As it happens, our change of variables formula in 2-variables has an analogous
statement in 3-variables. Specifically, if T is a transformation that maps a solid B
in the u, v, w-space one-to-one and onto the solid E in the x, y, z-space where the
components of T have continuous second partial derivatives, then
ZZZ
ZZZ
∂(x, y, z)
f dV =
f ◦ T (u, v, w) ·
du dv dw,
∂(u,
v, w)
E
B
where
xu
∂(x, y, z)
= yu
∂(u, v, w) zu
xv
yv
zv
xw yw zw in absolute value.
Let’s focus on a particularly useful example: spherical change of variables. Let
x = ρ sin φ cos θ
y = ρ sin φ sin θ
z = ρ cos φ.
In class, we computed the Jacobian
∂(x, y, z)
= ρ2 sin φ.
∂(θ, φ, ρ)
This transformation maps the box B = [0, 2π] × [0, π] × [0, R] in the θ, φ, ρ-space to
the sphere of radius R centered at the origin in x, y, z-space.
Example 1. Compute
RRR
E
sin((x2 + y2 + z2 )3/2 ) dV where E is the part of the
sphere of radius 1 centered at the origin in the 1st octant.
Solution. Let B be the box [0, π/2] × [0, π/2] × [0, 1]. Then the spherical change
of variables maps B one-to-one and onto E. So
ZZZ
Z π/2 Z π/2 Z 1
sin((ρ2 )3/2 )ρ2 sin φ dρ dφ dθ
sin((x2 + y2 + z2 )3/2 ) dV =
0
E
=
=
=
Example 2. Compute
RRR
π
2
0
Z π/2 Z 1
0
ρ2 sin(ρ3 ) sin φ dρ dφ
0
u = ρ3
0
Z
Z
π π/2 1
sin u sin φ du dφ
6 0
0
π
(1 − cos 1).
6
x + y + z dV where E is the region outside the sphere
p
of radius 1, inside the sphere of radius 2, and within the cone z = x2 + y2 .
E
Solution. E is easily described in spherical coordinates:
16ρ62
06φ6
1
π
4
0 6 θ 6 2π,
2
that is, the box [0, 2π] × [0, π/4] × [1, 2] is mapped one-to-one and onto E via the
spherical change of variables. So
ZZZ
Z 2π Z π/4 Z 2
(ρ sin φ cos θ + ρ sin φ sin θ + ρ cos φ) ρ2 sin φ dρ dφ dθ.
x+y+z dV =
0
E
0
1
Note that if we do the θ integration first (which is allowed since we’re now integrating over a rectangle) we can use the fact that
Z 2π
Z 2π
sin θ dθ =
cos θ dθ = 0
0
to get that
0
ZZZ
Z π/4 Z 2
ρ3 sin φ cos φ dρ dφ.
x + y + z dV = 2π
0
E
From there, it’s straight-forward:
Z π/4 Z 2
2π
ρ3 sin φ cos φ dρ dφ
0
=
1
=
=
Problem 1. Compute
z2 = z.
RRR
E
1
15π
2
Z π/4
sin φ cos φ dφ
u = sin φ
0
Z √
15π 1/ 2
u du
2 0
15π
.
8
zdV where E is the region within the sphere x2 + y2 +