求由曲线y=3/x和直线x+y=4所围平面图形饶x轴旋转所成旋转体的体积

Question

求由曲线y=3/x和直线x+y=4所围平面图形饶x轴旋转所成旋转体的体积

Answer 1

联立二方程,可得3/x = 4 - x,x² - 4x + 3 = 0,(x - 1)(x- 3) = 0
x = 1或x = 3,交点为A(1,3),B(3,1)
积分区间为[1,3]
在x处(1< x < 3),曲线y=3/x和直线x+y=4所围平面图形绕x轴旋转所成旋转体的截面为一个圆环,外径为4 - x,内径为3/x,截面积 = π[(4 - x)² - (3/x)²] = π[(x - 4)² - (3/x)²]
旋转体的体积V = ∫₁³π[(x - 4)² - (3/x)²]dx
= π[(x- 4)³/3 + 9/x]|₁³
= π(-1/3 + 3 + 27/3 - 9)
= 8π/3