SECTION 12.5

TRIPLE INTEGRALS

P212.5

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

P312.5

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.

P412.5

TRIPLE INTEGRALS

The planes through the endpoints of these subintervals parallel to the coordinate planes divide the box B into lmn sub-boxes

which are shown in Figure 1. Each sub-box Bijk has volume

ΔVijk = ΔxiΔyjΔzk

1 1

1

, ,

,

ijk i i j j

k k

B x x y y

z z

P512.5

TRIPLE INTEGRALS

Then, we form the triple Riemann sum

where the sample point 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.

* * *

1 1 1

, ,l m n

ijk ijk ijk ijki j k

f x y z V

* * *, ,ijk ijk ijkx y z

P612.5

The triple integral of f over the box B is:

if this limit exists.

Definition 3

Again, the triple integral always exists if f is continuous.

* * *

max , , 01 1 1

, ,

lim , ,i j k

B

l m n

ijk ijk ijk ijkx y z

i j k

f x y z dV

f x y z V

P712.5

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:

, ,

1 1 1

, , lim , , l m n

i j kl m ni j kB

f x y z dV f x y z V

P812.5

TRIPLE INTEGRALS

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

P912.5

If f is continuous on the rectangular box

B = [a, b] × [c, d] × [r, s],

then

FUBINI’S THEOREM FOR TRIPLE INTEGRALS

, ,

, ,

B

s d b

r c a

f x y z dV

f x y z dx dy dz

P1012.5

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

P1112.5

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:

, ,

, ,

B

b s d

a r c

f x y z dV

f x y z dy dz dx

P1212.5

Example 1

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

P1312.5

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 12 2

0 1 0

12 2 23 2 3 2

0 1 0 10

2 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 yz yzdy dz dy dz

y z z zdz dz

P1412.5

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.

P1512.5

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.

P1612.5

INTEGRAL OVER BOUNDED REGION

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 12.2).

, , , ,E B

f x y z dV F x y z dV

P1712.5

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.

P1812.5

TYPE 1 REGION

That is,

where D is the projection of E onto the xy-plane as shown in Figure 2.

1 2, , , , , ,E x y z x y D u x y z u x y

P1912.5

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).

P2012.5

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

2

1

,

, , , , ,

u x y

u x yE D

f x y z dV f x y z dz dA

P2112.5

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.

P2212.5

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 1 2 1 2, , , ( ) ( ), ( , ) ( , )E x y z a x b g x y g x u x y z u x y

P2312.5

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

P2412.5

TYPE 1 REGIONS

If, instead, D is a type II plane region (as in Figure 4), then

1 2 1 2, , , ( ) ( ), ( , ) ( , )E x y z c y d h y x h y u x y z u x y

P2512.5

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

P2612.5

Example 2

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

P2712.5

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)

P2812.5

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.

P2912.5

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.

P3012.5

Example 2 SOLUTION

This description of E as a type 1 region enables us to evaluate the integral as follows.

, , 0 1,0 1 ,0 1E x y z x y x z x y

121 1 1 1 1

0 0 0 0 00

1321 1 1

1 12 20 0 0

0

141 31

6 0

0

2

11

3

11 11

6 4 24

z x yx x y x

E z

y x

x

y

zz dV z dz dy dx dy dx

x yx y dy dx dx

xx dx

P3112.5

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.See Figure 7.

1 2, , , , ( , ) ( , )E x y z y z D u y z x u y z

P3212.5

TYPE 2 REGION

The back surface is x = u1(y, z).

The front surface is x = u2(y, z).

P3312.5

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

P3412.5

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. See Figure 8.

1 2, , , , ( , ) ,E x y z x z D u x z y u x z

P3512.5

TYPE 3 REGION

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

P3612.5

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).

P3712.5

Example 3

Evaluate

where E is the region bounded by the paraboloid y = x2 + z2 and the plane y = 4.

2 2

E

x z dV

P3812.5

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.

P3912.5

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

P4012.5

Example 3 SOLUTION

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

P4112.5

Example 3 SOLUTION

Therefore, the description of E as a type 1 region is:

Thus, we obtain:

Although this expression is correct, it is extremely difficult to evaluate.

2 2 2, , 2 2, 4,E x y z x x y y x z y x

2

2 2

2 42 2 2 2

2

y x

x y xE

x y dV x z dz dy dx

P4212.5

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.

P4312.5

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:

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

P4412.5

Example 3 SOLUTION

This integral could be written as:

However, it’s easier to convert to polar coordinates in the xz-plane:

x = r cos , z = r sin

2

2

2 4 2 2 2 2

2 44

x

xx z x z dz dx

P4512.5

Example 3 SOLUTION

That gives:

3

2 2

2 22 2 2 2 2

0 0

23 52 2 2 4

0 00

4 4

4 1284 2

3 5 15

E

D

x z dV

x z x z dA r r r dr d

r rd r r dr

Example 4

Express the iterated integral

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.

P4612.5

21

0 0 0, ,

x yf x y z dz dy dx

Example 4 SOLUTION

We can write

where

P4712.5

21

0 0 0, , , ,

x y

E

f x y z dz dy dx f x y z dV

2, , 0 1, 0 , 0E x y z x y x z y

Example 4 SOLUTION

This description of E enable us to write projections onto the three coordinate planes as follows：

P4812.5

21

2

23

on the -plane , 0 1, 0

, 0 1, 1

on the -plane , 0 1, 0

on the x -plane , 0 1, 0

xy D x y x y x

x y y y x

yz D x y y z y

z D x y x z x

Example 4 SOLUTION

From the resulting of the projections in Figure 12 we sketch the solid E in Figure 13.

P4912.5

Example 4 SOLUTION

We see that it is the solid enclosed by the planes and the parabolic cylinder .

If we integrate first with respect to x, then z, and then y, we use an alternate description of E：

Thus

P5012.5

0, 1, z x y z 2 (or )y x x y

, , 0 1, 0 , 1E x y z x z y y x

1 1

0 0, , , ,

y

yE

f x y z dV f x y z dx dz dy

P5112.5

APPLICATIONS OF TRIPLE INTEGRALS

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

P5212.5

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

P5312.5

APPLICATIONS OF TRIPLE INTEGRALS

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

P5412.5

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).

( , , )E

f x y z dV

P5512.5

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:

E

V E dV

P5612.5

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:

From Section 12.2, we know this represents the volume that lies between the surfaces

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

2

1

( , )

2 1( , )1 ( , ) ( , )

u x y

u x yE D D

dV dz dA u x y u x y dA

P5712.5

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

P5812.5

Example 5 SOLUTION

The tetrahedron T and its projection D on the xy-plane are shown in Figure 14 and 15.

P5912.5

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

P6012.5

Example 5 SOLUTION

So, we have:

This is obtained by the same calculation as in Example 4 in Section 12.2

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

P6112.5

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.

P6212.5

APPLICATIONS

All the applications of double integrals in Section 12.4 can be immediately extended to triple integrals.

P6312.5

APPLICATIONS

For example, suppose the density function of a solid object that occupies the region E is:

(x, y, z)

in units of mass per unit volume, at any given point (x, y, z).

Then, its mass is:

, ,E

m x y z dV

P6412.5

MOMENTS

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

P6512.5

CENTER OF MASS

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

P6612.5

MOMENTS OF INERTIA

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

P6712.5

TOTAL ELECTRIC CHARGE

As in Section 12.4, the total electric charge on a solid object occupying a region E and having charge density (x, y, z) is:

, ,E

Q x y z dV

P6812.5

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.

P6912.5

Example 6 SOLUTION

The solid E and its projection onto the xy-plane are shown in Figure 16.

P7012.5

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:

2, , 1 1, 1,0E x y z y y x z x

P7112.5

Example 6 SOLUTION

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

2

2

2

1 1

1 0

121 1 1

1 1

1 14 4

1 0

15

0

2

1 12

4

5 5

x

yE

x

yx y

m dV dz dx dy

xx dx dy dy

y dy y dy

yy

P7212.5

Example 6 SOLUTION

Due to the symmetry of E and about the xz-plane, we can immediately say that Mxz = 0, and therefore .

The other moments are calculated as follows.

0y

P7312.5

2

2

2

1 1

1 0

131 1 12

1 1

171 6

00

3

2 21

3 3 7

4

7

x

yz yE

x

yx y

M x dV x dz dx dy

xx dx dy dy

yy dy y

Example 6 SOLUTION

P7412.5

Example 6 SOLUTION

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

P7512.5

Example 6 SOLUTION

Therefore, the center of mass is:

, ,

, ,

5 5, 0,

7 14

yz xyxz

x y z

M MM

m m m