multiple integrals

MULTIPLE INTEGRALSMULTIPLE INTEGRALS

12

12.6Triple 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 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 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 ijki j k

f x y z V

* * *, ,ijk ijk ijkx y z

Equation 2 TRIPLE 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 ijkl 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 kl m ni 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 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


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 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.


3 2 12 2

0 1 0

12 23 2

0 10

23 2

0 1

1 32 2 2 33 3

0 01 0

2

2

3 27

4 4 4 4

B

x

x

y

y

xyz dV xyz dx dy dz

x yzdy dz

yzdy dz

y z z zdz 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 yE D

f x y z dV f x y z dz dA

Equation/Formula 6 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

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 7 TYPE 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 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 dVExample 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.


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

121 1 1 1 1

0 0 0 0 00

21 112 0 0

131

12 0

0

1 316 0

14

0

2

1

1

3

1

11 1

6 4 24

z x yx x y x

E z

x

y x

y

zz dV z dz dy dx dy dx

x y dy dx

x ydx

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 zD

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 zE D

f x y z dV f x y z dy dA

Equation 11 TYPE 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:

2z y x

2z y x 2z 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

42 2 2 2

2 2 2 24

x zE 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 44

x

xx 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 1282

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

af 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 dVAPPLICATIONS 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 yE 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 4 APPLICATIONS

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

13

2 2

x x y

xT

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 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 xyxzM MM

x y zm 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 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 dx dy

x dx dy

APPLICATIONS

2

121

1

1 4

1

1 4

0

15

0

2

12

1

4

5 5

x

x y

xdy

y dy

y dy

yy

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 dx dy

x dx dy


APPLICATIONS

2

131

1

1 6

0

17

0

3

21

3

2

3 7

4

7

x

x y

xdy

y dy

yy

Example 5

2

2

2

1 1

1 0

21 1

10

1 1 2

1

1 6

0

2

22

13 7

x

xy yE

z x

yz

y

M z dV z dz dx dy

zdx dy

x dx dy

y dy


Therefore, the center of mass is:

5 57 14

, , , ,

,0,

yz xyxzM MM

x y zm m m


multiple integrals

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