设弓形的底为 b, 高为 h 。有个弓形面积的近似计算公式是
S = 2bh/3 + h^3/(2b)
这个公式是怎么推导出来的呢？

解：

假设 h << b

圆的半径 R = (b^2+4h^2)/(8h)                      (1)
R-h = (b^2-4h^2)/(8h)                             (2)

弓形方程 y = R(1-(x/R)^2)^0.5 - (b^2-4h^2)/(8h)   (3)

弓形面积 S = 2Int(0,b/2)[ydx]
= 2RInt(0,b/2)[(1-(x/R)^2)^0.5dx] - (b^3-4bh^2)/(8h)
= 2RInt(0,b/2)[1-(x/R)^2/2 - (x/R)^4/8]dx - (b^3-4bh^2)/(8h)
= bR -(b/2)^3/(3R) - (b/2)^5/(20R^3) - (b^3-4bh^2)/(8h)
= (b^3+4bh^2)/(8h) -(b/2)^3/(3R) - (b/2)^5/(20R^3) - (b^3-4bh^2)/(8h)
= bh - hb^3/(3(b^2+4h^2)) - b^5/(20R^3 2^5)
= bh - hb/(3(1+4h^2/b^2)) - b^5/(20R^3 2^5)

= bh - (hb/3)(1-4h^2/b^2+16h^4/b^4) - b^5/(20R^3 2^5)
= 2bh/3 + 4h^3/(3b)-16h^5/(3b^3) - 4h^3/(5b(1+4h^2/b^2))
= 2bh/3 + 4h^3/(3b)-16h^5/(3b^3) - 4h^3/(5b)
= 2bh/3 + 8h^3/(15b)-16h^5/(3b^3)

S = 2bh/3 + 8h^3/(15b)-16h^5/(3b^3)               (4)

如果忽略高次项，

S = 2bh/3 + 8h^3/(15b)                            (5)

为了方便记忆，考虑到高次项是负的，把第二项减小成
          8h^3/(16b)
这样，我们就得到

S = 2bh/3 + h^3/(2b)