题目

二重积分∫∫(y+√x^2+y^2)dxdy D为x^2+Y^2

提问时间：2020-12-24

答案

2012-10-18 21:11区域为：(x-1)²+y²≤4,以(1,0)为圆心,2为半径的圆.
先积y,
∫∫ x²dxdy
=∫[-1---->3] dx ∫[-√(3-x²+2x)----->√(3-x²+2x)] x²dy
=2∫[-1---->3] x²√(3-x²+2x)dx
=2∫[-1---->3] x²√(4-(x-1)²)dx
令x-1=2sinu,则√(4-(x-1)²)=2cosu,dx=2cosudu,u：0---->π/2
=2∫[-π/2---->π/2] (2sinu+1)²*2cosu*2cosudu
=32∫[-π/2---->π/2] sin²ucos²udu + 8∫[-π/2---->π/2] cos²udu
=8∫[-π/2---->π/2] sin²2u du + 4∫[-π/2---->π/2] (1+cos2θ)du
=4∫[-π/2---->π/2] (1-cos4u) du + 4π
=4(u-(1/4)sin4u) + 4π [-π/2---->π/2]
=8π