计算积分∫D∫xydxdy,其中D：x=0,y=0,x+y=1围成.

Question

计算积分∫D∫xydxdy,其中D：x=0,y=0,x+y=1围成.

Answer 1

∫∫_D f(x,y) dσ
= ∫(0-1) dy ∫(0-->(1 - y)) xy dx
= ∫(0-1) dy · y · x²/2 |(0-->(1 - y))
= ∫(0-1) dy · y/2 · (1 - y)² dy
= (1/2)∫(0-->1) y · (1 - 2y + y²) dy
= (1/2)∫(0-->1) (y - 2y² + y³) dy
= (1/2) · [y²/2 - (2/3)y³ + (1/4)y⁴] |(0-->1)
= (1/2) · (1/2 - 2/3 + 1/4)
= 1/24