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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