关于二重积分的计算!1.∫∫D4y^2sin（xy）dxdy,y=√（π/2）,x=0,y=x,所围成区域,求二重积分2

1.∫∫D 4y^2sin(xy)dxdy
=∫[0,√(π/2)]4y^2dy*∫[0,y]sin(xy)dx
=∫[0,√(π/2)]4ydy*∫[0,y]sin(xy)d(xy)
=∫[0,√(π/2)]4y*[-cos(xy)]|[0,y]*dy
=∫[0,√(π/2)]4y*[1-cos(y^2)]*dy
=∫[0,√(π/2)]4ydy - ∫[0,√(π/2)]4y*cos(y^2)*dy
=[2y^2]|[0,√(π/2)] - ∫[0,√(π/2)]2*cos(y^2)*d(y^2)
=2*(π/2) - [2sin(y^2)]|[0,√(π/2)]
=π - [2*sin(π/2)-2*sin0]
=π-2
2.∫[0,a]∫[0,√(a^2-x^2)]√(a^2-x^2-y^2)dxdy
=∫∫D √(a^2-r^2)*rdrdθ
=∫[0,π/2]dθ*∫[0,a]r(a^2-r^2)^(1/2)dr
=(π/2)*(-1/2)∫[0,a](a^2-r^2)^(1/2)d(a^2-r^2)
=(-π/4)*[(2/3)*(a^2-r^2)^(3/2)]|[0,a]
=(-π/4)*[0-(2/3*a^3)]
=π/4*2/3*a^3
=(π/6)*a^3
表达习惯略有不同,