怎么用微积分求X平方+Y平方=4 与直线Y=1之间的面积

1个回答

  • 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