3重積分

質問

I=∭_D〖xyz xdxdydz 〗 D={0≤x≤1,0≤y≤1-x,0≤z≤2-y}

回答が知りたいです。

回答No.1

info33 · ベストアンサー

I=∭_D (xyz) xdxdydz 〗, ( D={0≤x≤1,0≤y≤1-x,0≤z≤2-y} )
=∫ [0,1] x^2 dx ∫ [0, 1-x] ydy ∫ [0, 2-y] zdz
=∫ [0,1] x^2 dx ∫ [0, 1-x] ydy { [z^2/2] [z:0,2-y]}
=∫ [0,1] x^2 dx ∫ [0, 1-x] y {(2-y)^2/2} dy
=∫ [0,1] x^2 dx ∫ [0, 1-x] y(y-2)^2/2 dy
=(1/2) ∫ [0,1] x^2 dx ∫ [0, 1-x] (y^3-4y^2+4y) dy
=(1/2) ∫ [0,1] x^2 dx {[y^4/4-4y^3/3+2y^2] [y:0, 1-x] }
=(1/24) ∫ [0,1] x^2 {(3(x-1)^4+16(x-1)^3+24(x-1)^2} dx
=(1/24) ∫ [0,1] (3x^6+4x^5-6x^4-12x^3+11x^2) dx
=(1/24) { [3x^7/7+2x^6/3-6x^5/5-3x^4+11x^3/3] [0,1] }
=(1/2520)*(45+70-126-315+385)
=59/2520