SECTION 12.

TRIPLE INTEGRALS
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.
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

12.5 P2
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 Δxi = xi – xi–1.
 [c, d] into m subintervals of width Δyj = yj – yj–1.
 [r, s] into n subintervals of width Δzk = zk – zk–1.

12.5 P3
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 
which are shown in Figure 1.
 Each sub-box Bijk has volume
ΔVijk = ΔxiΔyjΔzk
12.5 P4
TRIPLE INTEGRALS

Then, we form the triple Riemann sum

  ijk ijk ijk  Vijk

l m n
* * *
f x , y , z
i 1 j 1 k 1

where the sample point  xijk* *

, yijk *
, zijk  is in Bijk.
By analogy with the definition of a double
integral (Definition 5 in Section 12.1), we
define the triple integral as the limit of the triple
Riemann sums in Equation 2.

12.5 P5
Definition 3

The triple integral of f over the box B is:

 f  x, y, z  dV
B

f  xijk  Vijk
l m n
 lim
max xi , y j , zk 0

i 1 j 1 k 1
* *
, yijk *
, zijk

if this limit exists.

Again, the triple integral always exists if f is

continuous.
12.5 P6
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  V
l m n

 f  x, y, z  dV 
B
lim
l , m , n 
i 1 j 1 k 1
i j k

12.5 P7
TRIPLE INTEGRALS

Just as for double integrals, the practical method

for evaluating triple integrals is to express them
as iterated integrals, as follows.

12.5 P8
FUBINI’S THEOREM FOR TRIPLE
INTEGRALS

If f is continuous on the rectangular box

B = [a, b] × [c, d] × [r, s],
then
 f  x, y, z  dV
B

  f  x, y, z  dx dy dz
s d b

r c a

12.5 P9
FUBINI’S THEOREM FOR 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

12.5 P10
FUBINI’S THEOREM FOR 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

  f  x, y, z  dy dz dx
b s d

a r c

12.5 P11
Example 1

Evaluate the triple integral 

2
xyz dV
B

where B is the rectangular box

B   x, y, z  0  x  1,  1  y  2, 0  z  3

12.5 P12
Example 1 SOLUTION

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
3 2 1
 xyz dV    
2
xyz 2 dx dy dz
0 1 0
B
x 1
3 x yz 
2
2 2
3 2 yz
2
    dy dz    dy dz
0 1 0 1 2
 2  x 0
y 2 3
y z 
3
2 2
3 3z z  2
27 3
   dz   dz   
0
 4  y 1 0 4 4 0 4
12.5 P13
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 12.2.

12.5 P14
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.

12.5 P15
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 12.2).

12.5 P16
INTEGRAL OVER BOUNDED REGION

We restrict our attention to:

 Continuous functions f
 Certain simple types of regions
A solid region E is said to be of type 1 if it lies
between the graphs of two continuous functions
of x and y.

12.5 P17
TYPE 1 REGION

That is,
E  x, y, z   x, y   D, u  x, y   z  u  x, y 
1 2

where D is the projection of E onto the xy-plane

as shown in Figure 2.

12.5 P18
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).

12.5 P19
TYPE 1 REGIONS

By the same sort of argument that led to

Formula 3 in Section 12.2, it can be shown that,
if E is a type 1 region given by Equation 5, then
u2  x , y 

f  x, y, z  dV    f  x, y, z  dz  dA

E D
 u1  x , y  

12.5 P20
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.

12.5 P21
 TYPE 1 REGIONS

In particular, if the projection D of E onto the

xy-plane is a type I plane region (as in Figure 3),
then
E   x, y, z  a  x  b, g1 ( x)  y  g 2 ( x), u1 ( x, y)  z  u2 ( x, y)

12.5 P22
TYPE 1 REGIONS

Thus, Equation 6 becomes:

 f  x, y, z  dV
E

f  x, y, z  dz dy dx
b g2 ( x ) u2 ( x , y )
  
a g1 ( x ) u1 ( x , y )

12.5 P23
 TYPE 1 REGIONS

If, instead, D is a type II plane region (as in

Figure 4), then
E   x, y, z  c  y  d , h1 ( y )  x  h2 ( y ), u1 ( x, y )  z  u2 ( x, y )

12.5 P24
TYPE 1 REGIONS

Then, Equation 6 becomes:

 f  x, y, z  dV
E

f  x, y, z  dz dx dy
d h2 ( y ) u2 ( x , y )
  
c h1 ( y ) u1 ( x , y )

12.5 P25
Example 2

Evaluate  z dV
E

where E is the solid tetrahedron bounded by the

four planes
x = 0, y = 0, z = 0, x + y + z = 1

12.5 P26
Example 2 SOLUTION

When we set up a triple integral, it’s wise to

draw two diagrams:
 The solid region E (See Figure 5)
 Its projection D on the xy-plane (See Figure 6)

12.5 P27
Example 2 SOLUTION

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.

12.5 P28
Example 2 SOLUTION

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 in
Figure 6, and we have the
following equation.

12.5 P29
Example 2 SOLUTION

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.
z 1 x  y
1 1 x 1 x  y 1 1 x z 2

 z dV    
E
0 0 0
z dz dy dx  
0 0  2
  z 0
dy dx

y 1 x
1 1 x 2  1  x  y  3

1  x  y  dy dx  12 0  
1
 1
2 
0 0
 3

 y 0
dx

1
1  1  x  
4
1
  1  x  dx   
1
 
1 3
6
0 6  4  0 24
12.5 P30
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 2

where D is the projection of E onto the yz-plane.

See Figure 7.

12.5 P31
TYPE 2 REGION

The back surface is x = u1(y, z).

The front surface is x = u2(y, z).

12.5 P32
TYPE 2 REGION

Thus, we have:

 f  x, y, z  dV
E


  
u2 ( y , z )
f  x, y, z  dx  dA
 u1 ( y , z ) 
D

12.5 P33
TYPE 3 REGION

Finally, a type 3 region is of the form

E  x, y, z   x, z   D, u ( x, z)  y  u  x, z 
1 2

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.
 See Figure 8.
12.5 P34
TYPE 3 REGION

For this type of region, we have:


f  x, y, z  dV   
u 2( x , z )
f  x, y, z  dy  dA

E D
 u1 ( x , z ) 

12.5 P35
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).

12.5 P36
Example 3

Evaluate 
E
x 2  z 2 dV

where E is the region bounded by the paraboloid

y = x2 + z2 and the plane y = 4.

12.5 P37
Example 3 SOLUTION

The solid E is shown in Figure 9.

If we regard it as a type 1 region, then we need
to consider its projection D1 onto the xy-plane.

12.5 P38
Example 3 SOLUTION

That is the parabolic region shown in Figure 10.

 The trace of y = x2 + z2 in the plane z = 0 is the
parabola y = x2

12.5 P39
Example 3 SOLUTION

From y = x2 + z2, we obtain:

z   y  x2

 So, the lower boundary surface of E is:

z   y  x2
 The upper surface is:
z  y  x2

12.5 P40
Example 3 SOLUTION

Therefore, the description of E as a type 1

region is:

E   x, y , z  2  x  2, x 2
 y  4,  y  x 2
 z  y  x 2

Thus, we obtain:
2 4 y  x2
 E
x 2  y 2 dV  
2 x 2   y  x2
x 2  z 2 dz dy dx

 Although this expression is correct, it is extremely

difficult to evaluate.

12.5 P41
Example 3 SOLUTION

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 shown in Figure 11.

12.5 P42
Example 3 SOLUTION

Then, the left boundary of E is the paraboloid

y = x2 + z2.
The right boundary is the plane y = 4.
So, taking u1(x, z) = x2 + z2 and u2(x, z) = 4 in
Equation 11, we have:

x  y dV    2 2 x 2  z 2 dy  dA
 4

2 2
 x  z 
E D 3

   4  x  z
2 2
 x  z dA
2 2

D3

12.5 P43
Example 3 SOLUTION

This integral could be written as:

4  x 
2 4 x2
  z x  z dz dx
2 2 2 2
2  4 x 2

However, it’s easier to convert to polar

coordinates in the xz-plane:
x = r cos , z = r sin 

12.5 P44
Example 3 SOLUTION

That gives:

 x  z dV
2 2

   4  x  z    r r dr d
2 2
2 2
x  z dA  
2 2
 4  r 2
0 0
D3
2
 4r r  128
3 5
d   4r  r  dr  2 
2 2
 2 4
  
0 0
 3 5 0 15

12.5 P45
Example 4

Express the iterated integral

x2
   f  x, y, z  dz dy dx
1 y

0 0 0

as a triple integral and then rewrite it as iterated

integral in a different order, integrating first
with respect to x, then z, and then y.

12.5 P46
Example 4 SOLUTION

We can write

x2
   f  x, y, z  dz dy dx     f  x, y, z  dV
1 y

0 0 0
E

where
E   x, y , z  0  x  1, 0  y  x 2

, 0 z y

12.5 P47
Example 4 SOLUTION

This description of E enable us to write

projections onto the three coordinate planes as
follows：
on the xy -plane D1   x, y  0  x  1, 0  y  x 2

  x, y  0  y  1, y  x 1 
on the yz -plane D2   x, y  0  y  1, 0  z  y
on the xz -plane D3   x, y  0  x  1, 0  z  x2 
12.5 P48
Example 4 SOLUTION

From the resulting of the projections in Figure

12 we sketch the solid E in Figure 13.

12.5 P49
Example 4 SOLUTION

We see that it is the solid enclosed by the planes

z  0, x  1, y  z and the parabolic cylinder
y  x (or x  y ) .
2

If we integrate first with respect to x, then z, and

then y, we use an alternate description of E：
E  x, y, z  0  x  1, 0  z  y, y  x 1 
Thus

   f  x, y, z  dV     f  x, y, z  dx dz dy
1 y 1

0 0 y
E

12.5 P50
APPLICATIONS OF TRIPLE INTEGRALS

Recall that:
b
 If f(x) ≥ 0, then the single integral  f ( x) dx represents
a
the area under the curve y = f(x) from a to b.

 If f(x, y) ≥ 0, then the double integral  f ( x, y) dA

D
represents the volume under the surface z = f(x, y) and
above D.

12.5 P51
APPLICATIONS OF TRIPLE INTEGRALS

The corresponding interpretation of a triple

integral  f ( x, y, z ) dV, where f(x, y, z) ≥ 0, is

E

not very useful.

 It would be the “hypervolume” of a four-dimensional
(4-D) object.
 Of course, that is very difficult to visualize.

12.5 P52
APPLICATIONS OF TRIPLE INTEGRALS

Remember that E is just the domain of the

function f.
 The graph of f lies in 4-D space.

12.5 P53
APPLICATIONS OF TRIPLE INTEGRALS

Nonetheless, the triple integral   E f ( x, y, z ) dV

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).

12.5 P54
APPLNS. OF TRIPLE INTEGRALS

Let’s begin with the special case where

f(x, y, z) = 1 for all points in E.
Then, the triple integral does represent the
volume of E:

V  E    dV
E

12.5 P55
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  u ( x, y )  u ( x, y )  dA

E
1 dV  
D
 u1 ( x , y ) 
dz 
D
2 1

From Section 12.2, we know this represents the

volume that lies between the surfaces

z = u1(x, y) and z = u2(x, y)

12.5 P56
Example 5

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

12.5 P57
Example 5 SOLUTION

The tetrahedron T and its projection D on the

xy-plane are shown in Figure 14 and 15.

12.5 P58
Example 5 SOLUTION

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

12.5 P59
Example 5 SOLUTION

So, we have:
1 1 x / 2 2 x 2 y
V T    dV     dz dy dx
0 x/2 0
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 12.2

12.5 P60
Example 5 SOLUTION

Notice that it is not necessary to use triple

integrals to compute volumes.
 They simply give an alternative method for setting
up the calculation.

12.5 P61
APPLICATIONS

All the applications of double integrals in

Section 12.4 can be immediately extended to
triple integrals.

12.5 P62
APPLICATIONS

For example, suppose the density function of a

solid object that occupies the region E is:
r(x, y, z)
in units of mass per unit volume, at any given
point (x, y, z).
Then, its mass is:
m   r  x, y, z  dV
E

12.5 P63
MOMENTS

Its moments about the three coordinate planes

are:
M yz   x r  x, y, z  dV
E

M xz   y r  x, y, z  dV
E

M xy   z r  x, y, z  dV
E

12.5 P64
CENTER OF MASS

The center of mass is located at the point

 x , y , z  , where:
M yz M xz M xy
x y z
m m m
 If the density is constant, the center of mass of the
solid is called the centroid of E.

12.5 P65
MOMENTS OF INERTIA

The moments of inertia about the three

coordinate axes are:
I x    y 2  z 2  r  x, y, z  dV
E

I y    x  z
2 2
 r  x, y, z  dV
E

I z    x 2  y 2  r  x, y, z  dV
E

12.5 P66
TOTAL ELECTRIC CHARGE

As in Section 12.4, the total electric charge on

a solid object occupying a region E and having
charge density s(x, y, z) is:

Q   s  x, y, z  dV
E

12.5 P67
Example 6

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.

12.5 P68
Example 6 SOLUTION

The solid E and its

projection onto the xy-
plane are shown in Figure
16.

12.5 P69
Example 6 SOLUTION

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 
 x, y , z  1  y  1, y 2

 x  1,0  z  x

12.5 P70
Example 6 SOLUTION

Then, if the density is r(x, y, z) = r, the mass is:

1 1 x
m   r dV     r dz dx dy
1 y 2
0
E
x 1
1 1 x  1
2
 r   2 x dx dy  r    dy
1 y 1 2
  x y2
r
1  y  dy  r  1  y  dy
1 1
 
4 4

2 1 0

1
 y  4r 5
 r y   
 5 0 5
12.5 P71
Example 6 SOLUTION

Due to the symmetry of E and r about the xz-

plane, we can immediately say that Mxz = 0, and
therefore y  0.
The other moments are calculated as follows.

12.5 P72
Example 6 SOLUTION

1 1 x
M yz   x r dV     x r dz dx dy
1 y 2
0
E
x 1
1 1 x  1
3
 r   2 x dx dy  r    dy
2
1 y 1 3
  x y2
1
2r 2r  y 7

0 1  y  dy  3
1
 6
y  7 
3  0
4r

7
12.5 P73
Example 6 SOLUTION

1 1 x
M xy   z r dV     z r dz dx dy
1 y 2
0
E
zx
1 z 
1
2
 r   2   dx dy
1 y
 2  z 0
r 1 1

2 
 2
x dx dy
1 y 2

r 2r
  1  y  dy 
1
6

3 0 7

12.5 P74
Example 6 SOLUTION

Therefore, the center of mass is:

 x, y, z 
 M yz M xz M xy 
 , , 
 m m m 
5 5
  , 0, 
7 14 

12.5 P75