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Triple IntegralsMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Fall 2011

J. Robert Buchanan Triple Integrals

Riemann Sum Approach

Suppose we wish to integrate w = f (x , y , z), a continuousfunction, on the box-shaped region

Q = {(x , y , z) |a ≤ x ≤ b, c ≤ y ≤ d , m ≤ z ≤ n}.

1 Let P = {Qk}Nk=1 be a partition of Q into rectangular boxes.2 Let the dimensions of Qk be ∆xk , ∆yk , and ∆zk , then

∆Vk = ∆xk ∆yk ∆zk .

3 Let (uk , vk ,wk ) be any point in Qk .

J. Robert Buchanan Triple Integrals

Riemann Sum Approach

Riemann Sum:N∑

k=1

f (uk , vk ,wk )∆Vk .

If ‖P‖ is the length of the longest box diagonal in P, then wemay define the triple integral.

DefinitionFor any function f (x , y , z) defined on the rectangular box Q, wedefine the triple integral of f over Q by∫∫∫

Qf (x , y , z) dV = lim

‖P‖→0

N∑k=1

f (uk , vk ,wk )∆Vk ,

provided the limit exists and is the same for every choice ofevaluation points (uk , vk ,wk ) in Qk .

J. Robert Buchanan Triple Integrals

Riemann Sum Approach

Riemann Sum:N∑

k=1

f (uk , vk ,wk )∆Vk .

If ‖P‖ is the length of the longest box diagonal in P, then wemay define the triple integral.

DefinitionFor any function f (x , y , z) defined on the rectangular box Q, wedefine the triple integral of f over Q by∫∫∫

Qf (x , y , z) dV = lim

‖P‖→0

N∑k=1

f (uk , vk ,wk )∆Vk ,

provided the limit exists and is the same for every choice ofevaluation points (uk , vk ,wk ) in Qk .

J. Robert Buchanan Triple Integrals

Fubini’s Theorem

Theorem (Fubini’s Theorem)

Suppose that f (x , y , z) is continuous on the box Q defined by

Q = {(x , y , z) |a ≤ x ≤ b, c ≤ y ≤ d , m ≤ z ≤ n}.

Then we can write the triple integral over Q as the triple iteratedintegral:∫∫∫

Qf (x , y , z) dV =

∫ b

a

∫ d

c

∫ n

mf (x , y , z) dz dy dx .

There are five other equivalent orders of integration.

J. Robert Buchanan Triple Integrals

Fubini’s Theorem

Theorem (Fubini’s Theorem)

Suppose that f (x , y , z) is continuous on the box Q defined by

Q = {(x , y , z) |a ≤ x ≤ b, c ≤ y ≤ d , m ≤ z ≤ n}.

Then we can write the triple integral over Q as the triple iteratedintegral:∫∫∫

Qf (x , y , z) dV =

∫ b

a

∫ d

c

∫ n

mf (x , y , z) dz dy dx .

There are five other equivalent orders of integration.

J. Robert Buchanan Triple Integrals

Example (1 of 4)

Let Q = {(x , y , z) |0 ≤ x ≤ 1, −1 ≤ y ≤ 2, 0 ≤ z ≤ 3} andevaluate ∫∫∫

Qxyz2 dV .

J. Robert Buchanan Triple Integrals

Example (2 of 4)

∫∫∫Q

xyz2 dV =

∫ 2

−1

∫ 3

0

∫ 1

0xyz2 dx dz dy

=

∫ 2

−1

∫ 3

0

12

x2yz2∣∣∣∣10

dz dy

=

∫ 2

−1

∫ 3

0

12

yz2 dz dy =

∫ 2

−1

16

yz3∣∣∣∣30

dy

=

∫ 2

−1

92

y dy =94

y2∣∣∣∣2−1

=274

J. Robert Buchanan Triple Integrals

Example (3 of 4)

Let Q = {(x , y , z) |0 ≤ x ≤ 2, −3 ≤ y ≤ 0, −1 ≤ z ≤ 1} andevaluate ∫∫∫

Q(x2 + yz) dV .

J. Robert Buchanan Triple Integrals

Example (4 of 4)

∫∫∫Q

(x2 + yz) dV =

∫ 1

−1

∫ 0

−3

∫ 2

0(x2 + yz) dx dy dz

=

∫ 1

−1

∫ 0

−3

(13

x3 + xyz)∣∣∣∣2

0dy dz

=

∫ 1

−1

∫ 0

−3

(83

+ 2yz)

dy dz

=

∫ 1

−1

(83

y + y2z)∣∣∣∣0−3

dz

=

∫ 1

−18− 9z dz = 8z − 9

2z2∣∣∣∣1−1

= 16

J. Robert Buchanan Triple Integrals

Triple Integrals Over General Regions (1 of 2)

To develop of the triple integral of f (x , y , z) over a generalregion Q we must form an inner partition of Q.

x

y

z

J. Robert Buchanan Triple Integrals

Triple Integrals Over General Regions (2 of 2)

DefinitionFor a function f (x , y , z) defined in the bounded, solid region Q,the triple integral of f (x , y , z) over Q is∫∫∫

Qf (x , y , z) dV = lim

‖P‖→0

N∑k=1

f (uk , vk ,wk )∆Vk ,

provided the limit exists and is the same for every choice ofevaluation points (uk , vk ,wk ) in Qk .

J. Robert Buchanan Triple Integrals

Evaluating Triple Integrals

If region Q can be described as

Q = {(x , y , z) | (x , y) ∈ R, k1(x , y) ≤ z ≤ k2(x , y)}

then ∫∫∫Q

f (x , y , z) dV =

∫∫R

∫ k2(x ,y)

k1(x ,y)f (x , y , z) dz dA.

J. Robert Buchanan Triple Integrals

Example (1 of 7)

Integrate f (x , y , z) = z over the region bounded by the planex + y + z = 1 and the coordinate planes.

J. Robert Buchanan Triple Integrals

Example (2 of 7)

0.0

0.5

1.0

x

0.0

0.5

1.0

y

0.0

0.5

1.0

z

J. Robert Buchanan Triple Integrals

Example (3 of 7)

∫∫∫Q

z dV =

∫∫R

∫ 1−x−y

0z dz dA

=

∫∫R

12

(1− x − y)2 dA

=12

∫ 1

0

∫ 1−x

0(1− x − y)2 dy dx

= −16

∫ 1

0(1− x − y)3

∣∣∣1−x

0dx

=16

∫ 1

0(1− x)3 dx = − 1

24(1− x)4

∣∣∣10

=1

24

J. Robert Buchanan Triple Integrals

Example (4 of 7)

Integrate f (x , y , z) =√

x2 + z2 over the portion of theparaboloid y = x2 + z2 where y ≤ 4.

-2

-1

0

1

2

x

0

1

2

3

4

y

-2

-1

0

1

2

z

J. Robert Buchanan Triple Integrals

Example (5 of 7)

∫∫∫Q

√x2 + z2 dV =

∫∫R

∫ 4

x2+z2

√x2 + z2 dy dA

=

∫∫R

4√

x2 + z2 − (x2 + z2)3/2 dA

=

∫ 2π

0

∫ 2

0(4r − r3)r dr dθ

= 2π∫ 2

0(4r2 − r4) dr

= 2π(

43

r3 − 15

r5)∣∣∣∣2

0

=128π

15

J. Robert Buchanan Triple Integrals

Example (6 of 7)

Find the volume of the solid region in the positive orthantbounded by z = 2− y and x = 4− y2.

0

1

2

3

4

x

0.0

0.5

1.0

1.5

2.0

y

0.0

0.5

1.0

1.5

2.0

z

J. Robert Buchanan Triple Integrals

Example (7 of 7)

V =

∫∫∫Q

1 dV =

∫∫R

∫ 2−y

01 dz dA

=

∫∫R

(2− y) dA =

∫ 2

0

∫ 4−y2

0(2− y) dx dy

=

∫ 2

0(2− y)(4− y2) dy =

∫ 2

0(y3 − 2y2 − 4y + 8) dy

=

(14

y4 − 23

y3 − 2y2 + 8y)∣∣∣∣2

0

=203

J. Robert Buchanan Triple Integrals

Mass and Center of Mass

If ρ(x , y , z) denotes the density of material at (x , y , z) in regionQ, then the mass of the solid occupying region Q is

m =

∫∫∫Qρ(x , y , z) dV .

The moments of with respect to the coordinate planes are

Myz =

∫∫∫Q

xρ(x , y , z) dV

Mxz =

∫∫∫Q

yρ(x , y , z) dV

Mxy =

∫∫∫Q

zρ(x , y , z) dV

The center of mass is the point with coordinates:

(x , y , z) =

(Myz

m,Mxz

m,Mxy

m

)J. Robert Buchanan Triple Integrals

Example (1 of 4)

Find the mass and center of mass of the solid region boundedby the plane z = 4 and the paraboloid z = x2 + y2 whosedensity is described by ρ(x , y , z) = 3 + x .

J. Robert Buchanan Triple Integrals

Example (2 of 4)

-2-1

01

2x

-2

-1

0

1

2

y

0

1

2

3

4

z

J. Robert Buchanan Triple Integrals

Example (3 of 4)

m =

∫∫∫Q

(3 + x) dV =

∫∫R

∫ 4

x2+y2(3 + x) dz dA

=

∫∫R

[4(3 + x)− (x2 + y2)(3 + x)

]dA

=

∫ 2π

0

∫ 2

0

[4(3 + r cos θ)− r2(3 + r cos θ)

]r dr dθ

=

∫ 2π

0

∫ 2

0[12r − 3r3 + (4r2 − r4) cos θ] dr dθ

=

∫ 2π

0

[12 +

(323− 32

5

)cos θ

]dθ

= 24π

J. Robert Buchanan Triple Integrals

Example (4 of 4)

Myz =

∫∫∫Q

(3x + x2) dV =16π

3

Mxz =

∫∫∫Q

(3y + xy) dV = 0

Mxy =

∫∫∫Q

(3z + xz) dV = 64π

Thus

(x , y , z) =

(Myz

m,Mxz

m,Mxy

m

)=

(29,0,

83

).

J. Robert Buchanan Triple Integrals

Homework

Read Section 13.5.Exercises: 1–43 odd

J. Robert Buchanan Triple Integrals