先求出交点（2,-2）, 则当y0在0到-2上移动时， x会随之变化， 且大小为x1<=x<=x2, 所以原式可以化成两次定积分， 即化成：
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第1个回答 2012-12-19
求出积分区域,
y^2/2<=x<=y+4,
-2<=y<=4,
I=∫ [-2,4] dy ∫ [y^2/2,y+4] xydx
=∫ [-2,4] dy ( x^2/2) [y^2/2,y+4]
=(1/2)∫ [-2,4] (y+4)^2-y^4/4)ydy
=(1/2)∫ [-2,4] (y^3+8y^2+16y-y^5/4)
=(1/2)(y^4/4+8y^3/3+8y^2-y^6/24)[-2,4]
=90.本回答被提问者采纳
y^2/2<=x<=y+4,
-2<=y<=4,
I=∫ [-2,4] dy ∫ [y^2/2,y+4] xydx
=∫ [-2,4] dy ( x^2/2) [y^2/2,y+4]
=(1/2)∫ [-2,4] (y+4)^2-y^4/4)ydy
=(1/2)∫ [-2,4] (y^3+8y^2+16y-y^5/4)
=(1/2)(y^4/4+8y^3/3+8y^2-y^6/24)[-2,4]
=90.本回答被提问者采纳