376 CHAPTER 20. THE INTEGRAL IN OTHER COORDINATES
(xi+ yj+(z− z0)k)1(
x2 + y2 +(z− z0)2)3/2 Gα dV
Therefore, the total force is∫H(xi+ yj+(z− z0)k)
1(x2 + y2 +(z− z0)
2)3/2 Gα dV.
By the symmetry of the sphere, the i and j components will cancel out when the integralis taken. This is because there is the same amount of stuff for negative x and y as there isfor positive x and y. Hence what remains is
αGk∫
H
(z− z0)[x2 + y2 +(z− z0)
2]3/2 dV
as claimed. Now for the interesting part, the integral is evaluated. In spherical coordinatesthis integral is. ∫ 2π
0
∫ b
a
∫π
0
(ρ cosφ − z0)ρ2 sinφ(ρ2 + z2
0−2ρz0 cosφ)3/2 dφ dρ dθ . (20.1)
Rewrite the inside integral and use integration by parts to obtain this inside integral equals
12z0
∫π
0
(ρ
2 cosφ −ρz0) (2z0ρ sinφ)(
ρ2 + z20−2ρz0 cosφ
)3/2 dφ =
12z0
−2−ρ2−ρz0√(
ρ2 + z20 +2ρz0
) +2ρ2−ρz0√(
ρ2 + z20−2ρz0
)−∫
π
02ρ
2 sinφ√(ρ2 + z2
0−2ρz0 cosφ) dφ
. (20.2)
There are some cases to consider here.First suppose z0 < a so the point is on the inside of the hollow sphere and it is always
the case that ρ > z0. Then in this case, the two first terms reduce to
2ρ (ρ + z0)√(ρ + z0)
2+
2ρ (ρ− z0)√(ρ− z0)
2=
2ρ (ρ + z0)
(ρ + z0)+
2ρ (ρ− z0)
ρ− z0= 4ρ
and so the expression in 20.2 equals
12z0
4ρ−∫
π
02ρ
2 sinφ√(ρ2 + z2
0−2ρz0 cosφ) dφ
=1
2z0
4ρ− 1z0
∫π
0ρ
2ρz0 sinφ√(ρ2 + z2
0−2ρz0 cosφ) dφ