Multiple integrals

12.6
Triple Integrals
In this section, we will learn about:
Triple integrals and their applications.
MULTIPLE 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      
Equation 1TRIPLE 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
 Each sub-box has volume ΔV = Δx Δy Δz
TRIPLE INTEGRALS
   1 1 1, , ,ijk i i j j k kB x x y y z z  
    

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

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

The triple integral of f over the box B is:
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 3TRIPLE INTEGRAL

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:
   , ,
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

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

FUBINI’S TH. (TRIPLE INTEGRALS)
If f is 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

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

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

  

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

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.

3 2 1
2 2
0 1 0
12 2
3 2
0 1
0
2
3 2
0 1
1 32 2 2 3
3 3
0 0
1 0
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








 
  
 

  
     
  
   
 
 
 

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 11.3
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.

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 11.3).
   , , , ,
E B
f x y z dV F x y z dV 

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.

That is,
where D is the projection of E
onto the xy-plane.
TYPE 1 REGION
        1 2, , , , , ,E x y z x y D u x y z u x y   
Equation 5

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

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

Thus, Equation 6 becomes:
 
 
2 2
1 1
( ) ( , )
( ) ( , )
, ,
, ,
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 7TYPE 1 REGIONS

If, instead, D is a type II plane 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

Then, Equation 6 becomes:
 
 
2 2
1 1
( ) ( , )
( ) ( , )
, ,
, ,
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 8TYPE 1 REGIONS

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

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

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.

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.

 This description of E as 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

 
 
 
 
12
1 1 1 1 1
0 0 0 0 0
0
21 1
1
2 0 0
13
1
1
2 0
0
1 31
6 0
14
0
2
1
1
3
1
11 1
6 4 24
z x y
x x y x
E z
x
y x
y
z
z dV z dz dy dx dy dx
x y dy dx
x y
dx
x dx
x
  
   


 

 
   
 
  
  
  
  
 
 
   
  
     
 



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

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

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 10TYPE 2 REGION

TYPE 3 REGION
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.
      1 2, , , , ( , ) ,E x y z x z D u x z y u x z   

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

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).
TYPE 2 & 3 REGIONS

Evaluate
where E is the region bounded by
the paraboloid y = x2 + z2 and the plane y = 4.
2 2
E
x z dV
BOUNDED 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.

That is the parabolic
region shown here.
 The trace of y = x2 + z2
in the plane z = 0 is
the parabola y = x2

From y = x2 + z2, we obtain:
 So, the lower boundary surface of E is:
 The upper surface is:
2
z y x  
2
z y x  
2
z y x 

Therefore, the description of E as a type 1
region is:
  2 2 2
, , 2 2, 4,
E
x y z x x y y x z y x

         

Thus, we obtain:
 Though this expression is correct,
it is extremely difficult to evaluate.
2
2 2
2 2
2 4
2 2
2
E
y x
x y x
x y dV
x z dz dy dx

  

 

  

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.

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

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
2 4
2 2 2 2
2 4
4
x
x
x z x z dz dx

  
   

That gives:
 
 
 
3
2 2 2 2 2 2
2 2
2
0 0
2 2
2 4
0 0
23 5
0
4
4
4
4 128
2
3 5 15
E D
x z dV x z x z dA
r r r dr d
d r r dr
r r






    
 
 
 
   
 
 
 
 

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.
( )
b
a
f x dx
( , )
D
f x y dA
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.
( , , )
E
f x y z dV

Remember that E is just the domain
of the function f.
 The graph of f lies in 4-D space.

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.
( , , )
E
f x y z dV

Then, the triple integral does represent
the volume of E:
 
E
V E dV 
Equation 12APPLNS. 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:
 
2
1
( , )
( , )
2 1
1
( , ) ( , )
u x y
u x y
E D
D
dV dz dA
u x y u x y dA
 
  
 
  

APPLNS. OF TRIPLE INTEGRALS

From Section 11.3, we know this represents
the volume that lies between the surfaces
z = u1(x, y) and z = u2(x, y)
APPLNS. OF TRIPLE INTEGRALS

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
Example 4APPLICATIONS

The tetrahedron T and its projection D
on the xy-plane are shown.

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

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
2 2
x x y
x
T
x
x
V T dV dz dy dx
x y dy dx
  

 
  

   
 

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

All the applications of double integrals
in Section 11.5 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).
APPLICATIONS

Then, its mass is:
 , ,
E
m x y z dV 
Equation 13MASS

Its moments about 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 14MOMENTS

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.
 , ,x y z
yz xyxz
M MM
x y z
m m m
  
Equations 15CENTER OF MASS

The moments of inertia about the three
coordinate axes are:
   
   
   
2 2
2 2
2 2
, ,
, ,
, ,
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 16MOMENTS OF INERTIA

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

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

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
  
  
  

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.

The solid E and its projection onto
the xy-plane are shown.

The lower and upper
surfaces of E are
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       

Then, if the density is ρ(x, y, z),
the mass is:
2
2
1 1
1 0
1 1
1
E
x
y
y
m
dV
dz dxdy
xdxdy









  
 

APPLICATIONS
 
 
2
12
1
1
1
4
1
1
4
0
15
0
2
1
2
1
4
5 5
x
x y
x
dy
y dy
y dy
y
y









 
  
 
 
 
 
   
 



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.
0y 

2
2
1 1
1 0
1 1
2
1
yz
E
x
y
y
M
x dV
x dz dxdy
x dxdy









  
 

APPLICATIONS
 
2
13
1
1
1
6
0
17
0
3
2
1
3
2
3 7
4
7
x
x y
x
dy
y dy
y
y







 
  
 
 
 
  
 



Example 5

 
2
2
2
1 1
1 0
2
1 1
1
0
1 1
2
1
1
6
0
2
2
2
1
3 7
x
xy y
E
z x
y
z
y
M z dV z dz dxdy
z
dxdy
x dxdy
y dy
 


 





 
 
  
 

  
   
 
 


Therefore, the center of mass is:
 
 5 5
7 14
, , , ,
,0,
yz xyxz
M MM
x y z
m m m
 
  
 


Multiple integrals

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