Lecture 24: Triple Integrals in Rectangular
Coordinates
We continue with our exploration of multiple integrals and
progress from double integrals to triple integrals by seeking
to integrate a continuous function of the form f (x, y, z) over
a closed and bounded box B of R3. The development which
we use is the logical extension of that introduced in L20.
We begin by assuming the box B is defined by
B = {(x, y, z) | a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}
and we use the same concept to partition the box B.
Then we can define the triple integral of f (x, y, z) over B
as
ZZZ
f (x, y, z) dV =
B
L24 - 2
Fubini’s Theorem for Triple Integrals
If f is continuous on B = [a, b] × [c, d] × [r, s] , then
ZZZ Z sZ dZ b
f (x, y, z) dV = f (x, y, z) dx dy dz
r c a
B
ZZZ
ex. Evaluate (x + yz 2) dV , where B = [−1, 5] ×
B
[2, 4] × [0, 1].
L24 - 3
Next we expand the definition of the triple integral to com-
pute a triple integral over a more general bounded region
E in R3. A solid region E is type 1 if E lies between the
graphs of two continuous functions of x and y:
E = {(x, y, z) | (x, y) ∈ D, u1(x, y) ≤ z ≤ u2(x, y)},
where D is the projection of E onto the xy-plane.
Then
ZZZ
f (x, y, z) dV =
E
where the double integral can be evaluated in any of the
methods that we saw in the previous lectures.
ZZZ
We use a similar concept to evaluate f (x, y, z) dV
E
when the region E is a type 2 or type 3 region.
L24 - 4
ZZZ
ex. Write an iterated integral for f (x, y, z) dV ,
E
where E is the region in the first octant bounded by the
surface z = 9 − x2 − y 2 and the coordinate planes.
y
x
L24 - 5
ZZZ
ex. Express f (x, y, z) dV as iterated integral in six
E
different ways, where E is the solid bounded by the surfaces
y 2 + z 2 = 9, x = −2, and x = 2.
Z Z Z
f (x, y, z) dz dA
D1
L24 - 6
Z Z Z
f (x, y, z) dx dA
D2
Z Z Z
f (x, y, z) dy dA
D3
L24 - 7
Volume of Solids by Triple Integrals
ZZZ
The volume of a solid E is defined as V (E) = 1 dV .
E
ex. Describe the solid region whose volume is
Z 4 Z √16−x2 Z 4
dz dy dx
0 0 (x2 +y 2 )/4
and rewrite the triple integral in the order dy dx dz.
y
x
L24 - 8
ex. Find the volume of the cap with height 1 of the sphere
x2 + y 2 + z 2 = 4.
y
x
L24 - 9
Now You Try It (NYTI):
ZZZ
1. Evaluate z dV , where E is bounded by the cylinder y 2 +z 2 = 9
E
and the planes x = 0, y = 3x, and z = 0 in the first octant. 27/8
2. Find the volume of the solid enclosed by the cylinder y = x2 and
the planes z = 0 and y + z = 1. 8/15
L24 - 10
Z 2Z 2−y Z 4−y 2
3. (a) Sketch the solid whose volume is dx dz dy and
0 0 0
find its volume.
20/3
√
Z 4 Z 4−x Z 2−y
(b) Rewrite the integral in the order dz dy dx. dz dy dx
0 0 0
