∫(上限a,下限-a)√(a^2-x^2)dx(1/2)πa^2

Question

∫(上限a,下限-a) √(a^2-x^2) dx
(1/2)πa^2

Answer 1

令x=a*sint,
则√(a²-x²)= a*cost,dx=a*cost dt
所以
原积分
= ∫ (上限π/2,下限-π/2) a² *cos²t dt
= a² /2 * ∫ (上限π/2,下限-π/2) (cos2t +1) dt
而
∫ (上限π/2,下限-π/2) (cos2t +1) dt
= -0.5sin2t +t 代入上下限π/2和-π/2
= π
所以
∫(上限a,下限-a) √(a^2-x^2) dx
= a² /2 * ∫ (上限π/2,下限-π/2) (cos2t +1) dt
= (1/2)πa^2