# Section 13.8: Triple Integrals ```Section 13.8: Triple Integrals
Consider a function f of three variables defined on a closed box
B = [a, b] &times; [c, d] &times; [r, s] = {(x, y, z)|a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}.
Partition the intervals [a, b], [c, d], and [r, s] as
a = x0 &lt; x1 &lt; &middot; &middot; &middot; &lt; xi−1 &lt; xi &lt; &middot; &middot; &middot; &lt; xl = b,
c = y0 &lt; y1 &lt; &middot; &middot; &middot; &lt; yj−1 &lt; yj &lt; &middot; &middot; &middot; &lt; ym = d,
r = z0 &lt; z1 &lt; &middot; &middot; &middot; &lt; zk−1 &lt; zk &lt; &middot; &middot; &middot; &lt; zn = s.
This forms a partition, P , of the box B into lmn sub-boxes
Bijk = [xi−1 , xi ] &times; [yj−1 , yj ] &times; [zk−1 , zk ].
Let ∆xi = xi − xi−1 , ∆yj = yj − yj−1 , and ∆zk = zk − zk−1 . Then the area of the box Bijk is
∆Vijk = ∆xi ∆yj ∆zk .
∗
∗
) in each sub-box Bijk and form the triple Rie, zijk
Choose a representative point (x∗ijk , yijk
mann sum
l X
m X
n
X
∗
∗
f (x∗ijk , yijk
, zijk
)∆Vijk .
i=1 j=1 k=1
Let ||P || denote the norm of the partition P which is the length of the longest diagonal of
all sub-boxes Bijk .
Definition: The triple integral of f over B is
ZZZ
f (x, y, z)dV = lim
||P ||→0
B
l X
m X
n
X
∗
∗
f (x∗ijk , yijk
, zijk
)∆Vijk
i=1 j=1 k=1
provided that the limit exists. If the limit exists, then f is called integrable.
Note: The practical method for evaluating triple integrals is to express them as iterated
integrals.
Theorem: (Fubini’s Theorem)
If f is continuous on the box B = [a, b] &times; [c, d] &times; [r, s], then
ZZZ
Z sZ dZ b
f (x, y, z)dV =
f (x, y, z)dxdydz.
B
r
c
a
Note: There are five other possible orders of integration which all give the same result.
Example: Evaluate the triple integral
ZZZ
8x2 yz 3 dV,
B
where B = [−1, 2] &times; [0, 1] &times; [1, 2].
By Fubini’s Theorem,
ZZZ
2
3
8x yz dV
2
Z
Z
1
Z
2
8x2 yz 3 dzdydx
=
−1
2
B
Z
0
Z
1
1
=
Z
−1 0
2 Z 1
=
−1
2
Z
=
Z
−1
2
2
2x yz dydx
2
4
1
30x2 ydydx
0
1
15x y dx
2 2
0
15x2 dx
−1
2
= 5x3 −1
= 45.
=
Definition: There are three types of solid regions.
• A solid region E is said to be of type 1 if
E = {(x, y, z)|(x, y) ∈ D, g1 (x, y) ≤ z ≤ g2 (x, y)},
where D is the projection of E onto the xy-plane.
• A solid region E is said to be of type 2 if
E = {(x, y, z)|(y, z) ∈ D, g1 (y, z) ≤ x ≤ g2 (y, z)},
where D is the projection of E onto the yz-plane.
• A solid region E is said to be of type 3 if
E = {(x, y, z)|(x, z) ∈ D, g1 (x, z) ≤ y ≤ g2 (x, z)},
where D is the projection of E onto the xz-plane.
Theorem: Suppose that f is a continuous function over some region E.
1. If E is a type 1 region, then
ZZZ
Z Z &quot;Z
f (x, y, z)dV =
E
D
f (x, y, z)dz dA.
g1 (x,y)
2. If E is a type 2 region, then
ZZZ
Z Z &quot;Z
f (x, y, z)dV =
E
D
#
g2 (y,z)
f (x, y, z)dx dA.
g1 (y,z)
3. If E is a type 3 region, then
ZZZ
Z Z &quot;Z
f (x, y, z)dV =
E
#
g2 (x,y)
D
#
g2 (x,z)
f (x, y, z)dy dA.
g1 (x,z)
In particular, if E is a type 1 region and its projection D onto the xy-plane is a type I region,
then
ZZZ
Z Z
Z
b
h2 (x)
g2 (x,y)
f (x, y, z)dV =
f (x, y, z)dzdydx.
E
a
h1 (x)
g1 (x,y)
ZZZ
zdV , where E is the solid region bounded by
Example: Evaluate the triple integral
E
the planes x = 0, y = 0, z = 0 and 2x + y + z = 2.
The solid region E and its projection D onto the xy-plane are shown in the figure below.
The region E can be expressed as
E = {(x, y, z)|0 ≤ x ≤ 1, 0 ≤ y ≤ 2 − 2x, 0 ≤ z ≤ 2 − 2x − y}.
Then,
ZZZ
1
Z
zdV
Z
2−2x
Z
=
E
zdzdydx
0
1
=
2
=
=
=
=
2−2x−y
Z
0
0
1 Z 2−2x
(2 − 2x − y)2 dydx
0
0
2−2x
Z 1
1
3
−
(2 − 2x − y) dx
6 0
0
Z
1 1
(2 − 2x)3 dx
6 0
1
1
4
− (2 − 2x) 48
0
1
.
3
We could have also projected this region onto the xz- or yz-planes.
ZZZ p
Example: Evaluate the triple integral
y 2 + z 2 dV , where E is the solid region bounded
E
by the paraboloid x = y 2 + z 2 and the plane x = 9.
The solid region E and its projection D onto the yz-plane are shown in the figure below.
The region E can be expressed as
E = {(x, y, z)|(y, z) ∈ D, y 2 + z 2 ≤ x ≤ 9},
where D is the disk y 2 + z 2 ≤ 9. In polar coordinates,
D = {(r, θ)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3}.
Then,
Z Z Z
ZZZ p
y 2 + z 2 dV
9
p
=
E
Z
=
y 2 + z 2 dx dA
D
y 2 +z 2
2π Z 3
2
r (9 − r2 )drdθ
0
Z
0
3
(9r2 − r4 )dr
0
3
1 4 3
= 2π 9r − r 5
0
1944π
=
.
5
= 2π
We could have also projected this region onto the xy- or xz-planes.
Theorem: (Volume as a Triple Integral)
The volume of the solid region E is
ZZZ
dV.
V =
E
Example: Find the volume of the tetrahedron T bounded by the plane 2x + 2y + z = 12 and
the coordinate planes.
The tetrahedron T can be expressed as
T = {(x, y, z)|0 ≤ x ≤ 6, 0 ≤ y ≤ 6 − x, 0 ≤ z ≤ 12 − 2x − 2y}.
Thus, the volume of the tetrahedron is
Z 6 Z 6−x Z 12−2x−2y
1dzdydx
V = =
0
0
0
Z 6 Z 6−x
(12 − 2x − 2y)dydx
=
0
0
6−x
Z 6
2 [(12 − 2x)y − y ]
dx
=
0
0
Z 6
=
[(12 − 2x)(6 − x) − (6 − x)2 ]dx
Z0 6
(x2 − 12x + 36)dx
=
0
6
1 3
2
=
x − 6x + 36x
3
0
= 72.
Note: The applications of double integrals in Section 13.6 can be extended to triple integrals.
Theorem: (Center of Mass of a Solid)
Suppose a solid occupies some region E in space and its density at a point (x, y, z) ∈ E is
given by ρ(x, y, z). Then the mass of the solid is given by
ZZZ
m=
ρ(x, y, z)dV.
E
The coordinates (x̄, ȳ, z̄) of the center of mass of the solid are
ZZZ
ZZZ
ZZZ
1
1
1
xρ(x, y, z)dV
ȳ =
yρ(x, y, z)dV
z̄ =
zρ(x, y, z)dV.
x̄ =
m
m
m
E
E
E
Example: Find the mass and center of mass of the solid E bounded by the parabolic cylinder
z = 1 − y 2 and the planes x + 5z = 5, x = 0, and z = 0 with constant density ρ(x, y, z) = 2.
The solid region E can be described as
E = {(x, y, z)|0 ≤ x ≤ 5 − 5z, 0 ≤ z ≤ 1 − y 2 , −1 ≤ y ≤ 1}.
Thus, the mass of the solid is
1
Z
Z
1−y 2
Z
5−5z
dxdzdy
m = 2
−1
Z 1
0
Z
0
1−y 2
(5 − 5z)dzdy
= 2
−1
1−y2
1 2 z− z 10
dy
2
−1
0
Z 1
1
2
2
4
(1 − y ) − (1 − 2y + y ) dy
10
2
−1
Z 1
(1 − y 4 )dy
5
−1
1
5y − y 5 −1
8.
Z
=
=
=
=
=
0
1
Using the formulas from the previous theorem,
ZZZ
1
x̄ =
2xdV
m
E
Z Z
2 Z 5−5z
1 1 1−y
=
xdxdzdy
4 −1 0
0
Z Z
2
1 1 1−y
(5 − 5z)2 dzdy
=
8 −1 0
Z Z
2
25 1 1−y
(1 − z)2 dzdy
=
8 −1 0
1−y2
Z
25 1
3
= −
(1 − z) dy
24 −1
0
Z
25 1
(1 − y 6 )dy
=
24 −1
1
25
1 7 =
y− y 24
7
−1
=
25
.
14
Similarly,
and
1
ȳ =
4
ZZZ
1
z̄ =
4
ZZZ
ydV = 0,
E
2
zdV = .
7
E
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