Question

求由x²＋（y-5）²＝16绕x轴旋转所围成的旋转体的体积

Answer 1

由x²+(y-5)²=16
==>
(y-5)²=16-x²
==>
y-5=±√(16-x²)
==>
y=5±√(16-x²)
且，x∈[-4,4]
所以，旋转体的体积=∫<-4,4>[π(y2)²-π(y1)²]dx【其中y2＞y1】
=π∫<-4,4>[(y2+y1)(y2-y1)]dx
=π∫<-4,4>[10×2√(16-x²)]dx
=20π∫<-4,4>√(16-x²)dx
=40π∫<0,4>√(16-x²)dx
令x=4sinα，则dx=4cosαdα；且x=4时，α=π/2；x=0时，α=0
原式=40π∫<0,π/2>4cosα×4cosαdα
=640π∫<0,π/2>cos²αdα
=640π∫<0,π/2>[(cos2α+1)/2]dα
=320π∫<0,π/2>(cos2α+1)dα
=320π[(1/2)sin2α+α]|<0,π/2>
=320π[(1/2)(0-0)+(π/2)-0]
=160π²

Answer 2

解：所求旋转体的体积=2∫<0,4>{π[5+√(16-x²)]²-π[5-√(16-x²)]²}dx
(自己作图分析)
=40π∫<0,4>√(16-x²)dx
=40π∫<0,π/2>4cosθ*4cosθdθ
(令x=4sinθ)
=320π∫<0,π/2>[1+cos(2θ)]dθ
(应用余弦倍角公式)
=320π*(π/2)
=160π².