Page 439 - Advanced engineering mathematics

12.10 Curvilinear Coordinates 419

This is the reason for the formula for converting a triple integral from rectangular to
cylindrical coordinates:

f (x, y, z)dx dy dz = f (r cos(θ),r sin(θ), z)rdrdθ dz.
M M
Again, in 3-space we can recognize the factor of r as the Jacobian

cos(θ) −r sin(θ) 0

∂(x, y, z)
= sin(θ) r cos(θ) 0 =r.

∂(r,θ, z)
0 0 1
EXAMPLE 12.32
In spherical coordinates, q 1 = ρ, q 2 = θ, and q 3 = ϕ. From Example 12.29 we know x, y and z in
terms of ρ, θ and ϕ, so compute the partial derivatives to obtain

2 2 2

∂x ∂y ∂z
h ρ = + + = 1,
∂ρ ∂ρ ∂ρ

2 2 2

∂x ∂y ∂z
h θ = + + = ρ sin(θ),
∂θ ∂θ ∂θ

2 2 2
∂x ∂y ∂z

h ϕ = + + = ρ.
∂ϕ ∂ϕ ∂ϕ
Therefore, in spherical coordinates, the differential element of arc length, squared, is
(ds) = (dρ) + ρ sin (ϕ)(dθ) + ρ (dϕ) .
2
2
2
2
2
2
2
In general, if ds i is the differential element of arc length along the q i axis (in the q i direction),
then
ds i = h i dq i .
Therefore the differential elements of area are
ds i ds j = h i h j dq i dq j .
The differential element of volume is
ds 1 ds 2 ds 3 = h 1 h 2 h 3 dq 1 dq 2 dq 3 .
The formula for a differential volume element in spherical coordinates is
2
ds ρ ds θ ds ϕ = ρ sin(ϕ)dρ dθ dϕ.
This should look familiar. In calculus, we are told that when we convert a triple integral from
rectangular to spherical coordinates we obtain

2
f (x, y, z)dx dy dz = F(ρ,θ,ϕ)ρ sin(ϕ)dρ dθ dϕ,
M M ρ,θ,ϕ
in which F(ρ,θ,ϕ) is obtained by substituting for x, y, z in terms of spherical coordinates in
2
f (x, y, z), and M ρ,θ,ϕ is the region M defined in spherical coordinates. Notice that the ρ sin(ϕ)
has shown up in the differential element of volume. That is, in terms of differentials,
2
dx dy dz = ρ sin(ϕ)dρ dθ dϕ.

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October 14, 2010 14:53 THM/NEIL Page-419 27410_12_ch12_p367-424