Question

三重积分x^2+y^2+z^2=2az(a>0)及x^2+y^2=z^2(含有z轴部分)

Answer 1

求体积?

{ x² + y² + z² = 2az x² + y² + (z - a)² = a² --> z = a + √(a² - x² - y²)

{ x² + y² = z² --> z = √(x² + y²)

z² = 2az - z² --> z = a、x² + y² = z²

V = ∫∫∫ dxdydz

= ∫(0→a) dz ∫∫Dz dxdy + ∫(a→2a) dz ∫∫Dz dxdy

= ∫(0→a) πz² dz + ∫(a→2a) π(2az - z²) dz

= π(1/3)[ z³ ] |(0→a) + π[ az² - (1/3)z³ ] |(a→2a)

= π(1/3)a³ + π[ 4a³ - (8/3)a³ ] - π[ a³ - (1/3)a³ ]

= πa³

或

V = ∫∫∫ dxdydz

= ∫(0→2π) dθ ∫(0→a) r dr ∫(r→a) dz + ∫(0→2π) dθ ∫(0→a) r dr ∫(a→a + √(a² - r²) dz

= πa³