Outline

MATH1251 Calculus
Chapter 5: Double Integrals
1 Riemann Sums
A/Prof Thanh Tran

School of Mathematics and Statistics 2 Repeated Integrals

The University of New South Wales
Sydney, Australia

Red Centre Room 4061 3 Changes of Variables

Email: thanh.tran@unsw.edu.au

4 Applications

1 2

Motivation Special case: is a rectangle

If is a bounded subset of R2 , how to find the area of ? First we let be a rectangle R = [a, b] [c, d ] and let f : R R
be a bounded function.
If f is a continuous and positive function defined on a rectangle R,
then the equation z = f (x, y) represents a surface that lies Partition the base R into lots of smaller rectangles Rk ,
above R. What is the volume of the solid that is bounded below k = 1, . . . , n.
by R and above by the surface z = f (x, y)?
We denote the partition by P = {R1 , . . . , Rn } and let
To answer these questions, we study the integral of f : R. We
Ak := area (Rk ). For k = 1, . . . , n define
use the notation Z
f (x, y) dA, fkmin = inf f (x, y) and fkmax = max f (x, y).
(x,y )Rk (x,y )Rk

where dA stands for with respect to area. Then
fkmin f (x, y) fkmax (x, y) Rk .
This study is more difficult than what you have seen for one
variable functions in that apart from the difficulty of f being a If Vk (f ) denotes the volume under the graph of f over Rk , then
function of two variables, the fact that the base could be
fkmin Ak Vk (f ) fkmax Ak .
complicated is another issue.
3 Riemann Sums 4
The Riemann sums Example 2
Partition R = [0, 1] [0, 2] into 8 small squares of side length 1/2
Definition 1
Rjk = {(x, y) | j/2 x (j + 1)/2, k/2 y (k + 1)/2} ,
The lower Riemann sum corresponding to the partition
P = {R1 , . . . , Rn } of R is
j = 0, 1, k = 0, 1, 2. Let f (x, y) = x 2 y. Calculate the upper and lower
n
X Riemann sums for this partition P.
L(f , P) = fkmin Ak ,
k =1

which gives the volume of the set of biggest boxes with bases
in P, that you can fit underneath the graph of f above R.
The upper Riemann sum for P = {R1 , . . . , Rn } is
n
X
U(f , P) = fkmax Ak ,
k =1

the volume of the set of smallest boxes with bases in P, that fit
above the graph of f .
Riemann Sums 5 Riemann Sums 6

Riemann integrability
Definition 3
For every > 0 if there is a partition P such that

0 U(f , P ) L(f , P ) <

then we say that f is Riemann integrable over R and write

Z
f dA = sup L(f , P) = inf U(f , P).
R P P

If f is integrable then there are partitions P(n) where P(n+1) refines P(n)
(i.e., any element of P(n) is the union of some elements of P(n+1) ) such
that Z
f dA = lim L(f , P(n) ).
n

Riemann Sums 7 Riemann Sums 8

Theorem 4
If f is continuous on the rectangle R then f is Riemann integrable over
R.

Example 5
Many discontinuous functions are integrable too. But something nasty
like (
1, if x and y rational,
f (x, y) =
0, if x and y irrational,
is not integrable.

Riemann Sums 9 Riemann Sums 10

How to compute? Cross section

For a rectangle R = [a, b] [c, d ], the cross section of f : R R at

As with one variable functions, using Riemann sums to calculate x = x0 is defined by
integrals is horrible. Z d
Cf (x0 ) := f (x0 , y) dy,
c
Bad news: There is no Fundamental Theorem of Calculus to
replace the limiting process with a symbolic manipulation. provided that the integral exists. This is the area of the slice of the
volume under the graph of f at x = x0 .
Similarly, the cross section at y = y0 is defined by
Good news: We can calculate the integral by doing 2 one-variable Z d
integrals. Cf (y0 ) := f (x, y0 ) dx.
c

Riemann Sums 11 Repeated Integrals 12

Fubinis Theorem Example 7
Let R be the rectangle [0, 1] [0, 2]. Find the volume under the graph
Theorem 6 of f (x, y) = xy 2 over R, taking the integrals in both orders.
If f is continuous, then
!
Z Z b Z b Z d
f dA = Cf (x) dx = f (x, y) dy dx
R a a c
!
Z d Z d Z b
= Cf (y) dy = f (x, y) dx dy.
c c a

Notes
You need to match the right limits with the right dx or dy.
When you calculate this, you always start with the inner
cross-section integral.
RR R
We usually write R f (x, y) dx dy rather than R f dA.
Repeated Integrals 13 Repeated Integrals 14

A special example
Why did we bother with the Riemann sum definition if we only want to
find the volumes via iterated integrals using Fubinis Theorem?
Example 8
Compute the iterated integrals of
(
1/x 2 , 0 y x 1,
f (x, y) =
1/y 2 , 0 x < y 1.

What is wrong with this function?

Repeated Integrals 15 Repeated Integrals 16

 General case: is not a rectangle
Just add a zero-volume piece to the graph.
Definition 9
Let R be any rectangle containing . Then
ZZ ZZ
f (x, y) dx dy = f (x, y) dx dy,
R

where (
f (x, y) if (x, y) ,
f (x, y) =
0 if (x, y)
/ .

Note
You can use Fubinis Theorem to calculate the volume, but now you
have to worry about where the slices begin and end (corresponding to
using f in the integral).

Repeated Integrals 17 Repeated Integrals 18

Theorem 10 Example 11
Let Let f (x, y) = xy 2
Z Z and be the triangle with vertices (0, 0), (1, 1), and
= {(x, y) R2 |a x b, c(x) y d (x)}.
(1, 1). Find f (x, y) dx dy.
Then ZZ Z b Z d(x)
f (x, y) dx dy = f (x, y) dy dx.
a c(x)

Repeated Integrals 19 Repeated Integrals 20

 Example 12
Z 1Z 2
Calculate sin(y 2 ) dy dx.
0 2x

Repeated Integrals 21 Repeated Integrals 22

Example 13
Z 1 Z 2x 2
Interchange the order of the integral f (x, y) dy dx.
1 x2

Repeated Integrals 23 Repeated Integrals 24

 Summary

Always sketch the base region .

Decide the order of integration.

What is the range of the outside variable? Write down the outside
integral.

For a fixed value of the outside integral, what is the range of the
inner variable?

Do the integrals, starting from the inner integral first.

Repeated Integrals 25 Repeated Integrals 26

Changing the variables

Usually we describe the plane in the (x, y)-coordinates. For some

integrals it is better to use a different coordinates such as polar
coordinates (r , ). Let (x, y) = g(u, v) be an invertible change of variables. Then we
would like to write
ZZ ZZ
b
f (x, y) dx dy = f [g(u, v)] g (u, v) du dv.
Z
One variable: To find I = f (x) dx using the change of variables g 1 ()
a
x = g(t) we do
Z g 1 (b)
But what should g (u, v) mean for a function g : R2 R2 ?

I= f [g(t)]g (t) dt
g 1 (a)

to turn this into an integral in the t variable.

Two variables: We do basically the same thing.

Changes of Variables 27 Changes of Variables 28

Jacobian determinant Polar coordinates

In the polar coordinates we have

Definition 14
The Jacobian matrix of a function g : R2 R2 is the matrix (x, y) = g(r , ) = (r cos , r sin ).

x x Thus the Jacobian matrix is

u v
x x

Jg(u, v) =


r  
y y cos r sin

. Jg(r , ) = = .
u v y y sin r cos
.
k
The Jacobian determinant (or Jacobian for short) of a function
Hence the Jacobian determinant is
g : R2 R2 is the determinant of the Jacobian matrix, denoted by
|Jg(u, v)|. |Jg(r , )| = det(Jg(r , )) = r (cos2 + sin2 ) = r .

Changes of Variables 29 Changes of Variables 30

Change of variables Formula Example 15

Z 1Z x  q 
Find I = 1+ x 2 + y 2 dy dx using polar coordinates. (Note
0 0
that this is not nice to do directly!)
ZZ ZZ
f (x, y) dx dy = f [g(u, v)]|Jg(u, v)| du dv
g 1 ()

What about g 1 ()?

This is just fancy way of saying describe in the (u, v)
coordinate system.

To do this, you first sketch and then describe it using the u and
v variables.

Changes of Variables 31 Changes of Variables 32

 Areas
Example 16
Find the area of the portion of the unit disk that lies in the first
quadrant. (Of course, the area is /4.)

Changes of Variables 33 Applications 34

Centroids Example 17
Suppose that the mass density over the region
Let (x, y) denote the mass density over a plate .
The mass of the plate is = {(x, y) | |x| y 1}

is (x, y) = x 2 + y 2 . Find the centre of mass (

x , y ).
ZZ
M= (x, y) dx dy. (1)

x , y )
The centre of mass (or centroid) of the plate is the point (
defined by

1
ZZ
x = x(x, y) dx dy
M
Z Z (2)
1
y = y(x, y) dx dy
M

If (x, y) 1 then (1) equals the area of .

Applications 35 Applications 36
 Example 18
Consider the spiral r = , [0, 2]. Find the area between the spiral
and the y-axis.

Applications 37 Applications 38

Population and population centroid

If (x, y) is the population density over a region , then
formula (1) defines the total population, whereas formulas (2)
define the population centroid..

If ZR is a nice enough region, then the integral

Z
(x, y) dx dy is the population in the region R.
R

ZZ
If f (x, y) = (x, y)/M then f (x, y) dx dy = 1. Hence it is a

x , y ) is
(two-dimensional) probability density function. The point (
the expected value of the corresponding random variable.

The 3-D versions of formulas (1) and (2) can be written in the
same manner.
Applications 39 Applications 40