Triple IntegralsMath 212
Brian D. Fitzpatrick
Duke University
February 26, 2020
MATH
Overview
Triple IntegralsMotivationIterated IntegralsNonrectangular Regions
Triple IntegralsMotivation
Example
An object conforms to the shape of a solid W in R3.
P1
f (P1)=14 kg/m3
P2f (P2)=7 kg/m3
P3
f (P3)=9 kg/m3
P4 f (P4)=21 kg/m3
Suppose f ∈ C (R3) measures density (kg/m3) throughout W .
DefinitionThe triple integral of f on W is∫∫∫
Wf dV = mass of W (in kg)
Triple IntegralsMotivation
ObservationTracking units allows us to interpret double integrals.∫∫∫
Wf
mass units
volume unit
dV
volume unit
= mass of W
Triple IntegralsIterated Integrals
QuestionHow can we calculate a triple integral?
AnswerUse iterated integrals!
Triple IntegralsIterated Integrals
Example
Suppose f = xy − z ◦C/m3 measures density throughout
W = [1, 3]× [3, 5]× [−2, 2]
The mass of W is∫∫∫W
f dV =
∫ 3
1
∫ 5
3
∫ 2
−2xy − z dz dy dx =
∫ 3
132 x dx
=
∫ 3
1
∫ 5
3xyz − 1
2z2∣∣∣∣z=2
z=−2dy dx = 16 x2
∣∣x=3
x=1
=
∫ 3
1
∫ 5
34 xy dy dx = 128 ◦C
=
∫ 3
12 xy2
∣∣y=5
y=3dx
Triple IntegralsIterated Integrals
ObservationDouble integrals over rectangular regions come in two flavors∫ x2
x1
∫ y2
y1
f dy dx (x-slicing)
∫ y2
y1
∫ x2
x1
f dx dy (y -slicing)
Triple integrals over rectangular regions come in six flavors∫ x2
x1
∫ y2
y1
∫ z2
z1
f dz dy dx
∫ x2
x1
∫ z2
z1
∫ y2
y1
f dy dz dx (x-slicing)∫ y2
y1
∫ x2
x1
∫ z2
z1
f dz dx dy
∫ y2
y1
∫ z2
z1
∫ x2
x1
f dx dz dy (y -slicing)∫ z2
z1
∫ x2
x1
∫ y2
y1
f dy dx dz
∫ z2
z1
∫ y2
y1
∫ x2
x1
f dx dy dz (z-slicing)
Triple IntegralsNonrectangular Regions
QuestionHow do we compute
∫∫∫W f dV if W is not rectangular?
AnswerOur slicing method will depend on the shape of W .
Triple IntegralsNonrectangular Regions
Example
Consider the “first octant” part of x + 2 y + 3 z ≤ 6.
x
y
z
63
2
y
z
6−x2
6−x3
x + 2 y + 3 z = 6
Each x-slice leaves an imprint on the yz-plane.
∫∫∫W
f dV =
∫ 6
0
∫ 6−x2
0
∫ 6−x−2 y3
0f dz dy dx =
∫ 6
0
∫ 6−x3
0
∫ 6−x−3 z2
0f dy dz dx
Triple IntegralsNonrectangular Regions
Example
Consider the “first octant” part of x + 2 y + 3 z ≤ 6.
x
y
z
63
2
x
z
6− 2 y
6−2 y3
x + 2 y + 3 z = 6
Each y -slice leaves an imprint on the xz-plane.
∫∫∫W
f dV =
∫ 3
0
∫ 6−2 y
0
∫ 6−x−2 y3
0f dz dx dy =
∫ 3
0
∫ 6−2 y3
0
∫ 6−2 y−3 z
0f dx dz dy
Triple IntegralsNonrectangular Regions
Example
Consider the “first octant” part of x + 2 y + 3 z ≤ 6.
x
y
z
63
2
x
y
6− 3 z
6−3 z2
x + 2 y + 3 z = 6
Each z-slice leaves an imprint on the xy -plane.
∫∫∫W
f dV =
∫ 2
0
∫ 6−3 z
0
∫ 6−x−3 z2
0f dy dx dz =
∫ 2
0
∫ 6−3 z2
0
∫ 6−2 y−3 z
0f dx dy dz
Triple IntegralsNonrectangular Regions
Example
Consider the “first octant” part of x2 + y2 ≤ z ≤ 1.
x
y
z
y
z
√1− x2
x2
1
z = x2 + y2
Each x-slice leaves an imprint on the yz-plane.∫∫∫W
f dV =
∫ 1
0
∫ √1−x20
∫ 1
x2+y2
f dz dy dx =
∫ 1
0
∫ 1
x2
∫ √z−x20
f dy dz dx
Triple IntegralsNonrectangular Regions
Example
Consider the “first octant” part of x2 + y2 ≤ z ≤ 1.
x
y
z
x
y
√z
√z
z = x2 + y2
Each z-slice leaves an imprint on the xy -plane.
∫∫∫W
f dV =
∫ 1
0
∫ √z0
∫ √z−x20
f dy dx dz =
∫ 1
0
∫ √z0
∫ √z−y2
0f dx dy dz
Triple IntegralsNonrectangular Regions
Example
Consider the region 2 x2 + 2 z2 ≤ y ≤ x2 + z2 + 4.
y
y = x2 + z2 + 4
y = 2 x2 + 2 z2
x2 + z2 = 4
x
y
y = x2 + z2 + 4
y = 2 x2 + 2 z2
√4− z2−
√4− z2
Each z-slice leaves an imprint on the xy -plane.∫∫∫W
f dV =
∫ 2
−2
∫ √4−z2−√4−z2
∫ x2+z2+4
2 x2+2 z2f dy dx dz