3Integration in ℝ2 and ℝ2
IA Vector Calculus
3.4 Further generalizations
Integration in R
n
Similar to the above,
R
D
f
(
x
1
, x
2
, ···x
n
) d
x
1
d
x
2
···
d
x
n
is simply the integra-
tion over an n-dimensional volume. The change of variable formula is
Proposition.
Z
D
f(x
1
, x
2
, ···x
n
) dx
1
dx
2
··· dx
n
=
Z
D
0
f({x
i
(u)})|J| du
1
du
2
··· du
n
.
Change of variables for n = 1
In the
n
= 1 case, the Jacobian is
dx
du
. However, we use the following formula for
change of variables:
Z
D
f(x) dx =
Z
D
0
f(x(u))
dx
du
du.
We introduce the modulus because of our natural convention about integrating
over
D
and
D
0
. If
D
= [
a, b
] with
a < b
, we write
R
b
a
. But if
a 7→ α
and
b 7→ β
,
but
α > β
, we would like to write
R
α
β
instead, so we introduce the modulus in
the 1D case.
To show that the modulus is the right thing to do, we check case by case: If
a < b and α < β, then
dx
du
is positive, and we have, as expected
Z
b
a
f(x) dx =
Z
β
α
f(u)
dx
du
du.
If α > β, then
dx
du
is negative. So
Z
b
a
f(x) dx =
Z
β
α
f(u)
dx
du
du = −
Z
α
β
f(u)
dx
du
du.
By taking the absolute value of
dx
du
, we ensure that we always have the numerically
smaller bound as the lower bound.
This is not easily generalized to higher dimensions, so we don’t employ the
same trick in other cases.
Vector-valued integrals
We can define
R
V
F
(
r
) d
V
in a similar way to
R
V
f
(
r
) d
V
as the limit of a sum over
small contributions of volume. In practice, we integrate them componentwise. If
F(r) = F
i
(r)e
i
,
then
Z
V
F(r) dV =
Z
V
(F
i
(r) dV )e
i
.
For example, if a mass has density ρ(r), then its mass is
M =
Z
V
ρ(r) dV
and its center of mass is
R =
1
M
Z
V
rρ(r) dV.
Example.
Consider a solid hemisphere
H
with
r ≤ a
,
z ≥
0 with uniform
density ρ. The mass is
M =
Z
H
ρ dV =
2
3
πa
3
ρ.
Now suppose that
R
= (
X, Y, Z
). By symmetry, we expect
X
=
Y
= 0. We can
find this formally by
X =
1
M
Z
H
xρ dV
=
ρ
M
Z
a
0
Z
π/2
0
Z
2π
0
xr
2
sin θ dϕ dθ dr
=
ρ
M
Z
a
0
r
3
dr ×
Z
π/2
0
sin
2
θ dθ ×
Z
2π
0
cos ϕ dϕ
= 0
as expected. Note that it evaluates to 0 because the integral of
cos
from 0 to 2
π
is 0. Similarly, we obtain Y = 0.
Finally, we find Z.
Z =
ρ
M
Z
a
0
r
3
dr
Z
π/2
0
sin θ cos θ dθ
Z
2π
0
dϕ
=
r
M
a
4
4
1
2
sin
2
θ
π/2
0
2π
=
3a
8
.
So R = (0, 0, 3a/8).