Welcome to an intro to triple integrals.
The idea of a triple integral
is really just an extension of a double integral.
So let's start by reviewing what we know
about a double integral.
We said before that the limit as n approaches infinity
of this infinite sum of f of x, y, times delta A
represents the double integral over the region R
of f of x, y, with respect to A.
So this delta A here represented a small change in the area,
which gave us differential A,
and differential A can be expressed as dx, dy, or dy, dx.
So a small change in x times a small change in y,
or we could express this as dy, dx,
resulting in the same small change of area.
For a triple integral,
we have the limit as n approaches infinity
of an infinite sum of f of x, y, z,
times a change in volume, rather than a change in area.
So this gives us a triple integral over a solid region V,
where the small change in V
or differential V can be expressed as dx, dy, dz,
or any other combination of these three differentials.
So a triple integral integrates
over a three dimensional region V,
whereas in a double integral
it integrates over a two dimensional region which we call R.
Because triple integrals have three differentials,
there are six possible orders of integration.
In the next several videos I'll be taking a look
at how you can use triple integrals
to determine volume of solids as well as other applications.
But on this video we're going focus on
practicing evaluating triple integrals.
There's a couple things to mention before we do that.
First, if f is continuous
on a solid region V defined by x, y, and z as we see here,
and h1, h2, g1 and g2 are all continuous functions,
then the triple integral
over the solid region V of f of x, y, z
with respects to V expressed as we see here
where we integrate with respects to z first,
and then second with respect to y,
and then third with respect to x,
the limits of integration of the innermost integral
with respects to V must be functions
of the outer limits of the integration
or functions of x and y,
and if we integrate with respects to y second,
the limits of integration
with respect to y must be expressed as functions of x,
and then lastly the limits of integration for x
would be constants.
Now all of these could be constants,
which we'll see in just a minute,
but if they are expressed in functions,
and this is your order of integration,
they must be expressed in this manner.
Of course if the order changes,
because there are six possible orders,
this would affect how these functions would be expressed
in terms of x, y, and z.
Let's go and take a look at our first example.
Here we want to sketch
the solid region of integration and then evaluate.
Because we're integrating with respect to z first,
z is going to be on the closed interval from zero to three,
y will be on the closed interval from zero to four,
and then x will be on the closed interval from zero to two.
So the region of integration would be
the boxed form by these intervals.
Notice in the in the x-y plane
we would have a two-by-four rectangle, something like this.
But then z is on the closed interval from zero to three
so it would be all the way up to here,
and so the region of integration
would be the rectangular prism or the box formed here.
Keep in mind this triple integral
does not represent this volume,
this volume is the region of integration.
Let's go ahead and evaluate this.
So you first integrate with respect to z,
treating x and y as constants.
So we'd have x, y, z squared over two,
and evaluate this at zero and three,
keeping in mind we're replacing z with three and zero.
So when z is equal to three
we're going to have nine, x, y over two,
or nine halves, x, y,
and then when z is zero we'd have zero.
Now we're going to integrate with respects to y.
We're going to have nine halves,
x, y to the second over two.
Now we'll replace y with four and zero.
So when y is equal to four, so it'd be 16 divided by four,
that's four times nine, that's going to be 36, x.
And when y is zero this will all be zero.
So I'll have 36, x.
And then lastly we integrate with respect to x.
It'd give us 36 times x squared over two,
so when x is equal to two,
I'll have four divided by two, that's two.
Times 36, that's 72,
and then when x is zero this is zero.
So the result is 72.
Let's take a look at another example.
Notice we first integrate with respect to z,
so we'll have z squared over two or one half z squared.
We evaluate this at limits of integration.
This is nice, this is z squared
because when we set up in our upper limits of integration
we're just going to have
one half times y squared minus x squared.
And then when z is zero this would be zero.
I think I'll go ahead and distribute the one half.
Now we'll integrate with respects to x,
treating y as a constant.
What you want to have, is one half x, y squared,
here we'll have minus one half
times x to the third over three.
It's important to remember that we're replacing x
with y over two and zero.
So when x is y over two,
here we're going to have y to the third over four
minus one half times y over two cubed,
that's going to be y cubed over eight,
so we're going to have minus one half,
this would be y cubed over eight divided by three
is going to be y cubed over 24.
Let's go and get a common denominator.
Common denominator is going to be 48,
so I'll multiply this by 12.
So 12, y cubed minus one, y cubed will be 11, y cubed.
Let's go ahead an finish this on the next page.
So if you integrate with respects to y,
we're going to have 11/48ths
times y to the fourth over four.
So we're going to have 11,
y to the fourth all over 48 times four.
It's going to be 192.
So when y is equal to two,
we'll have 11 times two to the fourth, all over 192,
minus, and then when y is zero we'd have zero.
So I'll have 176 over 192
which simplifies to 11/12ths.
We'll have to stop here for this video.
We'll take a look at another example
of a triple integral in the next video.