Question

∫∫(X+Y)dxdy,其中D是由y=x^2,y==4x^2及y=1所围成的闭区域，求二重积分

Answer 1

简单计算一下即可，答案如图所示

Answer 2

原式=∫<-π/2,π/2>dθ∫<0,2cosθ>√(4-r²)rdr (作极坐标变换)
=∫<-π/2,π/2>[(8/3)(1-sin³θ)]dθ
=(8/3)∫<-π/2,π/2>[1-sinθ(1-cos²θ)]dθ
=(8/3)[θ+cosθ-cos³θ/3]│<-π/2,π/2>
=(8/3)[π/2-(-π/2)]
=8π/3。

Answer 3

(作极坐标变换)
=∫<-π/2,π/2>[(8/3)(1-sin³θ)]dθ
=(8/3)∫<-π/2,π/2>[1-sinθ(1-cos²θ)]dθ
=(8/3)[θ+cosθ-cos³θ/3]│<-π/2,π/2>
=(8/3)[π/2-(-π/2)]
=8π/3。

Answer 4

没看出这有什么难的啊，两条抛物线加一条直线围成的区域，用Y-型表示，然后计算就行