第1个回答 2013-03-20
y=x与y=x^2的交点为(0,0)(1,1)
∫∫xydxdy
=∫[0,1]∫[x^2,x]ydyxdx
=∫[0,1]y^2/2[x^2,x]*xdx
=∫[0,1](x^3/2-x^5/2)dx
=(x^4/8-x^6/12)[0,1]
=1/24本回答被提问者采纳
第2个回答 2013-03-20
曲线交点(0,0),(1,1)
∫∫xydxdy=∫(0,1)xdx∫(x^2,x)ydy
=∫(0,1)x[x^2-x^4]/2dx
=[x^3/3-x^6/6]/2 |(0,1)
=1/12