Multiple integrals

Multiple Integrals

1. MULTIPLE INTEGRALS
12

2. 12.6
Triple Integrals
In this section, we will learn about:
Triple integrals and their applications.
MULTIPLE INTEGRALS

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. Let’s first deal with the simplest case
where fis defined on a rectangular box:  , , , ,B x y z a x b c y d r z s      
Equation 1 TRIPLE INTEGRALS

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

6. The planes through
the endpoints of these
subintervals parallel to
the coordinate planes
divide the box Binto
lmnsub-boxes
Each sub-box has volume ΔV = ΔxΔyΔz
TRIPLE INTEGRALS   
1 1 1
, , ,
ijk i i j j k k
B x x y y z z
  
  


7. Then, we form the triple Riemann sum
where the sample point
is in B
ijk. 
* * *
1 1 1
,,
l m n
ijk ijk ijk
i j k
f x y z V
  
  
* * *
,,
ijk ijk ijk
x y z
Equation 2 TRIPLE INTEGRALS

8. By analogy with the definition of a double
integral (Definition 5 in Section 11.1),
we define the triple integral as the limit of
the triple Riemann sums in Equation 2.
TRIPLE INTEGRALS

9. The triple integralof fover the box Bis:
if this limit exists.
Again, the triple integral always exists if f
is continuous. 
 
* * *
,,
1 1 1
,,
lim , ,
B
l m n
ijk ijk ijk
l m n
i j k
f x y z dV
f x y z V

  



Definition 3 TRIPLE INTEGRAL

10. We can choose the sample point to be any
point in the sub-box.
However, if we choose it to be the point
(x
i, y
j, z
k) we get a simpler-looking expression:   
,,
1 1 1
, , lim , ,
l m n
i j k
l m n
i j kB
f x y z dV f x y z V

  

TRIPLE INTEGRALS

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

12. FUBINI’S TH. (TRIPLE INTEGRALS)
If fis continuous on the rectangular box
B= [a, b] x [c, d] x [r, s], then 
 
,,
,,
B
s d b
r c a
f x y z dV
f x y z dxdy dz

  
Theorem 4

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

14. 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: 
 
,,
,,
B
b s d
a r c
f x y z dV
f x y z dy dz dx

  
FUBINI’S TH. (TRIPLE INTEGRALS)

15. Evaluate the triple integral
where Bis the rectangular box2
B
xyz dV   , , 0 1, 1 2, 0 3B x y z x y z       
Example 1 FUBINI’S TH. (TRIPLE INTEGRALS)

16. 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.
Example 1 FUBINI’S TH. (TRIPLE INTEGRALS)

17. 3 2 1
22
0 1 0
1
22
32
01
0
2
32
01
13
2 2 2 3
33
00
10
2
2
3 27
4 4 4 4
B
x
x
y
y
xyz dV xyz dx dy dz
x yz
dy dz
yz
dy dz
y z z z
dz dz













  
   
  
  
   


 Example 1 FUBINI’S TH. (TRIPLE INTEGRALS)

18. Now, we define the triple integral over
a general bounded region Ein three-
dimensional space (a solid) by much the same
procedure that we used for double integrals.
See Definition 2 in Section 11.3
INTEGRAL OVER BOUNDED REGION

19. We enclose Ein a box Bof the type given
by Equation 1.
Then, we define a function Fso that it agrees
with fon Ebut is 0 for points in Bthat are
outside E.
INTEGRAL OVER BOUNDED REGION

20. By definition,
This integral exists if fis continuous and
the boundary of Eis “reasonably smooth.”
The triple integral has essentially the same
properties as the double integral (Properties 6–9
in Section 11.3).   , , , ,
EB
f x y z dV F x y z dV 
INTEGRAL OVER BOUNDED REGION

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

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 xand y.

23. That is,
where Dis the projection of E
onto the xy-plane.
TYPE 1 REGION    12
, , , , , ,E x y z x y D u x y z u x y    Equation 5

24. Notice that:
The upper boundary of the solid Eis the surface
with equation z= u
2(x, y).
The lower boundary is the surface z= u
1(x, y).
TYPE 1 REGIONS

25. By the same sort of argument that led to
Formula 3 in Section 11.3, it can be shown
that, if Eis a type 1 region given by
Equation 5, then   


2
1
,
,
, , , ,
u x y
u x y
ED
f x y z dV f x y z dz dA



  
Equation/Formula 6 TYPE 1 REGIONS

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

27. In particular, if
the projection Dof E
onto the xy-plane
is a type I plane
region, then   
1 2 1 2
, , , ( ) ( ), ( , ) ( , )
E
x y z a x b g x y g x u x y z u x y

     
TYPE 1 REGIONS

28. Thus, Equation 6 becomes: 
 
22
11
( ) ( , )
( ) ( , )
,,
,,
E
b g x u x y
a g x u x y
f x y z dV
f x y z dz dy dx

  
Equation 7 TYPE 1 REGIONS

29. If, instead, Dis a type IIplane region, then  
1 2 1 2
, , , ( ) ( ), ( , ) ( , )
E
x y z c y d h y x h y u x y z u x y

     
TYPE 1 REGIONS

30. Then, Equation 6 becomes: 
 
22
11
( ) ( , )
( ) ( , )
,,
,,
E
d h y u x y
c h y u x y
f x y z dV
f x y z dz dx dy

  
Equation 8 TYPE 1 REGIONS

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

32. When we set up a triple integral, it’s wise
to draw twodiagrams:
The solid region E
Its projection Don the xy-plane
Example 2 TYPE 1 REGIONS

33. 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 u
1(x, y) = 0
and u
2(x, y) = 1 –x–y
in Formula 7.
Example 2 TYPE 1 REGIONS

34. 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.
Example 2 TYPE 1 REGIONS

35. This description of Eas a type 1 region
enables us to evaluate the integral as follows.  , , 0 1,0 1 ,0 1
E
x y z x y x z x y

        
E. g. 2—Equation 9TYPE 1 REGIONS

36.  
 


1
2
1 1 1 1 1
0 0 0 0 0
0
2
11
1
2
00
1
3
1
1
2
0
0
1
3
1
6
0
1
4
0
2
1
1
3
1
111
6 4 24
z x y
x x y x
E z
x
yx
y
z
z dV z dz dy dx dy dx
x y dy dx
xy
dx
x dx
x
  
   








  






  


     


 Example 2 TYPE 1 REGIONS

37. TYPE 2 REGION
A solid region Eis of type 2if it is of the form
whereDis the projection of E
onto the yz-plane.  12
, , , , ( , ) ( , )E x y z y z D u y z x u y z   

38. The back surface is x= u
1(y, z).
The front surface is x= u
2(y, z).
TYPE 2 REGION

39. Thus, we have: 
 
2
1
( , )
( , )
,,
,,
E
u y z
u y z
D
f x y z dV
f x y z dx dA




 
Equation 10 TYPE 2 REGION

40. TYPE 3 REGION
Finally, a type 3region is of the form
where:
Dis the projection of E
onto the xz-plane.
y= u
1(x, z) is the left
surface.
y= u
2(x, z) is the right
surface.   12
, , , , ( , ) ,E x y z x z D u x z y u x z   

41. For this type of region, we have:   
1
2( , )
( , )
, , , ,
u x z
u x z
ED
f x y z dV f x y z dy dA


  
Equation 11 TYPE 3 REGION

42. In each of Equations 10 and 11, there may
be two possible expressions for the integral
depending on:
Whether Dis a type Ior type IIplane region
(and corresponding to Equations 7 and 8).
TYPE 2 & 3 REGIONS

43. Evaluate
where Eis the region bounded by
the paraboloid y= x
2
+ z
2
and the plane y= 4.22
E
x z dV
BOUNDED REGIONS Example 3

44. The solid Eis
shown here.
If we regard it as
a type 1 region,
then we need to consider its projection D
1
onto the xy-plane.
Example 3 TYPE 1 REGIONS

45. That is the parabolic
region shown here.
The trace of y= x
2
+ z
2
in the plane z= 0 is
the parabola y= x
2
Example 3 TYPE 1 REGIONS

46. From y= x
2
+ z
2
, we obtain:
So, the lower boundary surface of Eis:
The upper surface is:2
z y x   2
z y x   2
z y x
Example 3 TYPE 1 REGIONS

47. Therefore, the description of Eas a type 1
region is:  
2 2 2
, , 2 2, 4,
E
x y z x x y y x z y x

         
Example 3 TYPE 1 REGIONS

48. Thus, we obtain:
Though this expression is correct,
it is extremely difficult to evaluate.2
22
22
24
22
2
E
yx
x y x
x y dV
x z dz dy dx

  



  
Example 3 TYPE 1 REGIONS

49. So, let’s instead consider Eas a type 3
region.
As such, its projection D
3
onto the xz-plane is
the disk x
2
+ z
2
≤ 4.
Example 3 TYPE 3 REGIONS

50. Then, the left boundary of Eis the paraboloid
y= x
2
+ z
2
.
The right boundary is the plane y= 4.
Example 3 TYPE 3 REGIONS

51. So, taking u
1(x, z) = x
2
+ z
2
and u
2(x, z) = 4
in Equation 11, we have: 
22
3
3
4
2 2 2 2
2 2 2 2
4
xz
ED
D
x y dV x z dy dA
x z x z dA


  

   
  

Example 3 TYPE 3 REGIONS

52. 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 
2
2
24
2 2 2 2
24
4
x
x
x z x z dz dx

  
  
Example 3 TYPE 3 REGIONS

53. That gives: 
 
 
3
2 2 2 2 2 2
22
2
00
22
24
00
2
35
0
4
4
4
4 128
2
3 5 15
ED
x z dV x z x z dA
r r r dr d
d r r dr
rr






    



  


 


Example 3 TYPE 3 REGIONS

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

55. 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.( , , )
E
f x y z dV
APPLICATIONS OF TRIPLE INTEGRALS

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

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

58. Then, the triple integral does represent
the volume of E:
E
V E dV
Equation 12 APPLNS. OF TRIPLE INTEGRALS

59. For example, you can see this in the case
of a type 1 region by putting f(x, y, z) = 1
in Formula 6: 
2
1
( , )
( , )
21
1
( , ) ( , )
u x y
u x y
ED
D
dV dz dA
u x y u x y dA




  

APPLNS. OF TRIPLE INTEGRALS

60. From Section 11.3, we know this represents
the volume that lies between the surfaces
z= u
1(x, y) and z= u
2(x, y)
APPLNS. OF TRIPLE INTEGRALS

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

62. The tetrahedron Tand its projection D
on the xy-plane are shown.
Example 4 APPLICATIONS

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

64. So, we have:
This is obtained by the same calculation
as in Example 4 in Section 11.3
 
1 1 / 2 2 2
0 / 2 0
1 1 / 2
0 / 2
1
3
22
x x y
x
T
x
x
V T dV dz dy dx
x y dy dx
  


  

   

Example 4 APPLICATIONS

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

66. All the applications of double integrals
in Section 11.5 can be immediately
extended to triple integrals.
APPLICATIONS

67. 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).
APPLICATIONS

68. Then, its massis: ,,
E
m x y z dV
Equation 13 MASS

69. Its momentsabout the three coordinate
planes are: 
 
 
,,
,,
,,
yz
E
xz
E
xy
E
M x x y z dV
M y x y z dV
M z x y z dV









Equations 14 MOMENTS

70. The center of massis located at the point
, where:
If the density is constant, the center of mass
of the solid is called the centroidof E. ,,x y z yz xy xz
MM M
x y z
m m m
  
Equations 15 CENTER OF MASS

71. The moments of inertiaabout the three
coordinate axes are:  
  
  
22
22
22
,,
,,
,,
x
E
y
E
z
E
I y z x y z dV
I x z x y z dV
I x y x y z dV









Equations 16 MOMENTS OF INERTIA

72. As in Section 11.5, the total electric charge
on a solid object occupying a region E
and having charge density σ(x, y, z) is: ,,
E
Q x y z dV
TOTAL ELECTRIC CHARGE

73. 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:    , , , ,
E
P X Y Z E f x y z dV
JOINT DENSITY FUNCTION

74. In particular,
The joint density function satisfies:
f(x, y, z) ≥ 0 
 
,,
,,
b d s
a c r
P a X b c Y d r Z s
f x y z dz dy dx
     
  
JOINT DENSITY FUNCTION , , 1f x y z dz dy dx

  


75. Find the center of mass of a solid of constant
density that is bounded by the parabolic
cylinder x= y
2
and the planes x= z, z= 0,
and x= 1.
Example 5 APPLICATIONS

76. The solid Eand its projection onto
the xy-plane are shown.
Example 5 APPLICATIONS

77. The lower and upper
surfaces of Eare
the planes
z= 0 and z= x.
So,we describe E
as a type 1 region:  
2
, , 1 1, 1,0E x y z y y x z x       
Example 5 APPLICATIONS

78. Then, if the density is ρ(x, y, z),
the mass is:
Example 5 APPLICATIONS2
2
11
10
11
1
E
x
y
y
m
dV
dz dxdy
xdxdy









  


79. APPLICATIONS 
 
2
1
2
1
1
1
4
1
1
4
0
1
5
0
2
1
2
1
4
55
x
xy
x
dy
y dy
y dy
y
y
















  




 Example 5

80. Due to the symmetry of Eand ρabout
the xz-plane, we can immediately say that
M
xz= 0, and therefore .
The other moments are calculated
as follows.0y
Example 5 APPLICATIONS

81. 2
2
11
10
11
2
1
yz
E
x
y
y
M
x dV
x dz dxdy
x dxdy









  
 Example 5 APPLICATIONS

82. APPLICATIONS 
2
1
3
1
1
1
6
0
1
7
0
3
2
1
3
2
37
4
7
x
xy
x
dy
y dy
y
y


















 Example 5

83.  
2
2
2
11
10
2
11
1
0
11
2
1
1
6
0
2
2
2
1
37
x
xy
y
E
zx
y
z
y
M z dV z dz dxdy
z
dxdy
x dxdy
y dy















  
   


 Example 5 APPLICATIONS

84. Therefore, the center of mass is: 
 
55
7 14
, , , ,
,0,
yz xyxz
MMM
x y z
m m m




Example 5 APPLICATIONS