Triple Integrals - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Triple/main.pdf · Triple Integrals MATH 311, Calculus III J. Robert Buchanan Department of Mathematics

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 .



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 .



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∫∫∫


‖P‖→0

N∑k=1




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.


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.


Example (1 of 4)

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

Qxyz2 dV .


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


Example (3 of 4)

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

Q(x2 + yz) dV .


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


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


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∫∫∫


‖P‖→0

N∑k=1




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.


Example (1 of 7)

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


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


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


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


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


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


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


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 .


Example (2 of 4)

-2-1

01

2x

-2

-1

0

1

2

y

0

1

2

3

4

z


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


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

).


Homework

Read Section 13.5.Exercises: 1–43 odd


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Triple Integrals - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Triple/main.pdf · Triple Integrals MATH 311, Calculus III J. Robert Buchanan Department of Mathematics