∫∫（y⼀x）^2dxdy,D为曲线y=1⼀x,y=x,y=2所围成的区域计算二重积分 详细步骤

解：原式=∫<1,2>dy∫<1/y,y>(y/x)²dx
=∫<1,2>y²dy∫<1/y,y>(1/x²)dx
=∫<1,2>y²(y-1/y)dy
=∫<1,2>(y³-y)dy
=(y^4/4-y²/2)│<1,2>
=2^4/4-2²/2-1^4/4+1²/2
=4-2-1/4+1/2
=9/4。

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

∫∫（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