计算下列二重积分。三小题
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(1) I = 2∫<0,1>xdx∫<0,1-x^2>2ydy = 2∫<0,1>x(1-x^2)^2dx
= -∫<0,1>(1-x^2)^2d(1-x^2) = -(1/3)[(1-x^2)^3]<0,1> = 1/3.
(2) I =∫<0,π>dx∫<x,π>cos(x+y)dy =∫<0,π>dx[sin(x+y)]<y=x, y=π>
=∫<0,π>(-sinx-sin2x>dx = [cosx+(1/2)cos2x]<0,π> = -2.
(3) I =∫<0,1>xdx∫<x^2,√x>√ydy = (2/3)∫<0,1>xdx[y^(3/2)]<x^2,√x>
= (2/3)∫<0,1>x[x^(3/4)-x^3]dx = (2/3)∫<0,1>[x^(7/4)-x^4]dx
= (2/3)[(4/11)x^(11/4)-(1/5)x^5]<0,1> = (2/3)(4/11-1/5) = 6/55.
= -∫<0,1>(1-x^2)^2d(1-x^2) = -(1/3)[(1-x^2)^3]<0,1> = 1/3.
(2) I =∫<0,π>dx∫<x,π>cos(x+y)dy =∫<0,π>dx[sin(x+y)]<y=x, y=π>
=∫<0,π>(-sinx-sin2x>dx = [cosx+(1/2)cos2x]<0,π> = -2.
(3) I =∫<0,1>xdx∫<x^2,√x>√ydy = (2/3)∫<0,1>xdx[y^(3/2)]<x^2,√x>
= (2/3)∫<0,1>x[x^(3/4)-x^3]dx = (2/3)∫<0,1>[x^(7/4)-x^4]dx
= (2/3)[(4/11)x^(11/4)-(1/5)x^5]<0,1> = (2/3)(4/11-1/5) = 6/55.
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