>>522
x^2+y^2+4√3x=0 へ y=x^2 をぶち込んで、x^2+x^4+4√3x=x(x^3+x+4√3)=x(x+√3)(x^2-√3x+4)=0
交点はx=0と-√3だから、S=∫[x=-√3〜0] √{12-(x+2√3)^2} - x^2 dx、
x+2√3=2√3*sin(θ)とおくと、∫[x=-√3〜0] √{12-(x+2√3)^2}=12∫[θ=π/6〜π/2] cos^2(θ) dθ
=6∫[θ=π/6〜π/2] 1+cos(2θ) dθ=6{θ+sin(2θ)/2}_[θ=π/6〜π/2]=6{π/2-(π/6+√3/4)}=2π-(3√3/2)
よって、S=2π-(3√3/2)-√3=2π-(5√3/2)