multiple integrals

MULTIPLE INTEGRALSMULTIPLE INTEGRALS

15

MULTIPLE INTEGRALS

In this chapter, we extend the idea

of a definite integral to double and triple

integrals of functions of two or three

variables.

MULTIPLE INTEGRALS

These ideas are then used to compute

volumes, masses, and centroids of more

general regions than we were able to

consider in Chapters 6 and 8.

MULTIPLE INTEGRALS

We also use double integrals to calculate

probabilities when two random variables

are involved.

MULTIPLE INTEGRALS

We will see that polar coordinates

are useful in computing double integrals

over some types of regions.

MULTIPLE INTEGRALS

Similarly, we will introduce two coordinate

systems in three-dimensional space that

greatly simplify computing of triple integrals

over certain commonly occurring solid

regions.

Cylindrical coordinates

Spherical coordinates

MULTIPLE INTEGRALS

15.1Double Integrals

over Rectangles

In this section, we will learn about:

Double integrals and using them

to find volumes and average values.

DOUBLE INTEGRALS OVER RECTANGLES

Just as our attempt to solve the area problem

led to the definition of a definite integral,

we now seek to find the volume of a solid.

In the process, we arrive at the definition

of a double integral.

DEFINITE INTEGRAL—REVIEW

First, let’s recall the basic facts

concerning definite integrals of functions

of a single variable.

DEFINITE INTEGRAL—REVIEW

If f(x) is defined for a ≤ x ≤ b, we start by

dividing the interval [a, b] into n subintervals

[xi–1, xi] of equal width ∆x = (b – a)/n.

We choose sample points xi* in these

subintervals.

DEFINITE INTEGRAL—REVIEW

Then, we form the Riemann sum

*

1

( )n

ii

f x x

Equation 1

DEFINITE INTEGRAL—REVIEW

Then, we take the limit of such sums as

n → ∞ to obtain the definite integral of f

from a to b:

*

1

( ) lim ( )nb

ia ni

f x dx f x x

Equation 2

DEFINITE INTEGRAL—REVIEW

In the special case where f(x) ≥ 0,

the Riemann sum can be interpreted as

the sum of the areas of the approximating rectangles.

DEFINITE INTEGRAL—REVIEW

Then, represents the area

under the curve y = f(x) from a to b.

( )b

af x dx

VOLUMES

In a similar manner, we consider a function f

of two variables defined on a closed rectangle

R = [a, b] x [c, d]

= {(x, y) 2 | a ≤ x ≤ b, c ≤ y ≤ d

and we first suppose that f(x, y) ≥ 0.

The graph of f is a surface with equation z = f(x, y).

VOLUMESLet S be the solid that lies above R and

under the graph of f, that is,

S = {(x, y, z) 3 | 0 ≤ z ≤ f(x, y), (x, y) R}

Our goal is to find

the volume of S.

VOLUMES

The first step is to divide the rectangle R

into subrectangles.

We divide the interval [a, b] into m subintervals [xi–1, xi] of equal width ∆x = (b – a)/m.

Then, we divide [c, d] into n subintervals [yj–1, yj] of equal width ∆y = (d – c)/n.

VOLUMES Next, we draw lines

parallel to the coordinate axes through the endpoints of these subintervals.

VOLUMES Thus, we form the subrectangles

Rij = [xi–1, xi] x [yj–1, yj] = {(x, y) | xi–1 ≤ x ≤ xi, yj–1 ≤ y ≤ yj}

each with area ∆A = ∆x ∆y

VOLUMES

Let’s choose a sample point (xij*, yij*)

in each Rij.

VOLUMES

Then, we can approximate the part of S

that lies above each Rij by a thin rectangular box (or “column”)

with: Base Rij

Height f (xij*, yij*)

VOLUMES

Compare the figure

with the earlier one.

VOLUMES

The volume of this box is the height of the box times

the area of the base rectangle:

f(xij *, yij *) ∆A

VOLUMES

We follow this procedure for all

the rectangles and add the volumes

of the corresponding boxes.

VOLUMES

Thus, we get an approximation to the total

volume of S: * *

1 1

( , )m n

ij iji j

V f x y A

Equation 3

VOLUMES

This double sum means that:

For each subrectangle, we evaluate f at the chosen point and multiply by the area of the subrectangle.

Then, we add the results.

VOLUMES

Our intuition tells us that the approximation

given in Equation 3 becomes better as

m and n become larger.

So, we would expect that:

* *

,1 1

lim ( , )m n

ij ijm n

i j

V f x y A

Equation 4

VOLUMES

We use the expression in Equation 4 to define

the volume of the solid S that lies under

the graph of f and above the rectangle R.

It can be shown that this definition is consistent with our formula for volume in Section 6.2

VOLUMES

Limits of the type that appear in Equation 4

occur frequently—not just in finding volumes

but in a variety of other situations as well—

even when f is not a positive function.

So, we make the following definition.

DOUBLE INTEGRAL

The double integral of f over the rectangle R

is:

if this limit exists.

Definition 5

* *

,1 1

( , ) lim ( , )m n

ij ijm n

i jR

f x y dA f x y A

The precise meaning of the limit in Definition 5

is that, for every number ε > 0, there is

an integer N such that

for: All integers m and n greater than N Any choice of sample points (xij*, yij*) in Rij*

DOUBLE INTEGRAL

* *

1 1

( , ) ( , )m n

ij iji jR

f x y dA f x y A

A function f is called integrable if the limit

in Definition 5 exists.

It is shown in courses on advanced calculus that all continuous functions are integrable.

In fact, the double integral of f exists provided that f is “not too discontinuous.”

INTEGRABLE FUNCTION

In particular,

if f is bounded [that is, there is a constant M

such that |f(x, y)| ≤ for all (x, y) in R],

and

f is continuous there, except on a finite

number of smooth curves,

then

f is integrable over R.

INTEGRABLE FUNCTION

The sample point (xij*, yij*)

can be chosen to be any point

in the subrectangle Rij*.

DOUBLE INTEGRAL

However, suppose we choose it

to be the upper right-hand corner of Rij

[namely (xi, yj)].

DOUBLE INTEGRAL

Then, the expression for the double integral

looks simpler:

DOUBLE INTEGRAL

,1 1

( , ) lim ( , )m n

i im n

i jR

f x y dA x y A

Equation 6

By comparing Definitions 4 and 5,

we see that a volume can be written as

a double integral, as follows.

DOUBLE INTEGRAL

If f(x, y) ≥ 0, then the volume V of the solid

that lies above the rectangle R and below

the surface z = f(x, y) is:

DOUBLE INTEGRAL

( , )R

V f x y dA

The sum in Definition 5

is called a double Riemann sum.

It is used as an approximation to the value of the double integral.

Notice how similar it is to the Riemann sum in Equation 1 for a function of a single variable.

DOUBLE REIMANN SUM

* *

1 1

( , )m n

ij iji j

f x y A

If f happens to be a positive function,

the double Riemann sum:

Represents the sum of volumes of columns, as shown.

Is an approximation to the volume under the graph of f and above the rectangle R.

DOUBLE REIMANN SUM

Estimate the volume of the solid that lies

above the square R = [0, 2] x [0, 2] and

below the elliptic paraboloid z = 16 – x2 – 2y2.

Divide R into four equal squares and choose the sample point to be the upper right corner of each square Rij.

Sketch the solid and the approximating rectangular boxes.

DOUBLE INTEGRALS Example 1

The squares are shown here.

The paraboloid is the graph of f(x, y) = 16 – x2 – 2y2

The area of eachsquare is 1.

DOUBLE INTEGRALS Example 1

Approximating the volume by the Riemann

sum with m = n = 2, we have:

DOUBLE INTEGRALS

2 2

1 1

( , )

(1,1) (1,2) (2,2)

13(1) 7(1) 10(1) 4(1)

34

i ji j

V f x y A

f A f A f A

Example 1

DOUBLE INTEGRALS

That is

the volume of

the approximating

rectangular boxes

shown here.

Example 1

We get better approximations to

the volume in Example 1 if we increase

the number of squares.

DOUBLE INTEGRALS

DOUBLE INTEGRALS

The figure shows how, when we use 16, 64,

and 256 squares, The columns start to look more like the actual solid. The corresponding approximations get more accurate.

DOUBLE INTEGRALS

In Section 15.2, we will be able to show that

the exact volume is 48.

If R = {(x, y)| –1 ≤ x ≤ 1, –2 ≤ y ≤ 2},

evaluate the integral

It would be very difficult to evaluate this integral directly from Definition 5.

However, since , we can compute it by interpreting it as a volume.

DOUBLE INTEGRALS Example 2

21R

x dA

21 0x

If

then

x2 + z2 = 1

z ≥ 0

DOUBLE INTEGRALS

21z x Example 2

So, the given double integral represents

the volume of the solid S that lies:

Below the circular cylinder x2 + z2 = 1 Above the rectangle R

DOUBLE INTEGRALS Example 2

The volume of S is the area of a semicircle

with radius 1 times the length of the cylinder.

DOUBLE INTEGRALS Example 2

2 2121 (1) 4 2

R

x dA

The methods that we used for approximating

single integrals (Midpoint Rule, Trapezoidal

Rule, Simpson’s Rule) all have counterparts

for double integrals.

Here, we consider only the Midpoint Rule

for double integrals.

THE MIDPOINT RULE

This means that we use a double Riemann

sum to approximate the double integral,

where the sample point (xij*, yij*) in Rij

is chosen to be the center of Rij.

That is, is the midpoint of [xi–1, xi] and is the midpoint of [yj–1, yj] .

THE MIDPOINT RULE

( , )ij ijx y

( )ijx( )ijy

where:

is the midpoint of [xi–1, xi]

is the midpoint of [yj–1, yj].

MIDPOINT RULE FOR DOUBLE INTEGRALS

1 1

( , ) ( , )m n

i ji jR

f x y dA f x y A

ix

jy

Use the Midpoint Rule with m = n = 2 to

estimate the value of the integral

where

R = {(x, y) | 0 ≤ x ≤ 2, 1 ≤ y ≤ 2}

MIDPOINT RULE Example 3

2( 3 )R

x y dA

In using the Midpoint Rule with m = n = 2,

we evaluate f(x, y) = x – 3y2 at the centers

of the four subrectangles

shown here.

MIDPOINT RULE Example 3

Thus,

The area of each subrectangle is ∆A = ½.

MIDPOINT RULE

311 22 2

5 71 24 4

x x

y y

Example 3

Thus,

MIDPOINT RULE

2

2 2

1 1

1 1 22 11

2 2

( 3 )

( , )

( , ) ( , ) ( , )

( , )

R

i ji j

x y dA

f x y A

f x y A f x y A f x y A

f x y A

Example 3

MIDPOINT RULE

5 7 3 51 12 4 2 4 2 4

3 72 4

67 139 51 1231 1 1 116 2 16 2 16 2 16 2

958

, , ,

,

11.875

f A f f A

f A

Example 3

In Section 15.2, we will develop

an efficient method for computing

double integrals.

Then, we will see that the exact value of the double integral in Example 3 is –12.

NOTE

Remember that the interpretation of a double

integral as a volume is valid only when

the integrand f is a positive function.

The integrand in Example 3 is not a positive function.

So, its integral is not a volume.

NOTE

In Examples 2 and 3 in Section 15.2,

we will discuss how to interpret integrals

of functions that are not always positive

in terms of volumes.

NOTE

Suppose we keep dividing each of these

subrectangles into four smaller ones with

similar shape.

NOTE

Then, we get these Midpoint Rule

approximations.

Notice how these approximations approach the exact value of the double integral, –12.

NOTE

Recall from Section 6.5 that the average value

of a function f of one variable defined on

an interval [a, b] is:

AVERAGE VALUE

1( )

b

ave af f x dx

b a

Similarly, we define the average value

of a function f of two variables defined on

a rectangle R to be:

where A(R) is the area of R.

AVERAGE VALUE

1( , )

( )ave

R

f f x y dAA R

If f(x, y) ≥ 0, the equation

says that:

The box with base R and height fave has the same volume as the solid that lies under the graph of f.

AVERAGE VALUE

( ) ( , )ave

R

A R f f x y dA

Suppose z = f(x, y) describes a mountainous

region and you chop off the tops of

the mountains at height fave .

Then, you can use them to fill in the valleys so that the region becomes completely flat.

AVERAGE VALUE

The contour map shows the snowfall, in

inches, that fell on the state of Colorado on

December

20 and 21,

2006.

AVERAGE VALUE Example 4

The state is in the shape of a rectangle that

measures 388 mi west to east and 276 mi

south to

north.

AVERAGE VALUE Example 4

Use the map to estimate the average snowfall

for the entire state on those days.

AVERAGE VALUE Example 4

Let’s place the origin at the southwest corner

of the state.

AVERAGE VALUE Example 4

Then, 0 ≤ x ≤ 388, 0 ≤ y ≤ 276, and f(x, y)

is the snowfall, in inches, at a location x miles

to the east and y miles to the north of

the origin.

AVERAGE VALUE Example 4

If R is the rectangle that represents Colorado,

the average snowfall for the state on

December 20–21 was:

where A(R) = 388 · 276

AVERAGE VALUE Example 4

1( , )

( )ave

R

f f x y dAA R

To estimate the value of this double

integral, let’s use the Midpoint Rule with

m = n = 4.

AVERAGE VALUE Example 4

That is, we divide R into 16 subrectangles

of equal size.

AVERAGE VALUE

The area of each subrectangle

is:

∆A = (1/16) (388)(276)

= 6693 mi2

AVERAGE VALUE

Using the contour map to estimate the value

of f at the center of each subrectangle,

we get:

AVERAGE VALUE

4 4

1 1

( , ) ( , )

[0 15 8 7 2 25 18.5

11 4.5 28 17 13.5

12 15 17.5 13]

(6693)(207)

i ji jR

f x y dA f x y A

A

Therefore,

On December 20–21, 2006, Colorado received an average of approximately 13 inches of snow.

AVERAGE VALUE

(6693)(207)

(388)(276)

12.9

avef

We now list three properties of double

integrals that can be proved in the same

manner as in Section 5.2

We assume that all the integrals exist.

PROPERTIES OF DOUBLE INTEGRALS

These properties are referred to as

the linearity of the integral.

where c is a constant

PROPERTIES 7 & 8

[ ( , ) ( , )] ( , )

( , )

R R

R

f x y g x y dA f x y dA

g x y dA

( , ) ( , )

R R

cf x y dA c f x y dA

If f(x, y) ≥ g(x, y) for all (x, y) in R,

then

PROPERTY 9

( , ) ( , )R R

f x y dA g x y dA