设D是曲线y=√(2x-x^2)与直线x+y=2围成的平面区域,求I=∫∫(D为积分区域)(x+y)dxdy

Question

设D是曲线y=√(2x-x^2)与直线x+y=2围成的平面区域,求I=∫∫(D为积分区域)(x+y)dxdy

Answer 1

先解出交点：
{ y = √(2x - x²)
{ y = 2 - x
得(1,1)、(2,0)
I = ∫∫ (x + y) dxdy、这区域不宜用极坐标
= ∫(1→2) ∫(2 - x→√(2x - x²)) (x + y) dydx
= ∫(1→2) (xy + y²/2)：(2 - x→√(2x - x²)) dx
= ∫(1→2) [x - 2 + x√(2x - x²)] dx
= (x²/2 - 2x)：(1→2) + ∫(1→2) x√[1 - (x - 1)²] dx、Let x - 1 = sinθ
= - 1/2 + ∫(0→π/2) (1 + sinθ)cos²θ dθ
= - 1/2 + ∫(0→π/2) cos²θ dθ + ∫(0→π/2) sinθcos²θ dθ
= - 1/2 + 1/2 * π/2 - 1/3 * cos³θ：(0→π/2)
= - 1/2 + π/4 - 1/3 * (0 - 1)
= π/4 - 1/6
= (6π - 4)/24