∫∫（y/x）^2dxdy,D为曲线y=1/x,y=x,y=2所围成的区域计算二重积分 这个题的正确

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原式=∫dy∫(y/x)²dx
=∫y²dy∫(1/x²)dx
=∫y²(y-1/y)dy
=∫(y³-y)dy
=(y^4/4-y²/2)│
=2^4/4-2²/2-1^4/4+1²/2
=4-2-1/4+1/2
=9/4.======以下答案可供参考======
供参考答案1:
∫∫（y/x）^2dxdy
=∫(0,1)dy∫(1/y,y)（y/x）^2dx
=∫(0,1)-y^2/x|(1/y,y)dy
=∫(0,1)(y^3-y)dy
=1/4-1/2=-1/4
供参考答案2:
D: 1/x ≤ y ≤ x, 1 ≤ x ≤ 2
I = ∫[1,2] 1/x² dx ∫ [1/x, x] y² dy
= ∫[1,2] 1/x² * [ x³/3 ﹣1/(3x³) ] dx
= (1/3) ∫[1,2] [ x﹣ x^(-5) ] dx
= (1/3) [ x²/2 + (1/4) x^(-4) ] | [1,2]
= (1/3) [ 3/2 + (1/4) (1/16 ﹣1) ]
= 27/64