计算二重积分 ∫∫x^2dxdy 其中D是由椭圆 x^2/a^2+y^2/b^2=1 所围成的区域

先积y
∫∫x²dxdy
=∫[-a---->a] x² dx∫[-b√(1-x²/a²)----->b√(1-x²/a²)]dy
=(b/a)∫[-a---->a] x²√(a²-x²) dx
令x=asinu，则√(a²-x²)=acosu，dx=acosudu，u：-π/2---->π/2
=(b/a)∫[-π/2---->π/2] a²sin²u*acosu*acosu du
=a³b∫[-π/2---->π/2] sin²ucos²udu
=2a³b∫[0---->π/2] sin²ucos²udu
=(1/2)a³b∫[0---->π/2] sin²2udu
=(1/4)a³b∫[0---->π/2] (1-cos4u)du
=(1/4)a³b(u-(1/4)sin4u) |[0---->π/2]
=πa³b/8

∫∫x²dxdy=∫dθ∫(arcosθ)²*abrdr (作变换x=arcosθ,y=brsinθ.则dxdy=abrdθdr)
=a³b∫cos²θdθ∫r³dr
=(a³b/2)∫[1+cos(2θ)]dθ∫r³dr (应用倍角公式)
=(a³b/2){[θ+sin(2θ)/2]│}*[(r^4/4)│]
=(a³b/2)(2π+0-0-0)(1/4-0)
=πa³b/4。

一楼在做完第一个积分时少了个2倍,二楼的结果是正确的.不过一楼的方法更好些,二楼的方法一般的工科学生不会用.