scratch sphere integral
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Short worked calculation, dated 1.11.15, from the Appendix D Center of Gravity material in Phil's mechanics files. It evaluates an integral over theta with distance r1 from a point at radius b to a point on a sphere of radius r, using the substitution r1^2 = r^2 + b^2 - 2rb cos(theta). With limits |b-r| to b+r, the result is 2/b^2 when b>r and 0 when b<r. Some equation symbols in the text are garbled.
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I = !Syntax Error, Idθ den = (r12)3/2 = r13
Let
r12 = r2 + b2-2rbcosθ
2r1dr1 = +2rbsinθdθ sinθdθ = r1dr1/(rb)
rcosθ = (r2+b2- r12)/2b
b - rcosθ = b - (r2+b2- r12)/2b = (2b2 - r2-b2+r12)/(2b) = (b2-r2+r12)/2b
Then
I = ∫ (r1dr1/rb) * (b2-r2+r12)/2b * (1/r13)
= (2rb2)-1 ∫ (r1dr1 * (b2-r2+r12) * (1/r13)
= (2rb2)-1∫ dr1 * (b2-r2+r12) (1/r12)
= (2rb2)-1∫dr1 [ (b2-r2) / r12 + 1 ]
= (2rb2)-1 [ (b2-r2)∫dr1/r12 + ∫dr1 1 ]
What are the endpoints?
r12 = r2 + b2-2rbcosθ
r1low2 = r2 + b2-2rb = (r-b)2 r1lo = |b-r| = α
r1hi2 = r2 + b2+2rb = (r+b)2 r1hi = b+r = β
So Answer is
(2rb2)-1 [ (b2-r2)!Syntax Error, Idr1/r12 + !Syntax Error, Idr1 1 ]
= (2rb2)-1[ (b2-r2) {-(1/r1)}βα + (β-α) ]
= (2rb2)-1 [ (b2-r2) {-(1/β)+(1/α)} + (β-α) ]
= (2rb2)-1 [ (b2-r2) {-(1/(b+r))+(1/|b-r|)} + (b+r - |b-r| ) ]
= (2rb2)-1 [ (b2-r2) { - } + b+r - |b-r|) ]
= (2rb2)-1[ (b2-r2) {} + b+r - |b-r|) ) ]
= (2rb2)-1[ (b+r)(b-r) {} + b+r - |b-r|) ) ]
= (2rb2)-1[ (b-r) {} + b+r - |b-r|) ) ]
If b > r this says
= (2rb2)-1[ (b-r) {} + b+r - (b-r) ) ]
= (2rb2)-1[ 2r + 2r ] = 4r/(2rb2) = 2/b2
If b < r this says
= (2rb2)-1[ (b-r) {} + b+r + (b-r) ) ]
= (2rb2)-1[ -2b + 2b) ]
= 0