因为单位圆的对成型，所以对2x的积分是0.且∫∫x^2dx=∫∫y^2dy所以原积分=∫∫(x^2+2x+1+2y^2)dxdy=∫∫(x^2+1+2y^2)dxdy=(3/2)∫∫(x^2+y^2)dxdy+π=(3/2) ∫(0->2π)dθ ∫(0->1) r^3dr=3π/4
