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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
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