15
MULTIPLE INTEGRALS
MULTIPLE INTEGRALS
15.6
Triple Integrals
In this section, we will learn about:
Triple integrals and their applications.
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.
TRIPLE INTEGRALS
Equation 1
Let’s first deal with the simplest case
where f is defined on a rectangular box:
B   x, y, z  a  x  b, c  y  d , r  z  s
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.
TRIPLE INTEGRALS
The planes through
the endpoints of these
subintervals parallel to
the coordinate planes
divide the box B into
lmn sub-boxes
Bijk   xi 1 , xi    y j 1 , y j    zk 1 , zk 
 Each sub-box has volume ΔV = Δx Δy Δz
TRIPLE INTEGRALS
Equation 2
Then, we form the triple Riemann sum
 f  x
l
m
n
i 1 j 1 k 1
*
ijk

*
ijk
*
ijk
,y ,z
 V
*
*
*
where the sample point xijk
, yijk , zijk
is in Bijk.

TRIPLE INTEGRALS
By analogy with the definition of a double
integral (Definition 5 in Section 15.1),
we define the triple integral as the limit of
the triple Riemann sums in Equation 2.
TRIPLE INTEGRAL
Definition 3
The triple integral of f over the box B is:
 f  x, y, z  dV
B
 lim
l , m , n 
l
m
n

i 1 j 1 k 1
*
*
*
f  xijk
, yijk
, zijk
 V
if this limit exists.
 Again, the triple integral always exists if f
is continuous.
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:
 f  x, y, z  dV 
B
 f  x , y , z  V
l
lim
l , m , n 
m
n
i 1 j 1 k 1
i
j
k
TRIPLE INTEGRALS
Just as for double integrals, the practical
method for evaluating triple integrals is
to express them as iterated integrals, as
follows.
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
 f  x, y, z  dV
B

s
r
d
b
c
a
 
f  x, y, z  dx dy dz
FUBINI’S TH. (TRIPLE INTEGRALS)
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 y and z fixed)
2. With respect to y (keeping z fixed)
3. With respect to z
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:
 f  x, y, z  dV
B

b
a
  f  x, y, z  dy dz dx
s
d
r
c
FUBINI’S TH. (TRIPLE INTEGRALS) Example 1
Evaluate the triple integral
 xyz dV
2
B
where B is the rectangular box
B   x, y, z  0  x  1,  1  y  2, 0  z  3
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.
FUBINI’S TH. (TRIPLE INTEGRALS) Example 1
3
2
0
1 0
 xyz dV    
2
1
2
xyz dx dy dz
B
x 1
 x yz 
  
dy dz

0 1
 2  x 0
3
2
2
2
2
yz
 
dy dz
0 1 2
3
2
y 1
3
3 3z
y z 
z 
27
 
dz  
dz   

0
0 4
4
4 0 4

 y 1
3
2 2
2
3
INTEGRAL OVER BOUNDED REGION
Now, we define the triple integral over
a general bounded region E in threedimensional space (a solid) by much the same
procedure that we used for double integrals.
 See Definition 2 in Section 15.3
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.
INTEGRAL OVER BOUNDED REGION
By definition,
 f  x, y, z  dV   F  x, y, z  dV
E
B
 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).
INTEGRAL OVER BOUNDED REGION
We restrict our attention to:
 Continuous functions f
 Certain simple types of regions
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.
TYPE 1 REGION
Equation 5
That is,
E
 x, y, z   x, y   D, u  x, y   z  u
1
where D is the projection of E
onto the xy-plane.
2
 x, y 
TYPE 1 REGIONS
Notice that:
 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).
TYPE 1 REGIONS
Equation/Formula 6
By the same sort of argument that led to
Formula 3 in Section 15.3, it can be shown
that, if E is a type 1 region given by
Equation 5, then

E
u2  x , y 


f  x, y, z  dV   
f  x, y, z  dz dA
 u1  x , y 

D
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.
TYPE 1 REGIONS
In particular, if
the projection D of E
onto the xy-plane
is a type I plane
region, then
E
 x, y, z  a  x  b, g ( x)  y  g ( x), u ( x, y)  z  u ( x, y)
1
2
1
2
TYPE 1 REGIONS
Equation 7
Thus, Equation 6 becomes:
f  x, y, z  dV

E

b
a

g2 ( x )
g1 ( x )

u2 ( x , y )
u1 ( x , y )
f  x, y, z  dz dy dx
TYPE 1 REGIONS
If, instead, D is a type II plane region, then
E
 x, y, z  c  y  d , h ( y)  x  h ( y), u ( x, y)  z  u ( x, y)
1
2
1
2
TYPE 1 REGIONS
Equation 8
Then, Equation 6 becomes:
f  x, y, z  dV

E

d
c

h2 ( y )
h1 ( y )

u2 ( x , y )
u1 ( x , y )
f  x, y, z  dz dx dy
TYPE 1 REGIONS
Evaluate
Example 2
 z dV
E
where E is the solid tetrahedron
bounded by the four planes
x = 0, y = 0, z = 0, x + y + z = 1
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
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.
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.
TYPE 1 REGIONS
E. g. 2—Equation 9
E
 x, y, z  0  x  1, 0  y  1  x, 0  z  1  x  y
 This description of E as a type 1 region
enables us to evaluate the integral as follows.
TYPE 1 REGIONS
1 1 x
1 x  y
0 0
0
 z dV    
E
Example 2
z dz dy dx  
1 1 x

0 0

1
2
z 1 x  y
z 
2
  z 0
2
dy dx
2
1 1 x
  1  x  y  dy dx
0 0
 1  x  y 
 12   
0
3

1

1
6
 1  x 
1
3
0
1  1  x 
 
6 
4
3
y 1 x

dx

 y 0
dx
4
1

1
 
 0 24
TYPE 2 REGION
A solid region E is of type 2 if it is of the form
E
 x, y, z   y, z   D, u ( y, z)  x  u ( y, z)
1
where D is the projection of E
onto the yz-plane.
2
TYPE 2 REGION
The back surface is x = u1(y, z).
The front surface is x = u2(y, z).
TYPE 2 REGION
Equation 10
Thus, we have:

f  x, y, z  dV
E


  
f  x, y, z  dx dA
 u1 ( y , z )

D
u2 ( y , z )
TYPE 3 REGION
Finally, a type 3 region is of the form
E
 x, y, z   x, z   D, u ( x, z)  y  u
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.
1
2
 x, z 
TYPE 3 REGION
Equation 11
For this type of region, we have:

E
u 2( x , z )

f  x, y, z  dV   
f  x, y, z  dy  dA
 u1 ( x , z )

D
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).
BOUNDED REGIONS
Evaluate

Example 3
x 2  z 2 dV
E
where E is the region bounded by
the paraboloid y = x2 + z2 and the plane y = 4.
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.
TYPE 1 REGIONS
That is the parabolic
region shown here.
 The trace of y = x2 + z2
in the plane z = 0 is
the parabola y = x2
Example 3
TYPE 1 REGIONS
Example 3
From y = x2 + z2, we obtain:
z  yx
 So, the lower boundary surface of E is:
2
z   y  x2
 The upper surface is:
z
y  x2
TYPE 1 REGIONS
Example 3
Therefore, the description of E as a type 1
region is:
E

 x, y, z  2  x  2, x 2  y  4, 
y  x2  z 
y  x2

TYPE 1 REGIONS
Example 3
Thus, we obtain:

x  y dV
2
2
4
y  x2
2 x
 yx
E

2
 
2
x  z dz dy dx
2
2
 Though this expression is correct,
it is extremely difficult to evaluate.
2
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.
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.
TYPE 3 REGIONS
Example 3
So, taking u1(x, z) = x2 + z2 and u2(x, z) = 4
in Equation 11, we have:

E
2
2

x  y dV    2 2 x  z dy  dA
 x  z

D
2
4
2
3
   4  x  z
2
D3
2

x  z dA
2
2
TYPE 3 REGIONS
Example 3
This integral could be written as:
2
 
4 x2
2  4  x
2
4  x
2
z
2

x  z dz dx
2
2
However, it’s easier to convert to
polar coordinates in the xz-plane:
x = r cos θ, z = r sin θ
TYPE 3 REGIONS
Example 3
That gives:

E
x  z dV    4  x  z
2
2
2
2

x  z dA
2
D3
  4  r  r r dr d

  d   4r  r  dr
2
2
0
0

2
2
2
0
0
2
4
2
 4r r  128
 2 
  
5 0
15
 3
3
5
2
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.

b
a
f ( x) dx
 If f(x, y) ≥ 0, then the double integral
represents the volume under
the surface z = f(x, y) and above D.
 f ( x, y) dA
D
APPLICATIONS OF TRIPLE INTEGRALS
The corresponding interpretation of a triple
integral  f ( x, y, z ) dV, where f(x, y, z) ≥ 0,
E
is not very useful.
 It would be the “hypervolume”
of a four-dimensional (4-D) object.
 Of course, that is very difficult to visualize.
APPLICATIONS OF TRIPLE INTEGRALS
Remember that E is just the domain
of the function f.
 The graph of f lies in 4-D space.
APPLICATIONS OF TRIPLE INTEGRALS
Nonetheless, the triple integral
 f ( x, y, z) dV
E
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.
APPLNS. OF TRIPLE INTEGRALS
Equation 12
Then, the triple integral does represent
the volume of E:
V  E    dV
E
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:
u2 ( x , y )

 dA
1
dV

dz


 u1 ( x , y ) 
E
D
  u2 ( x, y )  u1 ( x, y )  dA
D
APPLNS. OF TRIPLE INTEGRALS
From Section 15.3, we know this represents
the volume that lies between the surfaces
z = u1(x, y)
and
z = u2(x, y)
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
APPLICATIONS
Example 4
The tetrahedron T and its projection D
on the xy-plane are shown.
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
APPLICATIONS
Example 4
So, we have:
V T    dV  
1 1 x / 2
0

x/2

2 x  2 y
0
dz dy dx
T

1 1 x / 2
0

x/2
 2  x  2 y  dy dx
 13
 This is obtained by the same calculation
as in Example 4 in Section 15.3
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.
APPLICATIONS
All the applications of double integrals
in Section 15.5 can be immediately
extended to triple integrals.
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).
MASS
Equation 13
Then, its mass is:
m     x, y, z  dV
E
MOMENTS
Equations 14
Its moments about the three coordinate
planes are:
M yz   x   x, y, z  dV
E
M xz   y   x, y, z  dV
E
M xy   z   x, y, z  dV
E
CENTER OF MASS
Equations 15
The center of mass is located at the point
 x, y, z , where:
x
M yz
m
M xz
y
m
z
M xy
m
 If the density is constant, the center of mass
of the solid is called the centroid of E.
MOMENTS OF INERTIA
Equations 16
The moments of inertia about the three
coordinate axes are:
I x    y  z
2
2
   x, y, z  dV
E
I y    x  z
2
   x, y, z  dV
I z    x  y
2
   x, y, z  dV
2
E
2
E
TOTAL ELECTRIC CHARGE
As in Section 15.5, the total electric charge
on a solid object occupying a region E
and having charge density σ(x, y, z) is:
Q     x, y, z  dV
E
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:
P   X , Y , Z   E    f  x, y, z  dV
E
JOINT DENSITY FUNCTION
In particular,
P  a  X  b, c  Y  d , r  Z  s 

b
a
  f  x, y, z  dz dy dx
d
s
c
r
The joint density function satisfies:
f(x, y, z) ≥ 0


  

  
f  x, y, z  dz dy dx  1
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.
APPLICATIONS
Example 5
The solid E and its projection onto
the xy-plane are shown.
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:
E


2
x
,
y
,
z

1

y

1,
y
 x  1, 0  z  x


APPLICATIONS
Example 5
Then, if the density is ρ(x, y, z),
the mass is:
m
   dV
E
1
x
1 y
0

1
 
 
1
2

1
1 y
2
 dz dx dy
x dx dy
APPLICATIONS
Example 5
x 1
x 
     dy
1 2
  x y2
1


2
1  y  dy


2
1
4
1
   1  y  dy
1
4
0
1

y 
4
  y   
5 0
5

5
APPLICATIONS
Example 5
Due to the symmetry of E and ρ about
the xz-plane, we can immediately say that
Mxz = 0, and therefore y  0 .
The other moments are calculated
as follows.
APPLICATIONS
M yz
  x  dV
E
1
x
1 y
0

1
 
 
1
2

1
1 y
2
x  dz dx dy
2
x dx dy
Example 5
APPLICATIONS
Example 5
x 1
x 
     dy
1 3
  x y2
3
1
2

3
1

y
dy



2

3

y 
y  7 

0
4

7
1
6
0
7
1
APPLICATIONS
Example 5
1
x
1 y
0
M xy   z  dV  
1
E
 
2
z  dz dx dy
zx
z 
    2   dx dy
1 y
 2  z 0
1


1
1


2
1
1 y

2
2
2
x dx dy
2
  1  y  dy 
3 0
7
1
6
APPLICATIONS
Example 5
Therefore, the center of mass is:

 M yz M xz M xy 
x, y , z  
,
,

m
m 
 m
5
5
  7 , 0, 14 
