怎么用微积分求X平方+Y平方=4与直线Y=1之间的面积

Question

怎么用微积分求X平方+Y平方=4 与直线Y=1之间的面积

Answer 1

先定义积分符号：S(a,b)f(x)dx 即为f(x)在x=(a,b)区间的定积分,下限为a,上限为b
x^2 + y^2 = 4
通过 x^2 + y^2 = 4 和 y = 1 解算出f(x)的区间,a = -√3,b = √3
下面做定积分：
从x取值区间判断出y大于0,所以：y = √(4 - x^2) = 2√[1 - (x/2)^2]
S = S(a,b)f(x)dx
这样做积分运算有点麻烦,可以变换原函数为参数方程
x = R*cost
y = R*sint
R = 2
a = -√3,对应 t1 = π/6
b = +√3,对应 t2 = π - π/6 = 5π/6
积分式变换为极坐标表述
微分面积单元扇形近似于三角形,此三角形高即为圆半径R,底边为 Rdt
dS = (R*Rdt)/2
S = (1/2)*S(t1,t2)(R^2)*dt
= (R^2/2)*t |(t1,t2)
= (R^2/2)*(t2 - t1)
= (R^2/2)*(5π/6 - π/6)
= (R^2/2)*(2π/3)
= (πR^2)/3 .R = 2
= 4π/3