正方形的边长为10,求阴影部分的面积

2025-04-05 07:33:38
推荐回答(1个)
回答1:

设上弧方程为x^2+(y-5)^2=25,
左弧方程为(x+5)^2+(y+5)^2=100,
两弧的右交点为(3,1).
右弧方程为(x-5)^2+(y+5)^2=100,
左右弧的交点为(0,5√3-5),
由对称性,阴影面积=2∫<0,3>{√[100-(x+5)^2]-5-[5-√(25-x^2)]}dx
=2∫<0,3>{√[100-(x+5)^2]+√(25-x^2)-10}dx
=2{[(x+5)/2√[100-(x+5)^2]+50arcsin[(x+5)/10]+(x/2)√(25-x^2)+(25/2)arcsin(x/5)}|<0,3>
=2{24-25√3/2+50(arcsin0.8-π/6)+6+(25/2)arcsin0.6]
=60+100arcsin0.8+25arcsin0.6-25√3-50π/3.