請問這兩題重積分怎麼積分(20點)？

1. (0,1),(1,0)連線為x+y=1 , y=1-x or x=1-y
∫[0~1]∫[0~1-x] f(x,y) dy dx or ∫[0~1]∫[0~1-y] f(x,y) dx dy
2. |x|+|y|=1
第1象限: x+y=1, x=1-y, y=0~1
第2象限: -x+y=1, x=y-1, y=0~1
第3象限: -x-y=1, x=-y-1, x=-1~0
第4象限: x-y=1, x=y+1, y=-1~0
so,∫∫(y-2x^2)dA
=∫[0~1]∫[y-1~1-y] (y-2x^2) dx dy+∫[-1~0]∫[-y-1~y+1] (y-2x^2)dx dy
=∫[0~1] 2y(1-y)-(4/3)(1-y)^3 dy+∫[-1~0] 2y(y+1)-(4/3)(y+1)^3 dy
= [y^2-(2/3)y^3+(1/3)(1-y)^4]|[0~1] + [(2/3)y^3+y^2-(1/3)(y+1)^4]|[-1~0]
= [ 1/3 - 1/3] +[-1/3- 1/3] = -2/3

2010-06-12 12:41:10 補充：
Q2:坐標變換亦可,設u=x-y, v=x+y, ∂(u,v)/∂(x,y)=2, so, ∂(x,y)/∂(u,v)=1/2, (x,y)=(u+v, v-u)/2
原式=∫[-1~1]∫[-1~1] [(v-u)/2-(u+v)^2 /2] *(1/2) du dv
=∫[-1~1]∫[-1~1] (-u^2-v^2)/4 du dv (已刪除奇函數部分)
=(-1/2)∫[-1~1] (1/3+ v^2) dv = -(v/3+ v^3/3) for v=0~1
= -4/3

2010-06-12 12:44:08 補充：
更正原作答:∫∫(y-2x^2)dA
=∫[0~1]∫[y-1~1-y] (y-2x^2) dx dy+∫[-1~0]∫[-y-1~y+1] (y-2x^2)dx dy
=∫[0~1] 2y(1-y)-(4/3)(1-y)^3 dy+∫[-1~0] 2y(y+1)-(4/3)(y+1)^3 dy
= [y^2-(2/3)y^3+(1/3)(1-y)^4]|[0~1] + [(2/3)y^3+y^2-(1/3)(y+1)^4]|[-1~0]
= [ -1/3 - 2/3] +[-1/3- 1/3] = -4/3

第二題用個Jacobian