3Integration in ℝ2 and ℝ2
IA Vector Calculus
3.3 Generalization to R
3
We will do exactly the same thing as we just did, but with one more dimension:
Definition
(Volume integral)
.
Consider a volume
V ⊆ R
3
with position vector
r
= (
x, y, z
). We approximate
V
by
N
small disjoint subsets of some simple
shape (e.g. cuboids) labelled by
I
, volume
δV
I
, contained within a solid sphere
of diameter `.
Assume that as
` →
0 and
N → ∞
, the union of the small subsets tend to
V . Then
Z
V
f(r) dV = lim
`→0
X
I
f(r
∗
I
)δV
I
,
where r
∗
I
is any chosen point in each small subset.
To evaluate this, we can take
δV
I
=
δxδyδz
, and take
δx →
0,
δy →
0 and
δz in some order. For example,
Z
V
f(r) dv =
Z
D
Z
Z
xy
f(x, y, z) dz
!
dx dy.
So we integrate
f
(
x, y, z
) over
z
at each point (
x, y
), then take the integral of
that over the area containing all required (x, y).
Alternatively, we can take the area integral first, and have
Z
V
f(r) dV =
Z
z
Z
D
Z
f(x, y, z) dx dy
dz.
Again, if we take f = 1, then we obtain the volume of V .
Often,
f
(
r
) is the density of some quantity, and is usually denoted by
ρ
. For
example, we might have mass density, charge density, or probability density.
ρ
(
r
)
δV
is then the amount of quantity in a small volume
δV
at
r
. Then
R
V
ρ(r) dV is the total amount of quantity in V .
Definition (Volume element). The volume element is dV .
Proposition. dV = dx dy dz.
We can change variables by some smooth, invertible transformation (
x, y, z
)
7→
(u, v, w). Then
Proposition.
Z
V
f dx dy dz =
Z
V
f|J| du dv dw,
with
J =
∂(x, y, z)
∂(u, v, w)
=
∂x
∂u
∂x
∂v
∂x
∂w
∂y
∂u
∂y
∂v
∂y
∂w
∂z
∂u
∂z
∂v
∂z
∂w
Proposition. In cylindrical coordinates,
dV = ρ dρ dϕ dz.
In spherical coordinates
dV = r
2
sin θ dr dθ dϕ.
Proof. Loads of algebra.
Example.
Suppose
f
(
r
) is spherically symmetric and
V
is a sphere of radius
a
centered on the origin. Then
Z
V
f dV =
Z
a
r=0
Z
π
θ=0
Z
2π
ϕ=0
f(r)r
2
sin θ dr dθ dϕ
=
Z
a
0
dr
Z
π
0
dθ
Z
2π
0
dϕ r
2
f(r) sin θ
=
Z
a
0
r
2
f(r)dr
h
− cos θ
i
π
0
h
ϕ
i
2π
0
= 4π
Z
a
0
f(r)r
2
dr.
where we separated the integral into three parts as in the area integrals.
Note that in the second line, we rewrote the integrals to write the differentials
next to the integral sign. This is simply a different notation that saves us from
writing r = 0 etc. in the limits of the integrals.
This is a useful general result. We understand it as the sum of spherical
shells of thickness δr and volume 4πr
2
δr.
If we take
f
= 1, then we have the familiar result that the volume of a sphere
is
4
3
πa
3
.
Example.
Consider a volume within a sphere of radius
a
with a cylinder of
radius b (b < a) removed. The region is defined as
x
2
+ y
2
+ z
2
≤ a
2
x
2
+ y
2
≥ b
2
.
a
b
We use cylindrical coordinates. The second criteria gives
b ≤ ρ ≤ a.
For the x
2
+ y
2
+ z
2
≤ a
2
criterion, we have
−
p
a
2
− ρ
2
≤ z ≤
p
a
2
− ρ
2
.
So the volume is
Z
V
dV =
Z
a
b
dρ
Z
2π
0
dϕ
Z
√
a
2
−ρ
2
−
√
a
2
−ρ
2
dz ρ
= 2π
Z
a
b
2ρ
p
a
2
− ρ
2
dρ
= 2π
2
3
(a
2
− ρ
2
)
3/2
a
b
=
4
3
π(a
2
− b
2
)
3/2
.
Example.
Suppose the density of electric charge is
ρ
(
r
) =
ρ
0
z
a
in a hemisphere
H of radius a, with z ≥ 0. What is the total charge of H?
We use spherical polars. So
r ≤ a, 0 ≤ ϕ ≤ 2π, 0 ≤ θ ≤
π
2
.
We have
ρ(r) =
ρ
0
a
r cos θ.
The total charge Q in H is
Z
H
ρ dV =
Z
a
0
dr
Z
π/2
0
dθ
Z
2π
0
dϕ
ρ
0
a
r cos θr
2
sin θ
=
ρ
0
a
Z
a
0
r
3
dr
Z
π/2
0
sin θ cos θ dθ
Z
2π
0
dϕ
=
ρ
0
a
r
4
4
a
0
1
2
sin
2
θ
π/2
0
[ϕ]
2π
0
=
ρ
0
πa
3
4
.