1. 16
MULTIPLE INTEGRALS

2. 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.

3. 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.

4. MULTIPLE INTEGRALS
We also use double integrals to calculate probabilities when two random variables are involved.

5. MULTIPLE INTEGRALS
We will see that polar coordinates are useful in computing double integrals over some types of regions.

6. 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

7. Double Integrals over Rectangles
MULTIPLE INTEGRALS
16.1
Double Integrals over Rectangles
In this section, we will learn about:
Double integrals and using them
to find volumes and average values.

8. 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.

9. DEFINITE INTEGRAL—REVIEW
First, let’s recall the basic facts concerning definite integrals of functions of a single variable.

10. 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.

11. DEFINITE INTEGRAL—REVIEW
Equation 1
Then, we form the Riemann sum

12. DEFINITE INTEGRAL—REVIEW
Equation 2
Then, we take the limit of such sums as n → ∞ to obtain the definite integral of f from a to b:

13. 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.
Fig , p. 987

14. DEFINITE INTEGRAL—REVIEW
Then, represents the area under the curve y = f(x) from a to b.
Fig , p. 987

15. and we first suppose that f(x, y) ≥ 0.
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) € R2 | 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).

16. S = {(x, y, z)  R3 | 0 ≤ z ≤ f(x, y), (x, y)  R}
VOLUMES
Let S be the solid that lies above R and under the graph of f, that is,
S = {(x, y, z)  R3 | 0 ≤ z ≤ f(x, y), (x, y)  R}
Our goal is to find the volume of S.
Fig , p. 987

17. The first step is to divide the rectangle R into subrectangles.
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.

18. VOLUMES
Next, we draw lines parallel to the coordinate axes through the endpoints of these subintervals.
Fig , p. 988

19. 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
Fig , p. 988

20. Let’s choose a sample point (xij*, yij*) in each Rij.
VOLUMES
Let’s choose a sample point (xij*, yij*) in each Rij.
Fig , p. 988

21. 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*)
Fig , p. 988

22. Compare the figure with the earlier one.
VOLUMES
Compare the figure with the earlier one.
Fig , p. 987
Fig , p. 988

23. VOLUMES
The volume of this box is the height of the box times the area of the base rectangle: f(xij *, yij *) ∆A
Fig , p. 988

24. VOLUMES
We follow this procedure for all the rectangles and add the volumes of the corresponding boxes.

25. Thus, we get an approximation to the total volume of S:
VOLUMES
Equation 3
Thus, we get an approximation to the total volume of S:
Fig , p. 988

26. This double sum means that:
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.

27. So, we would expect that:
VOLUMES
Equation 4
Our intuition tells us that the approximation given in Equation 3 becomes better as m and n become larger.
So, we would expect that:

28. 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

29. 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.

30. The double integral of f over the rectangle R is:
Definition 5
The double integral of f over the rectangle R is:
if this limit exists.

31. DOUBLE INTEGRAL
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*

32. A function f is called integrable if the limit in Definition 5 exists.
INTEGRABLE FUNCTION
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.”

33. 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.

34. DOUBLE INTEGRAL
The sample point (xij*, yij*) can be chosen to be any point in the subrectangle Rij*.

35. DOUBLE INTEGRAL
However, suppose we choose it to be the upper right-hand corner of Rij [namely (xi, yj)].
Fig , p. 988

36. Then, the expression for the double integral looks simpler:
Equation 6
Then, the expression for the double integral looks simpler:

37. DOUBLE INTEGRAL
By comparing Definitions 4 and 5, we see that a volume can be written as a double integral, as follows.

38. 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:

39. is called a double Riemann sum.
DOUBLE REIMANN SUM
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.

40. If f happens to be a positive function, the double Riemann sum:
DOUBLE REIMANN SUM
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.
Fig , p. 988

41. DOUBLE INTEGRALS
Example 1
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.

42. The squares are shown here.
DOUBLE INTEGRALS
Example 1
The squares are shown here.
The paraboloid is the graph of f(x, y) = 16 – x2 – 2y2
The area of each square is 1.
Fig , p. 990

43. Approximating the volume by the Riemann sum with m = n = 2, we have:
DOUBLE INTEGRALS
Example 1
Approximating the volume by the Riemann sum with m = n = 2, we have:

44. That is the volume of the approximating rectangular boxes shown here.
DOUBLE INTEGRALS
Example 1
That is the volume of the approximating rectangular boxes shown here.
Fig , p. 990

45. DOUBLE INTEGRALS
We get better approximations to the volume in Example 1 if we increase the number of squares.

46. The figure shows how, when we use 16, 64, and 256 squares,
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.
Fig , p. 990

47. In Section 16.2, we will be able to show that the exact volume is 48.
DOUBLE INTEGRALS
In Section 16.2, we will be able to show that the exact volume is 48.
Fig , p. 990

48. If R = {(x, y)| –1 ≤ x ≤ 1, –2 ≤ y ≤ 2}, evaluate the integral
DOUBLE INTEGRALS
Example 2
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.

49. DOUBLE INTEGRALS
Example 2
If then x2 + z2 = z ≥ 0

50. DOUBLE INTEGRALS
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
Fig , p. 991

51. DOUBLE INTEGRALS
Example 2
The volume of S is the area of a semicircle with radius 1 times the length of the cylinder.
Fig , p. 991

52. Here, we consider only the Midpoint Rule for double integrals.
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.

53. 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] .

54. MIDPOINT RULE FOR DOUBLE INTEGRALS
where:
is the midpoint of [xi–1, xi]
is the midpoint of [yj–1, yj].

55. MIDPOINT RULE
Example 3
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}

56. MIDPOINT RULE
Example 3
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.
Fig , p. 991

57. MIDPOINT RULE
Example 3
Thus,
The area of each subrectangle is ∆A = ½.

58. MIDPOINT RULE
Example 3
Thus,

59. MIDPOINT RULE
Example 3

60. NOTE
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.

61. 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.

62. 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.

63. NOTE
Suppose we keep dividing each of these subrectangles into four smaller ones with similar shape.
Fig , p. 991

64. Then, we get these Midpoint Rule approximations.
NOTE
Then, we get these Midpoint Rule approximations.
Notice how these approximations approach the exact value of the double integral, –12.
p. 992

65. AVERAGE VALUE
Recall from Section 6.5 that the average value of a function f of one variable defined on an interval [a, b] is:

66. where A(R) is the area of R.
AVERAGE VALUE
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.

67. If f(x, y) ≥ 0, the equation says that:
AVERAGE VALUE
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.

68. AVERAGE VALUE
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.
Fig , p. 992

69. AVERAGE VALUE
Example 4
The contour map shows the snowfall, in inches, that fell on the state of Colorado on December 20 and 21,
Fig , p. 992

70. 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.
Fig , p. 992

71. AVERAGE VALUE
Example 4
Use the map to estimate the average snowfall for the entire state on those days.
Fig , p. 992

72. Let’s place the origin at the southwest corner of the state.
AVERAGE VALUE
Example 4
Let’s place the origin at the southwest corner of the state.
Fig , p. 992

73. 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.

74. 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

75. AVERAGE VALUE
Example 4
To estimate the value of this double integral, let’s use the Midpoint Rule with m = n = 4.

76. That is, we divide R into 16 subrectangles of equal size.
AVERAGE VALUE
That is, we divide R into 16 subrectangles of equal size.
Fig , p. 993

77. The area of each subrectangle is: ∆A = (1/16) (388)(276) = 6693 mi2
AVERAGE VALUE
The area of each subrectangle is: ∆A = (1/16) (388)(276) = 6693 mi2

78. AVERAGE VALUE
Using the contour map to estimate the value of f at the center of each subrectangle, we get:

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

80. PROPERTIES OF DOUBLE INTEGRALS
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.

81. These properties are referred to as the linearity of the integral.
where c is a constant

82. If f(x, y) ≥ g(x, y) for all (x, y) in R, then
PROPERTY 9
If f(x, y) ≥ g(x, y) for all (x, y) in R, then