求圆（x-5)^2+y^2=16绕y轴旋转一周生成的旋转体的体积。（用定积分求旋转体的体积）

Answer 1

解法一：所求体积=2∫<1,9>2πx√[16-(x-5)²]dx
=4π∫<1,9>x√[16-(x-5)²]dx
=4π∫<-π/2,π/2>(4sint+5)*4cost*4costdt (令x=4sint+5)
=64π∫<-π/2,π/2>(4sint+5)cos²tdt
=640π∫<0,π/2>cos²tdt
=320π∫<0,π/2>[1+cos(2t)]dt
=320π[t+sin(2t)/2]│<0,π/2>
=320π(π/2+0)
=160π²；
解法二：所求体积=2∫<0,4>π[(5+√(16-y²))²-(5-√(16-y²))²]dy
=40π∫<0,4>√(16-y²)dy
=40π∫<0,π/2>4cost*4costdt (令y=4sint)
=320π∫<0,π/2>[1+cos(2t)]dt
=320π[t+sin(2t)/2]│<0,π/2>
=320π(π/2+0)
=160π²。

Answer 2

答：
x=5±√(16-y^2)
且关于x轴对称，所以V
=2π∫0到4 [(5+√(16-y^2))^2-(5-√(16-y^2))^2] dy
=2π∫0到4 20√(16-y^2) dy
=40π∫0到4 √(16-y^2) dy
令y=4sint,则t积分区域为0到π/2
则40π∫√(16-y^2) dy
=40π*16∫(cost)^2 dt
=40π*16(t/2+sin2t/4)|0到π/2
=160π^2

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