25.3. CURVILINEAR COORDINATES 471

Example 25.3.1 D≡ (0,∞)× (0,π)× (0,2π) and y1

y2

y3

=

 x1 sin(x2)

cos(x3)

x1 sin(x2)

sin(x3)

x1 cos(x2)

(We usually write this as  x

yz

=

 ρ sin(φ)cos(θ)ρ sin(φ)sin(θ)

ρ cos(φ)

where (ρ,φ ,θ) are the spherical coordinates. We are calling them x1,x2, and x3 to preservethe notation just discussed.) Thus

e1 (x) = sin(x2)cos

(x3)i1 + sin

(x2)sin

(x3)i2 + cos

(x2)i3,

e2 (x) = x1 cos(x2)cos

(x3)i1

+x1 cos(x2)sin

(x3)i2− x1 sin

(x2)i3,

e3 (x) =−x1 sin(x2)sin

(x3)i1 + x1 sin

(x2)cos

(x3)i2 +0i3.

It follows the metric tensor is

G =

 1 0 00(x1)2 0

0 0(x1)2 sin2 (x2

)= (gi j) = (ei ·e j) . (25.18)

Therefore, by Theorem 25.1.6G−1 =

(gi j)

=(ei,e j)=

 1 0 00(x1)−2 0

0 0(x1)−2 sin−2 (x2

) .

To obtain the dual basis, use Theorem 25.1.6 to write

e1 (x) = g1 je j (x) = e1 (x)

e2 (x) = g2 je j (x) =(x1)−2

e2 (x)

e3 (x) = g3 je j (x) =(x1)−2

sin−2 (x2)e3 (x) .

Note that ∂y

∂yk ≡ ek (y) = ik = ik where, as described,(

y1 · · · yn)

are the rectan-gular coordinates of the point in Rn.