Question

其中d由x=-2,y=2,x轴及曲线x=-根号2y-y^2围成

∫D∫ydxdy,其中D是直线X=-2,y=0,y=2,及曲线x=-根号下（2y-y的平方）所围成的平面区域.怎么算,

Answer 1

简单计算一下即可，答案如图所示

Answer 2

原式=∫(0,2)dy∫(-2,-√(2y-y²))ydx
=∫(0,2)y(2-√(2y-y²))dy
=∫(0,2)(2y-y√(2y-y²))dy
=(y²)│(0,2)-∫(0,2)y√(2y-y²)dy
=4-∫(0,2)y√(1-(1-y)²)dy
=4-∫(π/2,-π/2)cos²t(sint-1)dt (令1-y=sint)
=4+∫(π/2,-π/2)cos²td(cost)+∫(π/2,-π/2)(1+cos(2t))/2dt
=4+(cos³t/3+t/2+sin(2t)/4)│(π/2,-π/2)
=4+(-π/4-π/4)
=4-π/2.