6.1
Triple integrals
We will be interested in triple integrals in either cylindrical or spherical coordinates. However, let us quickly look at triple integrals in rectangular
coordinates. A triple integral is defined as an infinite limit of a sum by
ZZZ
f (x, y, z) dV = lim
n→∞
G
n
X
f (xk , yk , zk ) ∆Vk ,
(1)
k=1
The same rules that apply to double integrals apply here, except we now integrate over a solid G. In general, a triple integral in rectangular coordinates
will look like
#
ZZZ
Z Z "Z g2 (x,y)
f (x, y, z) dz dA ,
(2)
f (x, y, x) dx dy dz =
G
R
g1 (x,y)
where we treat the x and y coordinates as we did for double integrals. This
is certainly the case with cylindrical coordinates, but spherical coordinates
require different treatment. Using triple integrals, the volume of the solid G
is
ZZZ
dV .
(3)
Volume of G =
G
We will not have much need for rectangular coordinates: generally speaking, real problems in mechanics, electrostatics or fluids require cylindrical or
spherical coordinates.
6.1.1
Triple integrals in cylindrical coordinates
Consider the diagram in Figure 1. A small volume element in cylindrical
coordinates is a region between two radii ρ1 ≤ ρ ≤ ρ2 , two angles θ1 ≤ θ ≤ θ2 ,
and two heights z1 ≤ z ≤ z2 . If we want to find the volume of some solid G,
we can integrate over these “cylindrical wedges”
ZZZ
f (ρ, θ, z) dV = lim
n→∞
G
n
X
f (ρk , θk , zk ) ∆Vk ,
k=1
where ∆Vk = [area of base] · [height] = ρk ∆ρk ∆θk ∆zk .
1
(4)
2.0
1.5
1.0
z
0.5
0.0
0.5
x
0.0
1.0
1.0
0.5
0.0
y
Figure 1: Integrating in cylindrical coordinates.
Theorem: Let G be a solid, with upper surface z = g2 (ρ, θ) and lower
surface z = g1 (ρ, θ) in cylindrical polar coordinates. If the projection of G
onto the xy-plane is a plane is a simple polar region R, and if f (ρ, θ, z) is
continuous on G, then
#
ZZZ
Z Z "Z
g2 (ρ,θ)
f (ρ, θ, z) dz dA
f (ρ, θ, z) dV =
G
g1 (ρ,θ)
R
Z
θ2
"Z
ρ2 (θ)
"Z
g2 (ρ,θ)
#
#
(5)
f (ρ, θ, z)ρ dz dρ dθ .
=
θ1
g1 (ρ,θ)
ρ1 (θ)
This will be explained when we discuss Jacobians. Converting from rectangular to cylindrical polar coordinates, the triple integral becomes
ZZZ
ZZZ
f (x, y, x) dx dy dz =
f (ρ cos θ, ρ sin θ, z) ρ dz dρ dθ .
(6)
G
G
2
Example: Use cylindrical coordinates to calculate
3
Z
√
Z
−
−3
9−x2
√
Z
9−x2
9−x2 −y 2
x2 dz dy dx ,
0
shown in Figure 2.
y
2
x
-2
0
2
0
2
-2
2
z=9-x -y
8
6
z
4
2
0
x2 +y2 =9
Figure 2: The volume between x2 + y 2 = 9 for z = 0 and z = 9 − x2 − y 2 .
Solution: We see that z has lower limit 0, but the upper limit depends on x
and y via z = 9 − x2 − y 2 . Then, after integrating z, we see that the largest
values
are at the base of the solid, on the circle x2 + y 2 = 9, hence
√ y can take √
− 9 − x2 ≤ y ≤ 9 − x2 . Finally, the x variable will have limits when y = 0
3
giving x2 = 9, or −3 ≤ x ≤ 3. Converting to cylindrical coordinates
#
Z Z "Z
Z Z √ 2Z
2
2
2
3
−3
9−x −y
9−x
−
√
Z
9−ρ
ρ2 cos2 θ dz dA
x2 dz dy dx =
9−x2
2π
Z
0
3
0
R
Z
9−ρ2
2
2π
Z
2
3
Z
ρ cos θ ρ dz dρ dθ =
=
0
Z
2π
Z
=
0
Z
0
2π
0
0
3
3
(9ρ − ρ5 ) cos2 θ dρ dθ =
0
2π
Z
0
Z
z ρ3 cos2 θ
0
9−ρ2
z=0
dρ dθ
4
3
9ρ
ρ6
2
(
− ) cos θ
dθ
4
6
ρ=0
2π
243
1
243
cos2 θ dθ =
(1 + cos 2θ) dθ
4 0
4 0 2
2π 243π
243 1
1
=
(θ + sin 2θ)0 =
.
4 2
2
4
=
6.1.2
Triple integrals in spherical coordinates
The diagram Figure 3 shows spherical polar coordinates. Note that ρ =
x
0
1
2
2
1
z
0
2
1
y
0
Figure 3: Integrating in spherical coordinates.
constant gives a sphere, θ = constant gives a half-plane, φ = constant gives
4
a right circular cone, and φ = π/2 gives the xy-plane. Similarly to cylindrical coordinates, a small volume element in spherical coordinates is a region
between two radii ρ1 ≤ ρ ≤ ρ2 , two polar angles θ1 ≤ θ ≤ θ2 , and two
azimuth angles φ1 ≤ φ ≤ φ2 . If we want to find the volume of some solid
G, we can integrate over these “spherical wedges”
ZZZ
f (ρ, θ, φ) dV = lim
n→∞
G
n
X
f (ρk , θk , φk ) ∆Vk ,
(7)
k=1
where ∆Vk = [area of base] · [height] = ρ2k sin φk ∆ρk ∆θk ∆φk . The volume
integral is given by
ZZZ
ZZZ
f (ρ, θ, φ) dV =
f (ρ, θ, φ) ρ2 sin φ dρ dθ dφ .
(8)
G
appropriate limits
The limits are intentionally left unspecified because they can be quite involved and the order of integration may depend on how a problem is presented. Converting from rectangular to spherical coordinates, the triple integral becomes
ZZZ
f (x, y, x) dx dy dz
G
(9)
ZZZ
=
2
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ sin φ dρ dθ dφ .
G
Example: Use spherical coordinates to calculate
Z 2 Z √4−x2 Z √4−x2 −y2 p
z 2 x2 + y 2 + z 2 dz dy dx ,
√
−2
− 4−x2
0
which is shown in Figure 4.
Solution: We seepthat z has lower limit 0, but the upper limit depends on
x and y via z = 4 − x2 − y 2 . Then, after integrating z, we see that the
largest values
y can take√are at the base of the solid, on the circle x2 +y 2 = 4,
√
hence − 4 − x2 ≤ y ≤ 4 − x2 . Finally, the x variable will have limits when
5
y
2
0
z= 4 - x2 - y2
-2
2
z
1
x2 +y2 =4
0
-2
0
2
x
Figure 4: The volume between x2 + y 2 = 4 for z = 0 and z =
p
4 − x2 − y 2 .
y = 0 giving x2 = 4, or −2 ≤ x ≤ 2. Converting to spherical coordinates
Z 2 Z √4−x2 Z √4−x2 −y2 p
z 2 x2 + y 2 + z 2 dz dy dx
√
−2
− 4−x2 0
2π Z π/2
Z
Z
2
(ρ cos φ)2
=
0
Z
0
2π
Z
0
π/2
Z
=
0
Z
0
2π
Z
2
ρ5 cos2 φ sin φ dρ dφ dθ
0
π/2
=
0
p
ρ2 (ρ2 sin φ) dρ dφ dθ
0
2π
2
ρ6
cos2 φ sin φρ=0 dφ dθ
6
Z π/2
32
cos2 φ sin φ dφ dθ
=
3 0
0
Z Z
32 2π π/2
=
(− cos2 φ) d(cos φ) dθ
3 0
0
π/2
Z 2π Z
32
1
32 2π
64π
3
=
− cos θ
dθ =
dt =
.
3 0
3
9 0
9
φ=0
Z
6.1.3
Jacobians
What does it mean to change from rectangular coordinates to cylindrical or
spherical coordinates? It is in fact the same principle that we apply when
6
we use a change in variables to make integration easier. In short, we can
change the integration variable from x to u via x = g(u), assuming that g is
a differentiable function, giving
Z g−1 (b)
Z g−1 (b)
Z b
dx
f (g(u)) du =
f (g(u))g 0 (u) du .
(10)
f (x) dx =
du
g −1 (a)
g −1 (a)
a
This is something you know already. If we now have a double integral over x
and y, we can change to new coordinates u and v in the same way. We can
imagine that we have a vector-valued function that represents the original
variables and may be expressed through the new variables
r(u, v) = x(u, v)i + y(u, v)j .
(11)
Then, looking at the small area with vertices r(u0 , v0 ), r(u0 +∆u, v0 ), r(u0 , v0 +
∆v) and r(u0 + ∆u, v0 + ∆v), we see that we can express the area either by
∆A = ∆x∆y, the area as expressed through x and y, or alternatively, we
see that the parallelogram can be parameterised through u and v, in which
∂r
∂r
∆u and ∂v
∆v. The area element for
case, for a small region, the sides are ∂u
a small area is then
∂r
∂r ∂r ∂r
∆u∆v .
(12)
∆A ≈ ∆u ×
∆v = ×
∂u
∂v ∂u ∂v The cross product is
i
∂r ∂r ∂x
×
=
∂u ∂v ∂u
∂x
∂v
j
∂y
∂u
∂y
∂v
k
0
0
=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
k=
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
k.
(13)
This determinant is what we know as the Jacobian. A Jacobian is produced
when we change coordinates form the xy-plane to the uv-plane via equation
x = x(u, v) and y = y(u, v). We denote is by either J(u, v) or ∂(x, y)/∂(u, v)
and it is given by
∂x ∂x ∂x ∂y ∂y ∂x
∂(x, y) ∂u
∂v =
J(u, v) =
= ∂y ∂y
(14)
∂u ∂v − ∂u ∂v .
∂(u, v)
∂u
∂v
It appears through the are element
∂(x, y) ∆u∆v .
∆≈
∂(u, v) 7
(15)
In a double integral, this gives us
ZZ
ZZ
∂(x, y) dAuv ,
f (x, y) dAxy =
f x(u, v), y(u, v) ∂(u,
v)
S
(16)
R
where S is the region in the uv-plane corresponding to the region R in the
xy-plane. In a similar way for triple integrals, we have the Jacobian
∂x ∂x ∂x ∂v
∂w ∂u
∂(x, y, z)
∂y
∂y
∂y
,
=
(17)
J(u, v, w) =
∂v
∂w ∂(u, v, w) ∂u
∂z
∂z
∂z ∂u
∂v
∂w
which gives the change in the volume element
∂(x, y, z) ∆u∆v∆w ,
∆V ≈ ∂(u, v, w) (18)
and in the triple integral it becomes
ZZZ
ZZZ
∂(x, y, z) dVuvw ,
f x(u, v, w), y(u, v, w), z(u, v, w) f (x, y, z) dVxyz =
∂(u, v, w) H
G
(19)
where H is the region in the uvw-space corresponding to the region G in the
xyz-space. Here are some useful Jacobians:
• Polar coordinates:
x = ρ cos θ ,
y = ρ sin θ ,
which have Jacobian
∂(x, y) cos θ −ρ sin θ
=
∂(ρ, θ) sin θ ρ cos θ
= ρ(cos2 θ + sin2 θ) = ρ
⇒ dx dy = ρ dρ dθ .
(20)
• Cylindrical coordinates:
x = ρ cos θ ,
y = ρ sin θ ,
z,
which have Jacobian (check this!)
∂(x, y, z)
= ρ ⇒ dx dy dz = ρ dρ dθ dz .
∂(ρ, θ, z)
8
(21)
• Spherical coordinates:
x = ρ sin φ cos θ ,
y = ρ sin φ sin θ ,
z = ρ cos φ ,
which have Jacobian (check this!)
∂(x, y, z)
= ρ2 sin φ ⇒ dx dy dz = ρ2 sin φ dρ dθ dφ .
∂(ρ, θ, φ)
(22)
This explains why we previously used
dA = dx dy = ρ dρ dθ dz ,
(23)
when changing to cylindrical/polar coordinates,
dV = dx dy dz = ρ dρ dθ dz ,
(24)
when changing to spherical coordinates,
dV = dx dy dz = ρ2 sin φ dρ dθ dφ ,
when changing to spherical coordinates. Let’s look at examples:
Example: Find the Jacobian of the change of variabes
1
y = (u − v) .
2
1
x = (u + v) ,
2
Solution: The Jacobian is
∂x
∂(x, y) ∂u
=
∂(u, v) ∂y
∂u
∂x
∂v
∂y
∂v
=
1
2
1
2
1 1
1
=− .
1 = − −
−2
4 4
2
1
2
Example: Find the Jacobian of the change of variabes
x = au ,
Solution: The Jacobian is
∂(x, y, z)
= ∂(u, v, w) ∂x
∂u
∂y
∂u
∂z
∂u
y = bv ,
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
∂w
∂z
∂w
9
z = cw .
a 0 0
= 0 b 0
0 0 c
= abc .
(25)
6.1.4
Mass, Centre of Gravity, and Centroid of Solids
We previously discussed finding the mass and centre of gravity of a lamina,
with density function δ(x, y). If instead we have a solid, G, with density
function δ(x, y, z), we can easily generalise to get the mass of G
ZZZ
M=
δ(x, y, z) dV ,
(26)
G
and the centre of gravity (x̄, ȳ, z̄) where
ZZZ
1
x δ(x, y, z) dV ,
x̄ =
M
Z GZ Z
1
ȳ =
y δ(x, y, z) dV ,
M
ZG
ZZ
1
z δ(x, y, z) dV .
z̄ =
M
(27)
G
We can also define the centroid (x̃, ỹ, z̃) where
RRR
x dV
ZZZ
1
G
x̃ = RRR
x dV
=
V
dV
G
G
RRR
y dV
ZZZ
1
G
ỹ = RRR
y dV
=
V
dV
G
RRRG
z dV
ZZZ
1
G
z̃ = RRR
=
z dV .
V
dV
(28)
G
G
Example: Find the mass and centre of gravity of a cube with a square base
of length 2 in the xy-plane, centred on (1, 1), and of height 5, with density
function δ(x, y, z) = z.
10
Solution: The mass is given by
Z
ZZZ
δ(x, y, z) dV =
M=
2
Z
Z
Z
5
z dz dy dx
0
G
2
0
2
Z
0
2
1 2 5
(z ) 0 dy dx
0
0 2
Z Z
25 2 2
dy dx
=
2 0 0
Z 2
= 25
dx = 50 .
=
0
The x and y coordinates for the centre of gravity will clearly give the same
result due to the symmetry, so we just calculate one of them
Z 2Z 2Z 5
1
x̄ =
xz dz dy dx
50 0 0 0
Z 2Z 2
1
25
=
x dy dx
50 0 0
2
Z 2
1
2
=
xy y=0 dx
4 0
2
Z
1 2
x2 =
x dx = dx = 1 .
2 0
2 0
Similarly ȳ = 1. We also could have seen this by the the fact that the density
doesn’t vary in the xy-plane and so the centre of the square would coincide
with (x̄, ȳ). Finally,
Z 2Z 2Z 5
1
z 2 dz dy dx
z̄ =
50 0 0 0
Z 2 Z 2 3 5
1
z =
dy dx
50 0 0 3 0
Z 2Z 2
1
125
=
dy dx
50 0 0 3
Z Z
5 2 2
5
10
=
dy dx = (4) =
,
6 0 0
6
3
and so the centre of gravity is at 1, 1, 10
.
3
11