could take the volume of a very
small-- I don't want to say
area-- of a very small volume.
So let's say I wanted to take
the volume of a small cube.
Some place in this-- in the
volume under question.
And it'll start to make more
sense, or it starts to become a
lot more useful, when we have
variable boundaries and
surfaces and curves
as boundaries.
But let's say we want to
figure out the volume of this
little, small cube here.
That's my cube.
It's some place in this larger
cube, this larger rectangle,
cubic rectangle, whatever
you want to call it.
So what's the volume
of that cube?
Let's say that its width is dy.
So that length
right there is dy.
It's height is dx.
Sorry, no, it's
height is dz, right?
The way I drew it,
z is up and down.
And it's depth is dx.
This is dx.
This is dz.
This is dy.
So you can say that a small
volume within this larger
volume-- you could call that
dv, which is kind of the
volume differential.
And that would be equal to,
you could say, it's just
the width times the
length times the height.
dx times dy times dz.
And you could switch the
orders of these, right?
Because multiplication is
associative, and order
doesn't matter and all that.
But anyway, what can you
do with it in here?
Well, we can take the integral.
All integrals help us do is
help us take infinite sums of
infinitely small distances,
like a dz or a dx or
a dy, et cetera.
So, what we could do is we
could take this cube and
first, add it up in, let's
say, the z direction.
So we could take that cube and
then add it along the up and
down axis-- the z-axis--
so that we get the
volume of a column.
So what would that look like?
Well, since we're going up and
down, we're adding-- we're
taking the sum in
the z direction.
We'd have an integral.
And then what's the
lowest z value?
Well, it's z is equal to 0.
And what's the upper bound?
Like if you were to just take--
keep adding these cubes, and
keep going up, you'd run
into the upper bound.
And what's the upper bound?
It's z is equal to 2.
And of course, you would
take the sum of these dv's.
And I'll write dz first.
Just so it reminds us
that we're going to
take the integral with
respect to z first.
And let's say we'll do y next.
And then we'll do x.
So this integral, this value,
as I've written it, will
figure out the volume of a
column given any x and y.
It'll be a function of x and y,
but since we're dealing with
all constants here, it's
actually going to be
a constant value.
It'll be the constant value
of the volume of one
of these columns.
So essentially, it'll
be 2 times dy dx.
Because the height of one
of these columns is 2,
and then its with and
its depth is dy and dx.
So then if we want to figure
out the entire volume-- what
we did just now is we figured
out the height of a column.
So then we could take those
columns and sum them
in the y direction.
So if we're summing in the y
direction, we could just take
another integral of this
sum in the y direction.
And y goes from 0 to what?
y goes from 0 to 4.
I wrote this integral a
little bit too far to the
left, it looks strange.
But I think you get the idea.
y is equal to 0, to
y is equal to 4.
And then that'll give us the
volume of a sheet that is
parallel to the zy plane.
And then all we have left to do
is add up a bunch of those
sheets in the x direction, and
we'll have the volume
of our entire figure.
So to add up those sheets,
we would have to sum
in the x direction.
And we'd go from x is equal
to 0, to x is equal to 3.
And to evaluate this
is actually fairly
straightforward.
So, first we're taking the
integral with respect to z.
Well, we don't have anything
written under here, but we
can just assume that
there's a 1, right?
Because dz times dy times
dx is the same thing as
1 times dz times dy dx.
So what's the value
of this integral?
Well, the antiderivative
of 1 with respect to
z is just z, right?
Because the derivative
of z is 1.
And you evaluate
that from 2 to 0.
So then you're left with--
so it's 2 minus 0.
So you're just left with 2.
So you're left with 2, and you
take the integral of that from
y is equal to 0, to y is equal
to 4 dy, and then
you have the x.
From x is equal to 0,
to x is equal to 3 dx.
And notice, when we just took
the integral with respect to
z, we ended up with
a double integral.
And this double integral is the
exact integral we would have
done in the previous videos on
the double integral, where you
would have just said, well,
z is a function of x and y.
So you could have written, you
know, z, is a function of x
and y, is always equal to 2.
It's a constant function.
It's independent of x and y.
But if you had defined z in
this way, and you wanted to
figure out the volume under
this surface, where the surface
is z is equal to 2-- you
know, this is a surface, is z
is equal to 2-- we would
have ended up with this.
So you see that what we're
doing with the triple
integral, it's really,
really nothing different.
And you might be wondering,
well, why are we
doing it at all?
And I'll show you
that in a second.
But anyway, to evaluate
this, you could take the
antiderivative of this with
respect to y, you get 2y-- let
me scroll down a little bit.
You get 2y evaluating
that at 4 and 0.
And then, so you get 2 times 4.
So it's 8 minus 0.
And then you integrate
that from, with respect
to x from 0 to 3.
So that's 8x from 0 to 3.
So that'll be equal to
24 four units cubed.
So I know the obvious question
is, what is this good for?
Well, when you have a kind
of a constant value within
the volume, you're right.
You could have just done
a double integral.
But what if I were to tell you,
our goal is not to figure out
the volume of this figure.
Our goal is to figure out
the mass of this figure.
And even more, this volume--
this area of space or
whatever-- its mass
is not uniform.
If its mass was uniform, you
could just multiply its uniform
density times its volume,
and you'd get its mass.
But let's say the
density changes.
It could be a volume of some
gas or it could be even some
material with different
compounds in it.
So let's say that its density
is a variable function
of x, y, and z.
So let's say that the density--
this row, this thing that looks
like a p is what you normally
use in physics for density-- so
its density is a function
of x, y, and z.
Let's-- just to make it
simple-- let's make
it x times y times z.
If we wanted to figure out the
mass of any small volume, it
would be that volume times
the density, right?
Because density-- the units of
density are like kilograms
per meter cubed.
So if you multiply it times
meter cubed, you get kilograms.
So we could say that the mass--
well, I'll make up notation, d
mass-- this isn't a function.
Well, I don't want to write it
in parentheses, because it
makes it look like a function.
So, a very differential mass,
or a very small mass, is going
to equal the density at that
point, which would be xyz,
times the volume of that
of that small mass.
And that volume of that small
mass we could write as dv.
And we know that dv is the
same thing as the width times
the height times the depth.
dv doesn't always have to
be dx times dy times dz.
If we're doing other
coordinates, if we're doing
polar coordinates, it could be
something slightly different.
And we'll do that eventually.
But if we wanted to figure out
the mass, since we're using
rectangular coordinates, it
would be the density function
at that point times our
differential volume.
So times dx dy dz.
And of course, we can
change the order here.
So when you want to figure out
the volume-- when you want to
figure out the mass-- which I
will do in the next video, we
essentially will have to
integrate this function.
As opposed to just
1 over z, y and x.
And I'm going to do that
in the next video.
And you'll see that it's really
just a lot of basic taking
antiderivatives and avoiding
careless mistakes.
I will see you in
the next video.