Gravitation of hollow sphere

Let the radius of the hollow be 1,
the mass of the object be 1,
the object A be located at (0,0,a), |a|<1,
the area density of the planet be 1.

Let an area element P(1, θ, φ) (spherical coordinate) locate on the spherical, then the gravitation force acts on the object equals
-Gsinφ dθdφ (sinφcosθ, sinφsinθ, cosφ-a)/(a+1-2acosφ)^1.5,
θ=-π~π, φ=0~π.

To show the resutant force is 0, it suffices to show that
∫[0,π] (cosφ-a)sinφ/(a+1-2acosφ)^1.5 dφ = 0.

Set u= a+1-2acosφ, then the integral equals
1/(2a)∫[1-a,1+a] (1-a-u)/u du= 0


2009-12-18 14:32:39 補充：
the symbol of square is gone?
-Gsinφ dθdφ (sinφcosθ, sinφsinθ, cosφ-a)/(a²+1-2acosφ)^1.5, θ=-π~π, φ=0~π.
∫[0,π] (cosφ-a)sinφ/(a²+1-2acosφ)^1.5 dφ ( u²= a²+1-2acosφ)
=1/(2a²)∫[1-a,1+a] (1-a²-u²)/u² du= 0

2009-12-18 14:35:02 補充：
the symbol of square is gone! ???

呢到太強了~原來知識都有咁多高人~~~~~~

This is simple. You can refer to Physics Beyond 2000.

2009-12-18 14:53:05 補充：
高手... 在下佩服...

用spherical coordinate果然比cartesian快捷方便。