二重积分 ∫∫x^2dxdy D=x^2+y^2-2x

2个回答

  • 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