Question

高数定积分求面积

题目如图谢谢~

Answer 1

求y=√x³，x=0，y=8所围图形分别绕y轴及直线x=4旋转所得旋转体的体积。

Answer 2

y = x^(3/2), 即 x = y^(2/3)
绕 y 轴，Vy = π∫<0, 8>y^(4/3)dy = (3π/7)[y^(7/3)]<0, 8> = 384π/7
绕 x=4 ，方法一
V = 4^2·8π - π∫<0, 8>[4-y^(2/3)]^2 dy
= 128π - π∫<0, 8>[16-8y^(2/3)+y^(4/3)] dy
= 128π - π[16y - (24/5)y^(5/3) + (3/7)y^(7/3)]<0, 8>
= 128π - π[128 - 768/5 + 384/7] = (98+26/35)π
方法二 V = 4^2·8π - ∫<0, 4>2π(4-x)x^(3/2) dx =
= 128π - 2π∫<0, 4>[4x^(3/2)-x^(5/2)] dx
= 128π - 2π[(8/5)x^(5/2) - (2/7)x^(7/2)]<0, 4>
= 128π - 2π(256/5-256/7) = 128π - (29+9/35)π = (98+26/35)π

Answer 3

Answer 4