Hello!
For this video, we are going to be
talking about, well, we start talking
about multiple integration,
which is, um, pretty much what we're going
to be talking about for the rest of the
semester.
So we're going to start with double
integrals over rectangular regions,
and we're going to begin the way we did
in Calc I. We're going to begin with just
approximating
double integrals. First, let's start out
with a continuous function f of
x, y that's greater than or equal to zero
when it's defined on a
closed rectangle, [which] are defined by
the intervals [a, b] by
[c, d]; so that's defined as
all x, y in R^2 such that x
is between a and b and y is between c
and d.
Note that [a, b] by [c, d] is called a
Cartesian
product. The graph of f then represents a
surface
above the xy-plane with equation z
equals f of x, y,
and so z is going to be the height of
the surface at that point.
So we'll let S be the solid that lies
above R
and under the graph of f; that's visualized
in Figure 5.2 of the text, reproduced
here. So
this shaded region here under the curve
and above the rectangle, that
is our solid S. Just like we did with
functions of one variable, we're not
going to deal directly with the integral
just yet.
So what we're going to do is find the
volume of that solid
S; that is our goal. To do this, we're
going to divide the interval [a, b]
into m, as in Mary, sub-intervals and
divide
the inter- interval [c, d] into n, as in
Noel, sub intervals,
forming m times n rectangles, which we'll
denote by capital R
sub ij, i being, ranging from 1 to m
and j ranging from 1 to n. The area of
each sub rectangle,
delta A, is the product of delta x and
delta y,
where delta x is the difference between
adjoining
x points,
and that's b minus a, all divided by m;
and delta y is the difference between
adjacent y, um, grid points,
and that's d minus c, all divided by
n. Right. So, again, delta x
is x sub i minus x sub i minus
one; delta y is x [oops!! y] sub j
minus x [oops! y] sub j minus one. All right, so
we're going to choose a point
x sub ij star, y sub ij star
in each subrectangle
R sub ij. Then, we're going to find the
volume of the box with height f
evaluated at that point and the base is
delta x and delta y, and you can see that
from Figure 5.4 of the text, reproduced
here for your convenience.
So the volume of that single box is the
height, which is
f at the point, times delta A.
And so, if we compute the volume for all
of the rectangles and add them up,
that's going to give us the approximate
volume of our solid
S. And so we have a double sum because we have to sum over
all of the
i's and all of the j's. And then, as
in the single variable case, if we take
smaller and smaller subintervals,
which means that we're making m and n
get larger and larger,
you know like going to infinity, we
obtain
better approximations to the volume. And
so we see,
since we have a double sum, we represent
this with a double integral
over the rectangle R of our function
as the limit as m and n goes to infinity
of
that approximate volume, provided that
limit exists. So let's do an example.
You'll have to forgive me for drawing
the pictures ahead of time,
but I don't want to waste five minutes
just trying to get a rectangle to look
pretty.
Okay, so we're going to approximate the
volume under the surface z
equals f of x, y equals
x squared plus y and above the rectangle
R
equals [-1, 3] by [0, 4], 
and we're to divide our rectangle into
four squares with
m equals n equals two. We'll choose the
sample point
as the midpoint of our squares.
Okay, so we quickly can see that delta
x is going to be 3 minus negative 1 over
2, which is 2; and delta y is going to be
4 minus 0 over 2,
which is 2. And so that tells us then
that delta A
is 2 times 2, also known as 4.
The red dots here represent
our special points that are the
midpoints of each square,
and so our volume then is going to be
approximately
f of the first one, (0, 1),
plus f of (0, 3)
plus f of (2, 1) plus
f of (2, 3) [all] times delta A, which is 4,
and so that is 1 plus 3
plus 5 plus 7 [all] times 4,
which is 64. So our volume
is approximately 64.
Piece of cake, right? Isn't it nice when
there's only four.
All right, so some properties of double
integrals. If we have two integrable
functions
over, um, the rectangular region [a, b]
by [c, d], capital S and capital T are sub-
regions of that rectangle,
we have little m and capital M being
real numbers,
the sum of the two functions is
integrable and the integral of the sum
is the sum of the integrals.
If c is a real number, a constant times
our function is integrable
and the integral is equal to a constant
times the integral of the function.
If I can write R as the union of
S and T, where s and t don't intersect
except
on their boundaries, right, so, for example, this is S, this is T, then the integral
over the entire rectangle
is just the sum of the integrals over
those
sub-regions. Also, if I know that f
is greater than or equal to g over the
entire rectangle,
then hopefully it's clear that the
integral of f
is going to be greater than or equal to
the integral of g since
f and g at each point kind of represent
the height.
That should make sense. If I know that my
function
f is between little m and capital M,
I denote capital A of, of
R as the area of my region R,
then my integral is between little m
times
A and big M times A. And finally,
if I can write f of x, y as the product
of a function of x and a function of y,
then my integral can be computed as the
product of the integral
over x and the integral over y. So let's
do an example of that latter one.
Let's let R be the rectangle [1, 2]
by [0, 3] and find the double
integral of that,
of x y squared.
Okay, so f of x y
is x y squared, but we can write that, we
could think of that
as the product of two functions, g of x
and
h of y, where g
of x is x and h of y
is y squared.
And so, if we do that, then
our integral over our rectangle
is going to equal the product of the
integral
over x, which is the integral from 1 to 2
of the function
x, and the integral over y, which is the
integral from 0 to 3
of y squared.
So that's x squared over 2 from 1 to
2
times y cubed over 3,
0 to 3. That's gonna give me, uh,
three halves times
27 over 3, 9, which is 27
halves. Just so you can have some short
videos that really are
short ... Um, I think i will do....
Well, it's pretty short. Let's keep going.
The,
the last thing in this section is
iterated integrals.
And so what this is saying is: to compute
a double integral,
any double integral, over a rectangle, we
can break it up into single integrals by
either integrating first with respect to
x or
first with respect to y, and that's
called an iterated integral;
and we write it like this.
So when we integrate, let's say a
function of x, y
over x, we treat y as a constant.
So Fubini's theorem says that if f [of]
x, y is a continuous function on our
rectangle [a, b]
by [c, d], the double integral
of f over that
is, well, we can write that instead of d
A as dx dy,
and then we can choose to integrate
either with respect
to y first or
with respect to x first.
More generally, um, Fubini's Theorem
actually doesn't require f to be
continuous;
all it requires is that f can be
integrated
over that region. So let's do a couple
of quick examples. Let's integrate
our function for the following two
functions.
All right, so we're gonna integrate 6x y
squared over the rectangle [2, 4] by
[1, 2],
and so how are we going to do that?
So we can choose which way we want to do. So we can do
integral 2 to 4, integral 1 2. We can
choose
to do x first, and if we do that then
what we do when we integrate with
respect to x?
6x, that's going to be 3
x squared y squared, because we treat y
as a constant,
and so that's going to give me the
integral from 2 to 4
of, let's see. So that's
3 times 4 y squared is 3
one y squared dy,
so that's [the] integral from 2 to 4 of,
uh, 12 minus 3. And, wait ... what is
that?
Twelve minus three ... I'm, I'm having a
moment here.
What? I don't know what I'm doing right
now; I just had a brain thing.
Um 12 minus ....9,
9 y squared dy (Sheesh!),
which is the integral of 3y [oops!],
which is going to give me 3y cubed from 2
to 4. So I have 3
times 4 cubed minus 3 times 2 cubed,
which is going to give me, um, let's see
64 minus 8,
56 times 3,
18 ... 168,
if I didn't make any mistakes.
Well, let's check if I made any mistakes.
We'll do it the other way. So,
alternately,
we could have done
the integral 1 to 2, we could have
integrated over y
first.
And if we do that, we, when we integrate
over y,
integral 6y squared is 3y cubed x
from 2 to 4 dx.
So that's going to give me the integral
1 to 2 [of]]
3 times 4 cubed x minus 3
times 2 cubed x
dx, which is going to give me ...
Let's see, 64 times 3 is
192 and 3 times 8
is 24, so that's going to give me 8
268, so i get integral 1 to 268 [of ]x
dx, which is going to give me ...
Hold that thought; 6 divided by 3 is 2.
Oopsie!
My mistake. Well, that's certainly not
going to give you the right answer if we
make a mistake.
So you always want to be careful and
divide correctly.
So, uh, 64 times 2
is 128; 8 times 2 is 16; so that gives me
112
x dx, and then the anti-derivative of
that
is 56 x squared 1, from 1 to 2. So I
have 56 times 2 squared minus
56 times 1 squared,
so that is 56 times three,
basically, which is 168.
So whichever way we slice it, we get the
same integral.
All right, let's look at the second one.
Now, um,
I'm going to do the same. Let me do, let's
do y
first for the kicks, um, actually, not for
kicks.
Let's do, let's do integration with
respect to y first, and then we'll
discuss
why I did that and why I want to do that.
Okay, so now I'm looking at the integral
with respect to y of this.
So I don't have to deal with integration
by parts or anything -- 
x is just a constant. And it turns out
that that inner integral
is going to give me e to the xy,
evaluate at the limits the x, so I'm
going to get the integral from negative
one to two of e to the x
minus e to the zero dx,
or e to the x
minus 1, and the integral of that is e
to the x
minus x, evaluate at the limits. So I get e
squared minus 2
minus e to the minus 1 minus a negative
1.
So I get e to the 2 minus e to the
negative one
minus one.
Let's see, minus a minus a minus.
Wait, that's a plus. Oh, that's minus
three. Sorry.
Now, suppose
I wanted to do it the other way, right, I
wanted to do the integral
with respect to x first because I'm a
glutton for punishment.
We would have had to integrate by parts,
right, because we're looking at
this guy. So when we integrated by parts,
we would get
x over y e to the xy from negative 1
2 minus the integral negative 1 [to] 2 [of] 1 over
y e to the xy
dx dy.
And, uh, this
is less fun. In fact it even gets
way, way less time because we end up with
an
e to the xy divided by y to integrate,
and, like, what are we going to do with
that?
And so, in fact, this I think may even be
impossible to integrate this way.
And so that's the idea. Sometimes you
actually
do have to choose an approach; sometimes
you don't.
We can use double integrals to find the
area of a region,
which will be definitely more
interesting and useful when we're doing
non-rectangular regions, but the area of
a region
is given by the integral
of one.
We can also use double integrals to find
the volume of a solid, and that's what
we've been doing. But more application of
interest is to find the average value of
a function of two variables
over a region. And so, how we would do
this is, it's one over the area of our
region
times the integral of f
over our region. So we'll do
a rapid example.
Let's find the average of this, oh, of this
function over our region. So, first of all,
the area
is 1 minus zero times one minus zero,
which is one, and so the average is going
to be one over one [times the]
integral, well, from zero to one and zero
to one
of our function y over one plus x
squared.
Let's just get it over with and
integrate with respect to x
first, but you should test this doing
both.
So the integral with respect to x, that's
going to give me y
times the arctangent of x.
So I'm going to get the integral from
zero to one y times arctangent of one
minus
y times arctangent of zero
dy. Arctangent of zero is zero, 
arctangent of
one is pi over four,
so I get, uh, pi over eight
y squared, evaluated at the limits,
which gives me pi over eight. So the
average value of our function
is pi over eight. And look! Suggested
exercises.
Note I said one of them was a good
problem, which means i thought it was
kind of cool.
All right, so I am going to stop here, and
see you next time when we're going to
start talking about integrals
over general regions.
Have a great day!
