求由曲面x²=a²-az,x²+y²=(a/2)²,z=0(a>0)所
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求由曲面x²=a²-az,x²+y²=(a/2)²,z=0(a>0)所围立体的体积
一道大一高数题,重积分的
一道大一高数题,重积分的
![求由曲面x²=a²-az,x²+y²=(a/2)²,z=0(a>0)所](/uploads/image/z/19022498-26-8.jpg?t=%E6%B1%82%E7%94%B1%E6%9B%B2%E9%9D%A2x%26sup2%3B%3Da%26sup2%3B-az%2Cx%26sup2%3B%2By%26sup2%3B%3D%EF%BC%88a%EF%BC%8F2%EF%BC%89%26sup2%3B%2Cz%3D0%28a%EF%BC%9E0%29%E6%89%80)
∵x²=a²-az,x²+y²=(a/2)²,z=0(a>0)
∴所围立体是一个下底以a/2为半径的圆,上底以曲面az=y²+3a²/4为顶的圆柱体
故所围立体的体积=4∫(0,a/2)dx∫(0,√(a²/4-x²))[y²/a+3a/4]dy (∫(0,a/2)表示从0到a/2积分,其他类同)
=4∫(0,π/2)dθ∫(0,a/2)(r²sin²θ/a+3a/4)rdr (进行极坐标变换)
=4∫(0,π/2)dθ∫(0,a/2)(r³sin²θ/a+3ar/4)dr
=∫(0,π/2)dθ*(r^4*sin²θ/a+3ar²/2)│(0,a/2)
=(a³/16)∫(0,π/2)(sin²θ+6)dθ
=(a³/32)∫(0,π/2)(13-cos(2θ))dθ
=(a³/32)(13θ-sin(2θ))│(0,π/2)
=(a³/32)(13π/2)
=13πa³/64.
∴所围立体是一个下底以a/2为半径的圆,上底以曲面az=y²+3a²/4为顶的圆柱体
故所围立体的体积=4∫(0,a/2)dx∫(0,√(a²/4-x²))[y²/a+3a/4]dy (∫(0,a/2)表示从0到a/2积分,其他类同)
=4∫(0,π/2)dθ∫(0,a/2)(r²sin²θ/a+3a/4)rdr (进行极坐标变换)
=4∫(0,π/2)dθ∫(0,a/2)(r³sin²θ/a+3ar/4)dr
=∫(0,π/2)dθ*(r^4*sin²θ/a+3ar²/2)│(0,a/2)
=(a³/16)∫(0,π/2)(sin²θ+6)dθ
=(a³/32)∫(0,π/2)(13-cos(2θ))dθ
=(a³/32)(13θ-sin(2θ))│(0,π/2)
=(a³/32)(13π/2)
=13πa³/64.
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