Question

计算三重积分

Answer 1

V：x≥0,y≥0,z≥0,x+y+z≤1
∫∫∫v 1/(x+y+z+1)^3 dxdydz
=∫(0→锋册1)dx∫(0→1-x)dy∫(0→1-x-y)1/(x+y+z+1)^3 dz
=∫(0→1)dx∫(0→1-x)[(-1/2)*1/(x+y+z+1)^2]|[0,1-x-y] dy
=∫(0→1)dx∫(0→1-x)(-1/2)*[1/2^2-1/(x+y+z+1)^2]dy
=(-1/2)∫(0→1)dx∫(0→1-x)[1/4-1/(x+y+z+1)^2]dy
=(-1/2)∫(0→1)[(1/4)y+1/(x+y+z+1)]|[0,1-x] dx
=(-1/2)∫(0→1)[(1/4)(1-x)+1/2-1/(x+1)]dx
=(-1/2)∫(0→1)[-1/(x+1)+3/4-(1/4)x]dx
=(-1/2)*[-ln(x+1)+(3/4)x-(1/8)x^2]|[0,1]
=(-1/2)*(-ln2+3/4-1/8)
=(1/2)ln2-5/16

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