解：过点$D$作$DE\perp AB$于点$E，$因为$AB// DC，$$AB\perp BC，$所以四边形$BCDE$是矩形，则$BE = CD = 2，$$DE = BC = 4，$$AE=AB - BE=5 - 2 = 3。$
根据勾股定理$AD=\sqrt{AE^{2}+DE^{2}}=\sqrt{3^{2}+4^{2}} = 5。$
该几何体是由一个圆柱和一个圆锥组成，圆柱底面半径$r = BC = 4，$高$h_1 = CD = 2，$圆锥母线长$l = AD = 5，$高$h_2 = AE = 3。$
圆柱侧面积$S_{柱侧}=2\pi rh_1=2\pi\times4\times2 = 16\pi，$圆锥侧面积$S_{锥侧}=\pi rl=\pi\times4\times5 = 20\pi，$圆柱底面积$S_{底}=\pi r^{2}=\pi\times4^{2}=16\pi。$
所以几何体表面积$S = S_{柱侧}+S_{锥侧}+S_{底}=16\pi+20\pi+16\pi = 52\pi。$
