Question Number 10495 by ajfour last updated on 14/Feb/17

∫_0 ^(2π) (√(R^2 +r^2 −2Rrcos θ)) dθ

Answered by robocop last updated on 14/Feb/17

todo en funcion de θ    0≤(√(R^2 +r^2 −2Rrcosθ))  ≤2π  (√(R^2 +r^2 −2Rrcosθ   )) ≤2π  R^2 +r^2 −2Rrcosθ≤4π^2   −2Rrcosθ≤4π^2 −R^2 −r^2   2Rrcosθ≥R^2 +r^2 −4π^2   cosθ≥((R^2 +r^2^  −4π^2 )/(2Rr))    (√(R^2 +r^2 −2Rrcosθ))  ≥0  R^2 +r^2 −2Rrcosθ≥0  cosθ≤((R^2 +r^2 )/(2Rr))    ∫_((R^2 +r^2 −4π^(2  ) )/(2Rr)) ^((R^2 +r^2 )/(2Rr)) (√(R^2 +r^2 −2Rrcosθ)) dθ=(2/3)(R^2 +r^2 −4π^2 )+C

Commented byajfour last updated on 16/Feb/17

not satisfied.