Triple Integrals
Math 21a
1
Spring, 2009
Evaluate the following triple integrals as iterated integrals.
ZZZ
xy dV , where E = [0, 1] × [0, 2] × [0, 3].
(a)
E
ZZZ
xy 3 z 2 dV , where E = [−1, 1] × [−3, 3] × [0, 3].
(b)
E
ZZZ
y 2 z cos(xyz) dV , where E = [0, π] × [0, 1] × [0, 2].
(c)
E
Hint: Try using different orders of integration.
2
RRR
For each of the following regions E, write the triple integral
f (x, y, z) dV as an iterated
E
integral. There may be up to six different ways to do this, depending on whether you write it
with dx dy dz or dz dy dx or dx dz dy or. . .
(a) The tetrahedron bounded by the planes x + y + z = 1, x = 0, y = 0, and z = 0.
(b) The (solid) sphere x2 + y 2 + z 2 = a2 .
(c) The region between the paraboloid x = 1 − y 2 − z 2 and the yz-plane.
(d) The region bounded by the surface z = 3xy + 1 and the planes z = 0, x = 0, x = 1,
y = 0, and y = x.
(e) The region bounded by the cylinder x2 + z 2 = 1 and the planes y = 0 and y + z = 2.
(f) The pyramid whose base is the square [−1, 1] × [−1, 1] in the xy-plane and whose vertex
is the point (0, 0, 1).
3
Evaluate the following integrals.
ZZZ
1 dV , where E is the region in Problem 2(a).
(a)
E
ZZZ
(b)
z dV , where E is the region in Problem 2(d).
E
4
Find the volume of the pyramid described in Problem 2(f).
5
Consider a brick in the region [0, 1]×[0, 2]×[0, 1] whose density at a point (x, y, z) is ρ(x, y, z) =
2 + xy − 2z. Find the mass of the brick.
Triple Integrals – Solutions
1
(a) The integral is
1
Z
Z
2
3
Z
1
Z
Z
2
3xy dy dx
xy dz dy dx =
0
0
0
0
0
1
Z
3x
=
0
22
− 0 dx
2
1
Z
6x dx = 3.
=
0
(b) The integral is
Z 3Z
3
1
Z
Z
3 2
3
3
Z
3 2
xy z dx dy dz =
−3
0
−1
y z
−3
0
12 (−1)2
−
2
2
dy dz = 0.
Note that we can put the three integrals in whatever order we want, and putting the
integral with respect to x first makes the computation easier.
(c) If we write the integral as
Z
π
0
Z
1
0
Z
2
y 2 z cos(xyz) dz dy dx,
0
we have to use integration by parts and the integral is lots of work. However, if we write
it as
Z Z Z
1
2
π
y 2 z cos(xyz) dx dz dy,
0
0
0
it is a lot easier. In the inner integral, we can note that
∂
y sin(xyz) = y 2 z cos(xyz)
∂x
so we get
Z
1
Z
2
π
Z
1
Z
2
Z
2
(y sin(πyz) − y sin(0)) dz dy
y z cos(xyz) dx dz dy =
0
0
0
0
0
1
Z
Z
=
2
y sin(πyz) dz dy.
0
0
We then similarly have
∂
cos(xyz) = −y sin(xyz)
∂z
so
Z
0
1
Z
0
2
−1
y sin(πyz) dz dy =
π
Z
−1
=
π
Z
1
(cos(2πy) − cos(0)) dy
0
1
(cos(2πy) − 1) dy =
0
1
.
π
2
We only give one possible way to express each integral; there are others that are equally
correct.
Z 1 Z 1−x Z 1−x−y
ZZZ
f (x, y, z) dz dy dx.
f (x, y, z) dV =
(a)
0
0
0
E
ZZZ
Z Z √
Z √
(b)
a
a2 −x2
−a
√
− a2 −x2
f (x, y, z) dV =
E
a2 −x2 −y 2
−
f (x, y, z) dz dy dx.
√
a2 −x2 −y 2
(c) This region is defined by the inequality 0 ≤ x ≤ 1 − y 2 − z 2 . For 1 − y 2 − z 2 to be
nonnegative, we also need (y, z) to lie on the disk D bounded by y 2 + z 2 = 1. Thus we
get
ZZZ
1−y 2 −z 2
ZZ Z
f (x, y, z) dV =
f (x, y, z) dx dA
E
D
Z
1
0
Z √1−y2 Z
=
−1
√
−
1−y 2 −z 2
f (x, y, z) dx dz dy.
1−y 2
0
(d) This region is defined by the inequalities 0 ≤ z ≤ 3xy + 1, 0 ≤ x ≤ 1, and 0 ≤ y ≤ x, so
we get
Z 1 Z x Z 3xy+1
ZZZ
f (x, y, z) dz dy dx.
f (x, y, z) dV =
0
E
0
0
(e) Being between the planes y = 0 and y + z = 2 says that 0 ≤ y ≤ 2 − z, and being inside
the cylinder says that (x, z) is in the disk D bounded by x2 + z 2 = 1. Thus we get
ZZZ
ZZ Z
2−z
f (x, y, z) dV =
Z
1
f (x, y, z) y dA =
E
D
−1
0
√
Z
1−x2
√
− 1−x2
Z
2−z
f (x, y, z) dy dz dx.
0
(f) In the pyramid, z ranges from 0 at the base to 1 at the top. The cross-section of the
pyramid given by a fixed value of z is the square [−1 + z, 1 − z] × [−1 + z, 1 − z]. Thus
we get
ZZZ
Z Z
Z
1
1−z
1−z
−1+z
−1+z
f (x, y, z) dV =
E
3
f (x, y, z) dx dy dz.
0
(a) The integral is
Z
1
Z
1−x
Z
1−x−y
Z
1
1−x
Z
(1 − x − y) dy dx
1 dz dy dx =
0
0
0
0
Z
0
1
=
0
Z
=
0
1
(1 − x)2
(1 − x) −
2
2
(1 − x)2
1
dx = .
2
6
dx
(b) The integral is
Z
1
x
Z
Z
3xy+1
1
Z
x
(3xy + 1)2
dy dx
2
Z
x
Z
z dz dy dx =
0
0
0
0
1
=
2
1
2
Z
=
4
0
Z
0
1
1−z
Z
Z
1−z
(9x2 y 2 + 6xy + 1) dy dx
0
1
0
The volume of a region E is given by the integral
Z
1
7
(3x5 + 3x3 + x) dx = .
8
RRR
E
Z
1
Z
1 dV . In this case, that integral is
1−z
(2 − 2z) dy dz
1 dx dy dz =
−1+z
0
−1+z
Z
=
0
5
−1+z
0
1
4
(2 − 2z)2 dz = .
3
The mass is given by integrating the density over the region, so the mass is
Z
1
Z
2
Z
1
Z
1
Z
(2 + xy − 2z) dz dy dx =
0
0
0
2
(2 + xy − 1) dy dx
0
Z
0
1
(4 + 2x − 2) dy dx
=
0
= 4 + 1 − 2 = 3.