3重積分

質問

(1) I=∭_D〖(yz+zx)xdxdydz 〗 D={x^2+y^2+z^2≤4 ,y≥0,z≥0}
(2) I=∭_D〖2zx xdxdydz 〗 D={0≤x≤1,0≤y≤x,0≤z≤√(2-x^2-y^2 )}
回答が知りたいです。

回答No.2

info33 · ベストアンサー

(1)
I=∭[D](yz+zx)xdxdydz , D={(x,y,z)|x^2+y^2+z^2≤4 ,0≤y,0≤z}
=∭[D] ((x^2)dxdy zdz +xdx ydy zdz)
=∫ [-2,2] x^2 dx ∫ [0,(4-x^2)^(1/2)] dy ∫ [0,(4-x^2-y^2)^(1/2)] zdz
+∫ [-2,2] xdx ∫ [0,(4-x^2)^(1/2)] ydy ∫ [0,(4-x^2-y^2)^(1/2)] zdz
=∫ [-2,2] x^2 dx ∫ [0,(4-x^2)^(1/2)] dy {[(1/2)z^2] [0,(4-x^2-y^2)^(1/2)] }
+∫ [-2,2] xdx ∫ [0,(4-x^2)^(1/2)] ydy {[(1/2)z^2] [0,(4-x^2-y^2)^(1/2)] }
=∫ [-2,2] x^2 dx ∫ [0,(4-x^2)^(1/2)] dy {(1/2)(4-x^2-y^2) }
+∫ [-2,2] xdx ∫ [0,(4-x^2)^(1/2)] ydy {(1/2)(4-x^2-y^2) }
=(1/2) ∫ [-2,2] x^2 dx ∫ [0,(4-x^2)^(1/2)] (4-x^2-y^2)dy
+(1/2) ∫ [-2,2] xdx ∫ [0,(4-x^2)^(1/2)] (4-x^2-y^2) ydy
=(1/2) ∫ [-2,2] x^2 dx {[(4-x^2)y-(1/3)y^3] [0,(4-x^2)^(1/2)]}
+(1/2) ∫ [-2,2] xdx {[(4-x^2)y^2/2-(1/4)y^4] [0,(4-x^2)^(1/2)]}
=(1/2) ∫ [-2,2] x^2 dx {(4-x^2)-(1/3)(4-x^2)}(4-x^2)^(1/2)
+(1/8) ∫ [-2,2] xdx {[2(4-x^2)-(4-x^2)}(4-x^2)
=(1/3) ∫ [-2,2] x^2 (4-x^2)^(3/2) dx
+(1/8) ∫ [-2,2] (4-x^2)^2 xdx
=(2/3) ∫ [0,2] x^2 (4-x^2)^(3/2) dx
=(1/9)[24arcsin(x/2)-x(x^4-7x^2+6)(4-x^2)^(1/2)] [0,2]
=4π/3

(2)
I=∭_[D] 2zx xdxdydz , D={(x,y,z)|0≤x≤1,0≤y≤x,0≤z≤√(2-x^2-y^2 )}
=2 ∫ [0,1] x^2 dx ∫ [0, x] dy ∫ [0, (2-x^2-y^2)^(1/2)] z^2 dz
=2 ∫ [0,1] x^2 dx ∫ [0, x] dy {[(1/3)z^3] [0, (2-x^2-y^2)^(1/2)] }
=(2/3) ∫ [0,1] x^2 dx ∫ [0, x] (2-x^2-y^2)^(3/2) dy
=(√2)/15

お礼

ありがとうございます

回答No.1

EH1026TOYO

∭₍𝐃₎{(yz+zx)x}dxdydz 　　　D={x²+y²+z²≤4 ,y≥0,z≥0}
=(8+4π)/3

∭₍𝐃₎{2zx²}dxdydz 〗 　　　　D={0≤x≤1,0≤y≤x,0≤z≤√(2-x²-y² )}
= 5/18

お礼

ありがとうございます