1. 16
MULTIPLE INTEGRALS

2. 16.6 Triple Integrals In this section, we will learn about:
MULTIPLE INTEGRALS
16.6
Triple Integrals
In this section, we will learn about:
Triple integrals and their applications.

3. TRIPLE INTEGRALS
Just as we defined single integrals for functions of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables.

4. TRIPLE INTEGRALS
Equation 1
Let’s first deal with the simplest case where f is defined on a rectangular box:

5. The first step is to divide B into sub-boxes—by dividing:
TRIPLE INTEGRALS
The first step is to divide B into sub-boxes—by dividing:
The interval [a, b] into l subintervals [xi-1, xi] of equal width Δx.
[c, d] into m subintervals of width Δy.
[r, s] into n subintervals of width Δz.
Fig a, p. 1026

6. TRIPLE INTEGRALS
The planes through the endpoints of these subintervals parallel to the coordinate planes divide the box B into lmn sub-boxes
Each sub-box has volume ΔV = Δx Δy Δz
Fig b, p. 1026

7. Then, we form the triple Riemann sum
TRIPLE INTEGRALS
Equation 2
Then, we form the triple Riemann sum
where the sample point is in Bijk.

8. TRIPLE INTEGRALS
By analogy with the definition of a double integral (Definition 5 in Section 16.1), we define the triple integral as the limit of the triple Riemann sums in Equation 2.

9. The triple integral of f over the box B is:
Definition 3
The triple integral of f over the box B is:
if this limit exists.
Again, the triple integral always exists if f is continuous.

10. We can choose the sample point to be any point in the sub-box.
TRIPLE INTEGRALS
We can choose the sample point to be any point in the sub-box.
However, if we choose it to be the point (xi, yj, zk) we get a simpler-looking expression:

11. TRIPLE INTEGRALS
Just as for double integrals, the practical method for evaluating triple integrals is to express them as iterated integrals, as follows.

12. FUBINI’S TH. (TRIPLE INTEGRALS)
Theorem 4
If f is continuous on the rectangular box B = [a, b] x [c, d] x [r, s], then

13. FUBINI’S TH. (TRIPLE INTEGRALS)
The iterated integral on the right side of Fubini’s Theorem means that we integrate in the following order:
With respect to x (keeping y and z fixed)
With respect to y (keeping z fixed)
With respect to z

14. FUBINI’S TH. (TRIPLE INTEGRALS)
There are five other possible orders in which we can integrate, all of which give the same value.
For instance, if we integrate with respect to y, then z, and then x, we have:

15. FUBINI’S TH. (TRIPLE INTEGRALS)
Example 1
Evaluate the triple integral where B is the rectangular box

16. FUBINI’S TH. (TRIPLE INTEGRALS)
Example 1
We could use any of the six possible orders of integration.
If we choose to integrate with respect to x, then y, and then z, we obtain the following result.

17. FUBINI’S TH. (TRIPLE INTEGRALS)
Example 1

18. INTEGRAL OVER BOUNDED REGION
Now, we define the triple integral over a general bounded region E in three-dimensional space (a solid) by much the same procedure that we used for double integrals.
See Definition 2 in Section 15.3

19. INTEGRAL OVER BOUNDED REGION
We enclose E in a box B of the type given by Equation 1.
Then, we define a function F so that it agrees with f on E but is 0 for points in B that are outside E.

20. INTEGRAL OVER BOUNDED REGION
By definition,
This integral exists if f is continuous and the boundary of E is “reasonably smooth.”
The triple integral has essentially the same properties as the double integral (Properties 6–9 in Section 15.3).

21. INTEGRAL OVER BOUNDED REGION
We restrict our attention to:
Continuous functions f
Certain simple types of regions

22. TYPE 1 REGION
A solid region is said to be of type 1 if it lies between the graphs of two continuous functions of x and y.

23. That is, where D is the projection of E onto the xy-plane.
TYPE 1 REGION
Equation 5
That is, where D is the projection of E onto the xy-plane.
Fig , p. 1027

24. Notice that: TYPE 1 REGIONS
The upper boundary of the solid E is the surface with equation z = u2(x, y).
The lower boundary is the surface z = u1(x, y).
Fig , p. 1027

25. TYPE 1 REGIONS
Equation/Formula 6
By the same sort of argument that led to Formula 3 in Section 16.3, it can be shown that, if E is a type 1 region given by Equation 5, then

26. TYPE 1 REGIONS
The meaning of the inner integral on the right side of Equation 6 is that x and y are held fixed.
Therefore,
u1(x, y) and u2(x, y) are regarded as constants.
f(x, y, z) is integrated with respect to z.

27. TYPE 1 REGIONS
In particular, if the projection D of E onto the xy-plane is a type I plane region, then
Fig , p. 1028

28. Thus, Equation 6 becomes:
TYPE 1 REGIONS
Equation 7
Thus, Equation 6 becomes:

29. If, instead, D is a type II plane region, then
TYPE 1 REGIONS
If, instead, D is a type II plane region, then
Fig , p. 1028

30. Then, Equation 6 becomes:
TYPE 1 REGIONS
Equation 8
Then, Equation 6 becomes:

31. TYPE 1 REGIONS
Example 2
Evaluate where E is the solid tetrahedron bounded by the four planes x = 0, y = 0, z = 0, x + y + z = 1

32. When we set up a triple integral, it’s wise to draw two diagrams:
TYPE 1 REGIONS
Example 2
When we set up a triple integral, it’s wise to draw two diagrams:
The solid region E
Its projection D on the xy-plane
Fig , p. 1028
Fig , p. 1028

33. TYPE 1 REGIONS
Example 2
The lower boundary of the tetrahedron is the plane z = 0 and the upper boundary is the plane x + y + z = 1 (or z = 1 – x – y).
So, we use u1(x, y) = 0 and u2(x, y) = 1 – x – y in Formula 7.
Fig , p. 1028

34. TYPE 1 REGIONS
Example 2
Notice that the planes x + y + z = 1 and z = 0 intersect in the line x + y = 1 (or y = 1 – x) in the xy-plane.
So, the projection of E is the triangular region shown here, and we have the following equation.
Fig , p. 1028

35. TYPE 1 REGIONS
E. g. 2—Equation 9
This description of E as a type 1 region enables us to evaluate the integral as follows.

36. TYPE 1 REGIONS
Example 2

37. TYPE 2 REGION
A solid region E is of type 2 if it is of the form where D is the projection of E onto the yz-plane.
Fig , p. 1029

38. The back surface is x = u1(y, z).
TYPE 2 REGION
The back surface is x = u1(y, z).
The front surface is x = u2(y, z).
Fig , p. 1029

39. TYPE 2 REGION
Equation 10
Thus, we have:

40. Finally, a type 3 region is of the form
where:
D is the projection of E onto the xz-plane.
y = u1(x, z) is the left surface.
y = u2(x, z) is the right surface.
Fig , p. 1029

41. For this type of region, we have:
TYPE 3 REGION
Equation 11
For this type of region, we have:
Fig , p. 1029

42. TYPE 2 & 3 REGIONS
In each of Equations 10 and 11, there may be two possible expressions for the integral depending on:
Whether D is a type I or type II plane region (and corresponding to Equations 7 and 8).

43. BOUNDED REGIONS
Example 3
Evaluate where E is the region bounded by the paraboloid y = x2 + z2 and the plane y = 4.

44. The solid E is shown here.
TYPE 1 REGIONS
Example 3
The solid E is shown here.
If we regard it as a type 1 region, then we need to consider its projection D1 onto the xy-plane.
Fig , p. 1030

45. That is the parabolic region shown here.
TYPE 1 REGIONS
Example 3
That is the parabolic region shown here.
The trace of y = x2 + z2 in the plane z = 0 is the parabola y = x2
Fig , p. 1030

46. From y = x2 + z2, we obtain: So, the lower boundary surface of E is:
TYPE 1 REGIONS
Example 3
From y = x2 + z2, we obtain:
So, the lower boundary surface of E is:
The upper surface is:

47. Therefore, the description of E as a type 1 region is:
TYPE 1 REGIONS
Example 3
Therefore, the description of E as a type 1 region is:

48. TYPE 1 REGIONS
Example 3
Thus, we obtain:
Though this expression is correct, it is extremely difficult to evaluate.

49. So, let’s instead consider E as a type 3 region.
TYPE 3 REGIONS
Example 3
So, let’s instead consider E as a type 3 region.
As such, its projection D3 onto the xz-plane is the disk x2 + z2 ≤ 4.
Fig , p. 1030

50. Then, the left boundary of E is the paraboloid y = x2 + z2.
TYPE 3 REGIONS
Example 3
Then, the left boundary of E is the paraboloid y = x2 + z2.
The right boundary is the plane y = 4.

51. TYPE 3 REGIONS
Example 3
So, taking u1(x, z) = x2 + z2 and u2(x, z) = 4 in Equation 11, we have:

52. This integral could be written as:
TYPE 3 REGIONS
Example 3
This integral could be written as:
However, it’s easier to convert to polar coordinates in the xz-plane: x = r cos θ, z = r sin θ

53. TYPE 3 REGIONS
Example 3
That gives:

54. APPLICATIONS OF TRIPLE INTEGRALS
Recall that:
If f(x) ≥ 0, then the single integral represents the area under the curve y = f(x) from a to b.
If f(x, y) ≥ 0, then the double integral represents the volume under the surface z = f(x, y) and above D.

55. APPLICATIONS OF TRIPLE INTEGRALS
The corresponding interpretation of a triple integral , where f(x, y, z) ≥ 0, is not very useful.
It would be the “hypervolume” of a four-dimensional (4-D) object.
Of course, that is very difficult to visualize.

56. APPLICATIONS OF TRIPLE INTEGRALS
Remember that E is just the domain of the function f.
The graph of f lies in 4-D space.

57. APPLICATIONS OF TRIPLE INTEGRALS
Nonetheless, the triple integral can be interpreted in different ways in different physical situations.
This depends on the physical interpretations of x, y, z and f(x, y, z).
Let’s begin with the special case where f(x, y, z) = 1 for all points in E.

58. APPLNS. OF TRIPLE INTEGRALS
Equation 12
Then, the triple integral does represent the volume of E:

59. APPLNS. OF TRIPLE INTEGRALS
For example, you can see this in the case of a type 1 region by putting f(x, y, z) = 1 in Formula 6:

60. APPLNS. OF TRIPLE INTEGRALS
From Section 16.3, we know this represents the volume that lies between the surfaces z = u1(x, y) and z = u2(x, y)

61. APPLICATIONS
Example 4
Use a triple integral to find the volume of the tetrahedron T bounded by the planes
x + 2y + z = 2
x = 2y
x = 0
z = 0

62. The tetrahedron T and its projection D on the xy-plane are shown.
APPLICATIONS
Example 4
The tetrahedron T and its projection D on the xy-plane are shown.
Fig , p. 1031
Fig , p. 1031

63. The lower boundary of T is the plane z = 0.
APPLICATIONS
Example 4
The lower boundary of T is the plane z = 0.
The upper boundary is the plane x + 2y + z = 2, that is, z = 2 – x – 2y
Fig , p. 1031

64. So, we have: APPLICATIONS Example 4
This is obtained by the same calculation as in Example 4 in Section 16.3

65. APPLICATIONS
Example 4
Notice that it is not necessary to use triple integrals to compute volumes.
They simply give an alternative method for setting up the calculation.

66. APPLICATIONS
All the applications of double integrals in Section 16.5 can be immediately extended to triple integrals.

67. APPLICATIONS
For example, suppose the density function of a solid object that occupies the region E is: ρ(x, y, z) in units of mass per unit volume, at any given point (x, y, z).

68. MASS
Equation 13
Then, its mass is:

69. Its moments about the three coordinate planes are:
Equations 14
Its moments about the three coordinate planes are:

70. The center of mass is located at the point , where:
Equations 15
The center of mass is located at the point , where:
If the density is constant, the center of mass of the solid is called the centroid of E.

71. The moments of inertia about the three coordinate axes are:
Equations 16
The moments of inertia about the three coordinate axes are:

72. TOTAL ELECTRIC CHARGE
As in Section 16.5, the total electric charge on a solid object occupying a region E and having charge density σ(x, y, z) is:

73. JOINT DENSITY FUNCTION
If we have three continuous random variables X, Y, and Z, their joint density function is a function of three variables such that the probability that (X, Y, Z) lies in E is:

74. JOINT DENSITY FUNCTION
In particular,
The joint density function satisfies: f(x, y, z) ≥ 0

75. APPLICATIONS
Example 5
Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = y2 and the planes x = z, z = 0, and x = 1.

76. The solid E and its projection onto the xy-plane are shown.
APPLICATIONS
Example 5
The solid E and its projection onto the xy-plane are shown.
Fig a, p. 1033
Fig b, p. 1033

77. The lower and upper surfaces of E are the planes z = 0 and z = x.
APPLICATIONS
Example 5
The lower and upper surfaces of E are the planes z = 0 and z = x.
So, we describe E as a type 1 region:
Fig a, p. 1033

78. Then, if the density is ρ(x, y, z), the mass is:
APPLICATIONS
Example 5
Then, if the density is ρ(x, y, z), the mass is:

79. APPLICATIONS
Example 5

80. The other moments are calculated as follows.
APPLICATIONS
Example 5
Due to the symmetry of E and ρ about the xz-plane, we can immediately say that Mxz = 0, and therefore
The other moments are calculated as follows.

81. APPLICATIONS
Example 5

82. APPLICATIONS
Example 5

83. APPLICATIONS
Example 5

84. Therefore, the center of mass is:
APPLICATIONS
Example 5
Therefore, the center of mass is: