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solve-I-0-pi-r-R-cos-sin-R-2-r-2-2Rr-cos-3-2-d-for-r-lt-R-and-r-gt-R-respectively-

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Question Number 33823 by 33 last updated on 25/Apr/18
solve : I = ∫_0 ^π (((r−R cosθ) sin θ )/((R^(2 ) + r^2 − 2Rr cos θ)^(3/2) )) dθ for r < R and r > R respectively.
solve:I=∫π0(r−Rcosθ)sinθ(R2+r2−2Rrcosθ)3/2dθforr<Randr>Rrespectively.
Answered by MJS last updated on 26/Apr/18
∫u′v=uv−∫uv′ u′=((sin θ)/((r^2 +R^2 −2rRcos θ)^(3/2) )); v=r−Rcos θ u=−(1/(rR(r^2 +R^2 −2rRcos θ)^(1/2) )); v′=Rsin θ [((∫((sin θ)/((r^2 +R^2 −2rRcos θ)^(3/2) ))dθ=)),(( [t=r^2 +R^2 −2rRcos θ → dx=(dt/(2rRsin θ))])),((=(1/(2rR))∫(1/t^(3/2) )dt=−(1/(rRt^(1/2) )))) ] uv=−((r−Rcos θ)/(rR(r^2 +R^2 −2rRcos θ)^(1/2) )) −∫uv′=∫((sin θ)/(r(r^2 +R^2 −2rRcos θ)^(1/2) ))= [with the same substitution as above] =(((r^2 +R^2 −2rRcos θ)^(1/2) )/(r^2 R)) ∫(((r−Rcos θ)sin θ)/((r^2 +R^2 −2rRcos θ)^(3/2) ))= =−((r−Rcos θ)/(rR(r^2 +R^2 −2rRcos θ)^(1/2) ))+(((r^2 +R^2 −2rRcos θ)^(1/2) )/(r^2 R))+C= =((R−rcos θ)/(r^2 (√(r^2 +R^2 −2rRcos θ))))+C=F(θ) F(π)−F(0)=((R+r)/(r^2 ∣R+r∣))−((R−r)/(r^2 ∣R−r∣)) r, R>0 r<R ⇒ F(π)−F(0)=0 r>R ⇒ F(π)−F(0)=(2/r^2 )
∫u′v=uv−∫uv′u′=sinθ(r2+R2−2rRcosθ)32;v=r−Rcosθu=−1rR(r2+R2−2rRcosθ)12;v′=Rsinθ[∫sinθ(r2+R2−2rRcosθ)32dθ=[t=r2+R2−2rRcosθ→dx=dt2rRsinθ]=12rR∫1t32dt=−1rRt12]uv=−r−RcosθrR(r2+R2−2rRcosθ)12−∫uv′=∫sinθr(r2+R2−2rRcosθ)12=[withthesamesubstitutionasabove]=(r2+R2−2rRcosθ)12r2R∫(r−Rcosθ)sinθ(r2+R2−2rRcosθ)32==−r−RcosθrR(r2+R2−2rRcosθ)12+(r2+R2−2rRcosθ)12r2R+C==R−rcosθr2r2+R2−2rRcosθ+C=F(θ)F(π)−F(0)=R+rr2∣R+r∣−R−rr2∣R−r∣r,R>0r<R⇒F(π)−F(0)=0r>R⇒F(π)−F(0)=2r2

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