Triple Integrals
Triple Integrals
Triple integrals over a box: E  ( x, y, z ) a  x  b, c  y  d , r  z  s
b d
s
f ( x, y, z )dV    

a c r
E
f ( x, y, z )dzdydz
By Fubini’s Theorem we can
change the order of integration in
six possible ways.
Applications:
1.  1dV is the volume of the solid E.
E
2.
If 𝑓(𝑥, 𝑦, 𝑧) is the mass density of the solid E (e.g. in kg/m3), then
 f ( x, y, z )dV is the total mass of the solid (in kg.)
E
Triple integrals
y
2
xe
 sin zdV where E is the solid defined by
Example: Evaluate
E
E  ( x, y , z ) 1  x  2, 0  y  1, 0  z   
2 1 
2 1
y

2
xe
sin
zdV

2
xe
sin
zdzdydx

2
xe
[

cos
z
]
0 dydx



y
y
E
1 0 0
2 1
   2 xe
1 0
1 0
2 1
y
  cos   cos 0  dydx  4   xe
1 0
2
y
dydx  4  x[e y ]10 dx
1
 x   4( e  1)  2  1   6( e  1)
 4( e  1)  xdx  4( e  1)  


2


1
 2 1
2
2
2
By Fubini’s Theorem we can change the order of integration:
2  1
 2 xe sin zdV     2 xe sin zdydzdx
y
E
y
1 0 0
 2 1
y
2
xe
sin
zdV

2
xe

   sin zdydxdz
y
E
0 1 0
 1 2
y
y
2
xe
sin
zdV

2
xe

   sin zdxdydz
E
0 0 1
1  2
y
2
xe
sin
zdV

2
xe

   sin zdxdzdy
y
E
0 0 1
1 2 
y
y
2
xe
sin
zdV

2
xe

   sin zdzdxdy
E
0 1 0
All these integrals give the same result.
Triple Integrals Type 1
Triple integrals over more general regions:
There are six different orders of integration possible in a triple iterated integral.
Type 1: Let D be the projection of the solid on the xy-plane and let 𝑧 = 𝑢1 (𝑥, 𝑦)
and 𝑧 = 𝑢2 (𝑥, 𝑦) be the surfaces forming the “bottom” and the “top” of the solid
respectively.
u2 ( x , y )
f ( x, y, z )dV   

u ( x, y )
V
D
f ( x, y, z )dzdA
1
There are two different orders of
integration on D. For instance, as a type I
region we obtain the integral:
b g2 ( x ) u2 ( x , y )
f ( x, y, z )dV   

a g ( x ) u ( x , y )
V
1
1
f ( x, y, z )dzdydx
Triple Integrals Type 2
Let D be the projection of the solid on the yz-plane and let 𝑥 = 𝑢1 (𝑦, 𝑧) and
𝑥 = 𝑢2 (𝑦, 𝑧) be the surfaces forming the “back” and “front” of the solid.
u2 ( y , z )
f ( x, y, z )dV   

u ( y ,z )
V
D
f ( x, y, z )dxdA
1
There are two different orders of
integration on D. For instance, as a
type I region we obtain the integral:
b g2 ( y ) u2 ( y , z )
f ( x, y, z )dV   

a g ( y ) u ( y , z )
V
1
1
f ( x, y, z )dxdzdy
Triple Integrals Type 3
Let D be the projection of the solid on the xz-plane and let 𝑦 = 𝑢1 (𝑥, 𝑧) and
𝑦 = 𝑢2 (𝑥, 𝑧) be the surfaces forming the “left” and “right” sides of the solid.
u2 ( x , z )
f ( x, y, z )dV   

u ( x,z )
V
D
f ( x, y, z )dydA
1
There are two different orders of
integration on D. For instance, as a type I
region we obtain the integral:
b g2 ( x ) u2 ( x , z )
f ( x, y, z )dV   

a g ( x ) u ( x , z )
V
1
f ( x, y, z )dydzdx
1
Remarks:
1. The limits of integration for the middle integral can involve only the
outmost variable of integration.
2. The outside limits must be constant.
Triple Integrals Example 1
Evaluate
E xdV
planes and
where E is the tetrahedron bounded by the coordinate
x y z
  1
2 3 4
x y

z

4
1

 
The solid is bounded below by 𝑧 = 0 and above by

 2 3
Let D be the projection of the tetrahedron on the xy-plane.


41 x  y 
 2 3 x
D 0
E xdV   
2
 x
 x y
31  41  
 2
 2 3
0
0
x
y


41  
2
3


0
dzdA  0 

2
0
The arrow enters and exits the solid
at the limit of integration for z
 x
31 
 2
0

xz

x dzdydx
dydx  
0
 x
3 1 
 2
0

x y

4 x  1    dydx
 2 3
 x
3 1 
 2
 
x  y2 
  4 x  y 1    
0
  2 6 0
2
dx
  x 2 9  x 2 
  4 x  3  1     1    dx
0
  2 6 2 
2
2
x

  6 x  1   dx  2
0
 2
2
D
2
Triple Integrals Example 2
Use a triple integral to find the volume of the solid bounded by the paraboloid
𝑥 = 4𝑦 2 + 4𝑧 2 and the plane x = 4.
A line parallel to the x-axis intersects the solid at 𝑥 = 4𝑦 2 + 4𝑧 2 and at x = 4.
These are the limits of integration for x.
𝒙𝟐 + 𝒛𝟐 = 𝟏
Let D be the projection of the solid in the
yz-plane.
The surfaces 𝑥 = 4𝑦 2 + 4𝑧 2 and x = 4 intersect
on a curve C: 4y2 + 4z2 = 4 → y2 + z2 = 1
The circle is the boundary of the region D:
 1  y2  z  1  y2
V   dV  
D 4 y 2 4 z 2
E

1

4


1 y 2

4
1  1 y 2 4 y 2 4 z 2
polar coordinates:
1 y 2
1  1 y
dxdA
1
(4  4( y  z ))dzdy 
2
2
2
1 2
0 0
1  y  1
dxdzdy
1
(4  4r )rd dr 8  ( r  r 3 )dr  2
2
0
Triple Integrals - Example 3
Let E be the solid bounded by 𝑧 = 0, 𝑧 = 𝑦 and 𝑦 = 9 − 𝑥 2 .
Express
a.
f ( x, y, z )dV in the form:

E
f ( x, y, z )dzdydx

E
Limits for z: 0 ≤ 𝑧 ≤ 𝑦
Let D be the projection of the solid
on the 𝑥𝑦-plane.
y
f ( x, y, z )dzdydx   

0
E
D

3

f ( x, y, z )dzdA
9 x 2
3 0
y
0
f ( x, y, z )dzdydx
Triple Integrals - Example 3 continued
b.
f ( x, y, z )dydzdx

E
A line parallel to the y-axis intersects the solid on the
surface 𝑦 = 𝑧 (left surface) and on the surface
𝑦 = 9 − 𝑥 2 (right surface). These are the limits of
integration for the y-variable.
Let D be the projection of the solid on the 𝑥𝑧-plane. The arrow enters and exits
9 x 2
f ( x, y, z )dydzdx   

z
E
D
the solid
at the limit of integration for y
f ( x, y, z )dydA
The surfaces z = y and y = 9 − x2
intersect in a curve C. The projection of
this curve on the xz plane has equation
z = 9 – x2 and it is the boundary of the
domain D
9 x 2
D z
f ( x, y, z )dydA  
3

9 x 2 9 x 2
3 0
z
f ( x, y, z )dydzdx
Triple Integrals - Example 3 continued
c.
f ( x, y, z )dxdzdy

E
A line parallel to the x-axis intersects the solid on
the surface 𝑥 = − 9 − 𝑦 (back surface) and on
the surface 𝑥 = 9 − 𝑦 (front surface). These
are the limits of integration for the x-variable.
Let D be the projection of the solid on the 𝑦𝑧-plane.
9 y
f ( x, y, z )dxdydz   
f ( x, y, z )dxdA

 9 y
E
D
9 y
 0 0  9 y f ( x, y, z )dxdzdy
9
y
y
The arrow enters and
exits the solid at the limit
of integration for x