Multiple Integrals
In this section, we will learn about:
Multiple integrals and their applications.
MULTIPLE INTEGRALS
Introduction
Multiple integrals arise in a number of areas of science and engineering, including computations of
a) Area of a 2D region b) Volume
c) Mass of 2D plates d) Force on a 2D plate e) Average of a function
f) Center of Mass and Moment of Inertia g) Surface Area
In this chapter, we shall concentrate on computations of area and volume of a given region.
2
Double integrals - Area
Consider a function of one variable y = f(x) and R is the region of integration on the xy-plane.
For f(x)>0 , the definite integral is equal to the area A under the line y = f(x) and above the x-axis in the region R as
shown.
b
a b
a
x b f
a
x f
R
dx x
f dx
y dx
dy dA
A ( ) 0 ( ) ( )
0
Double integral
•The double integral consists of an inner (first evaluated) and an outer integrals.
•It is also called iterated integral.
3
Double integrals
over a rectangular region
} ,
: ) ,
{(x y a x b c y d
R
) )(
(
) (
) (
a b
c d
x c d
dx c
d
dx y
dx dy dA
A
b a b
a b a
d c b
a d R c
4
Double integrals over a nonrectangular region
b a
b a b
a
x g y
x g R y
dx dx
x g x
g dx
dy dA
A
function lower
function upper
) ( )
) (
( )
( 2 1
2 1
Type 1 region: is a region bounded in the interval axb
between continuous curves y = g1(x) and y = g2(x), where g1(x) g2(x) for a xb.
5
Double integrals over a non rectangular region
Type 2 region: is a region bounded in the interval c yd between continuous curves x = h1(y) and x = h2(y), where h1(y) h2(y) for c yd.
d c
d c d
c
y h x
y h R x
dy dy y h y
h dy
dx dA
A
function left
function right
) ( )
) (
( )
( 2 1
2 1
6
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 1 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
Each sub-box has volume ΔV = Δx Δy Δz TRIPLE INTEGRALS
1,
1,
1,
ijk i i j j k k
B 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
xijk* , yijk* , zijk*
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 3 TRIPLE 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 k
B
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 dx dy 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
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:
, ,
, ,
B
b s d a r c
f x y z dV
f x y z dy dz dx
FUBINI’S TH. (TRIPLE INTEGRALS)
Evaluate the triple integral where B is the rectangular box
2 B
xyz dV
, , 0 1, 1 2, 0 3
B x y z x y z
Example 1 FUBINI’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.
Example 1 FUBINI’S TH. (TRIPLE INTEGRALS)
3 2 1
2 2
0 1 0
2 2 1 3 2
0 1
0 3 2 2
0 1
1 3
2 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
Example 1 FUBINI’S TH. (TRIPLE INTEGRALS)
We restrict our attention to:
Continuous functions f
Certain simple types of regions
INTEGRAL OVER BOUNDED REGION
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 8 TYPE 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 2 TYPE 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
Example 2 TYPE 1 REGIONS
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.
Example 2 TYPE 1 REGIONS
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
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 9 TYPE 1 REGIONS
2 1
1 1 1 1 1
0 0 0 0 0
0 1 1 2
12 0 0
3 1 1 1
2 0
0
1 3
1 6 0
4 1
0
2 1 1
3 1
1 1 1
6 4 24
z x y
x x y x
E z
x
y x
y
z dV z dz dy dx z dy dx
x y dy dx
x y dx
x dx x
Example 2 TYPE 1 REGIONS
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 10 TYPE 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 11 TYPE 3 REGION
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.
Example 3 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
From y = x
2+ z
2, we obtain:
So, the lower boundary surface of E is:
The upper surface is:
z y x
2z y x 2
z y x 2
Example 3 TYPE 1 REGIONS
Therefore, the description of E as a type 1 region is:
E x y z, , 2 x 2, x2 y 4, y x 2 z y x 2
Example 3 TYPE 1 REGIONS
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
Example 3 TYPE 1 REGIONS
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.
Example 3 TYPE 3 REGIONS
Then, the left boundary of E is the paraboloid y = x2 + z2.
The right boundary is the plane y = 4.
Example 3 TYPE 3 REGIONS
So, taking u1(x, z) = x2 + z2 and u2(x, z) = 4 in Equation 11, we have:
2 2
3
3
2 2 4 2 2
2 2 2 2
4
x z
E D
D
x y dV x z dy dA
x z x z dA
Example 3 TYPE 3 REGIONS
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 x
4
x
x z x z dz dx
Example 3 TYPE 3 REGIONS
That gives:
3
2 2 2 2 2 2
2 2 2
0 0
2 2 2 4
0 0
3 5 2
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
Example 3 TYPE 3 REGIONS
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
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
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
APPLICATIONS OF TRIPLE INTEGRALS
Then, the triple integral does represent the volume of E:
E
V E dV
Equation 12 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:
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
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 4 APPLICATIONS
The tetrahedron T and its projection D on the xy-plane are shown.
Example 4 APPLICATIONS
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
Example 4 APPLICATIONS
So, we have:
This is obtained by the same calculation as in Example 4 in Section 15.3
1 1 / 2 2 2
0 / 2 0
1 1 / 2 0 / 2 13
2 2
x x y
T x
x x
V T dV dz dy dx
x y dy dx
Example 4 APPLICATIONS
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 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).
APPLICATIONS
Then, its mass is:
, ,
E
m x y z dV
Equation 13 MASS
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 14 MOMENTS
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 xz xy
M M M
x y z
m m m
Equations 15 CENTER 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 16 MOMENTS OF INERTIA
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:
, ,
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.
Example 5 APPLICATIONS
The solid E and its projection onto the xy-plane are shown.
Example 5 APPLICATIONS
The lower and upper surfaces of E are
the planes
z = 0 and z = x.
So, we describe E as a type 1 region:
, , 1 1, 2 1,0
E x y z y y x z x
Example 5 APPLICATIONS
Then, if the density is ρ(x, y, z), the mass is:
Example 5 APPLICATIONS
2
2
1 1
1 0
1 1
1 E
x y
y
m
dV
dz dx dy x dx dy
APPLICATIONS
2
2 1 1
1
1 4
1
1 4
0
5 1
0
2 2 1
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.
0 y
Example 5 APPLICATIONS
2
2
1 1
1 0
1 1 2
1 yz
E
x y
y
M
x dV
x dz dx dy x dx dy
Example 5 APPLICATIONS
APPLICATIONS
2
3 1 1
1
1 6
0
7 1
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
1 1 2
1 0
1 1 2
1
1 6
0
2
2
1 2
3 7
x
xy y
E
z x
y z
y
M z dV z dz dx dy
z dx dy
x dx dy y dy
Example 5 APPLICATIONS
Therefore, the center of mass is:
75 145
, , , ,
,0,
yz xz xy
M M M
x y z
m m m
Example 5 APPLICATIONS
Appendix
76
Appendix
77
Appendix
78
Appendix
79
