27.3. CYLINDRICAL AND SPHERICAL COORDINATES 545

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 . (27.1)

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

 . (27.2)

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 27.2 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

=

12z0

(4ρ − 2ρ

z0

(ρ

2 + z20 −2ρz0 cosφ

)1/2 |π0)