Question

求曲线y=e^x与y=2及x=0围成的图形的面积A,和绕y轴旋转一周的体积、

Answer 1

联立y = e^x 和y = 2,可得二者的交点为(ln2,2)

x = 0为y轴,三者所围的图形的面积为f(x) = 2 - e^x在0和ln2之间的定积分

F(x) = ∫(2-e^x)dx = 2x - e^x + C

A = F(ln2) - F(0) = (2ln2 - 2) - (0 - 1) = 2ln2 -1

计算绕y轴旋转一周的体积时,用y作为自变量更简便.y = e^x,x = lny; 在y处的半径为lny,截面积为π(lny)²,积分区间为[1,2]

G(y) = π∫(lny)²dy = π(yln²y -2ylny + 2y) +C

G(2) = π(2ln²2 - 4ln2+ 4) +C

G(1) = 2π + C

V = G(2) - G(1) = 2π(ln²2 - 2ln2 + 1) = 2π(ln2 -1)²