令x=rcosθ,y=rsinθ,则0＜r＜R,0＜θ＜2π.所以原积分=∫(0到2π)dθ∫(0到R)(6-3rcosθ-2rsinθ)rdr
=∫(0到2π)[(3r^2-r^3cosθ-2/3×r^3sinθ)(r=R)-(3r^2-r^3cosθ-2/3×r^3sinθ)(r=0)]dθ
=R^2∫(0到2π)[(3-Rcosθ-2/3×Rsinθ)dθ
=R^2×(3θ-Rsinθ+2/3×Rcosθ)(θ=2π)-R^2×(3θ-Rsinθ+2/3×Rcosθ)(θ=0)
=6πR^2.