∫ ∫ ∫ ∫ ∫ ∫ (r+R)/|r+R|³ dxdydz dXdYdZ
where vectors r and R are
r = (x,y,z)
R = (X,Y,Z)
and |r+R| is absolute value of vector r+R
|r+R| = √[(x+X)² +(y+Y)² + (z+Z)²]
Integration is done inside upper half of unit sphere, that is
0 < z, Z
0 < R, r < 1
1 Answer
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Changing the variables (X, Y, Z) -> (-X, -Y, -Z), one reduces the integral to
I = ∫ ∫ f(r, R) dr dR, f(r,R) = (z-Z) / |r – R|^3,
where r lies in the upper hemisphere z>0, and R lies in the lower hemisphere Z<0. Integral of f over r and R lying in the same area is equal to zero, because swapping r and R in this case gives
∫ ∫ f(r, R) dr dR = ∫ ∫ f(R, r) dR dr = – ∫ ∫ f(r, R) dr dR.
Hence, one can perform the integration over dR inside the unit sphere. For r<1, this integral is equal to the vertical electric field inside a sphere with a unit charge density,
g = ∫ f(r, R) dR = (4 π /3) z.
To get the answer, integrate g over the upper hemisphere,
I = ∫ g dr =
= (4 π /3) ∫ ρ^3 cosθ sinθ dρ dθ dφ =
= π^2 / 3.
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