作业帮三重积分比较I1,I2大小
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/10/07 02:12:26
三重积分比较I1,I2大小
设Ω由平面x+y+z+1=0,x+y+z+2=0,x=0,y=0,z=0围成,I1=∫∫∫[ln(x+y+z+3)]²dV,I2=∫∫∫(x+y+z)²dV,比较I1,I2大小
设Ω由平面x+y+z+1=0,x+y+z+2=0,x=0,y=0,z=0围成,I1=∫∫∫[ln(x+y+z+3)]²dV,I2=∫∫∫(x+y+z)²dV,比较I1,I2大小
设Ω由平面x+y+z+1=0,x+y+z+2=0,x=0,y=0,z=0围成,L₁=∫∫∫[ln(x+y+z+3)]²dV；L₂=∫∫∫(x+y+z)²dV,
比较L₁,L₂的大小.
上下限的书写很麻烦,我在积分符号前面用加黑的中括号【a,b】表示,前面的a是
下限,后面的b是上限.
L₁=【Ω】∫∫∫[ln(x+y+z+3)]²dV=【Ω】2∫∫∫ln(x+y+z+3)dxdydz
=2【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫ln(x+y+z)dz
=2【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫ln(x+y+z)d(x+y+z)
=2【1,2】∫dx【1-x,2-x】∫{(x+y+z)[ln(x+y+z)-1]}【1-x-y,2-x-y】dy
=2【1,2】∫dx【1-x,2-x】∫[2(ln2-1)+1]dy
=2【1,2】∫dx【1-x,2-x】∫(2ln2-1)dy
=2(2ln2-1)【1,2】∫dx【1-x,2-x】∫dy
=2(2ln2-1)【1,2】∫dx=2(2ln2-1)=0.7726
L₂=【Ω】∫∫∫(x+y+z)²dV=【Ω】∫∫∫(x+y+z)²dxdydz
=【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫(x+y+z)²dz
=【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫(x+y+z)²d(x+y+z)
=【1,2】∫dx【1-x,2-x】∫[(x+y+z)³/3]【1-x-y,2-x-y】dy
=【1,2】∫dx【1-x,2-x】∫[8/3-1/3)dy
=(7/3)【1,2】∫dx【1-x,2-x】∫dy
=(7/3)【1,2】∫dx=7/3=2.3333
故L₂>L₁.
比较L₁,L₂的大小.
上下限的书写很麻烦,我在积分符号前面用加黑的中括号【a,b】表示,前面的a是
下限,后面的b是上限.
L₁=【Ω】∫∫∫[ln(x+y+z+3)]²dV=【Ω】2∫∫∫ln(x+y+z+3)dxdydz
=2【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫ln(x+y+z)dz
=2【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫ln(x+y+z)d(x+y+z)
=2【1,2】∫dx【1-x,2-x】∫{(x+y+z)[ln(x+y+z)-1]}【1-x-y,2-x-y】dy
=2【1,2】∫dx【1-x,2-x】∫[2(ln2-1)+1]dy
=2【1,2】∫dx【1-x,2-x】∫(2ln2-1)dy
=2(2ln2-1)【1,2】∫dx【1-x,2-x】∫dy
=2(2ln2-1)【1,2】∫dx=2(2ln2-1)=0.7726
L₂=【Ω】∫∫∫(x+y+z)²dV=【Ω】∫∫∫(x+y+z)²dxdydz
=【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫(x+y+z)²dz
=【1,2】∫dx【1-x,2-x】∫dy【1-x-y,2-x-y】∫(x+y+z)²d(x+y+z)
=【1,2】∫dx【1-x,2-x】∫[(x+y+z)³/3]【1-x-y,2-x-y】dy
=【1,2】∫dx【1-x,2-x】∫[8/3-1/3)dy
=(7/3)【1,2】∫dx【1-x,2-x】∫dy
=(7/3)【1,2】∫dx=7/3=2.3333
故L₂>L₁.