explain the following solution step by step please:

(a) PY = V . D = -(PD.) +- Pp = 4(=+1)coso - (=+1)cos+0 Py = 3(=+1)coso #C/m' (b) Omme = [ ppdv = [[[ 3(= +1)coso pdo d pd= = 3/ pop/ (= + 1)az [ cospag= 3 ( = p +=(sing = 3(2)(8+ 4)(1-0) = 72//C (C) Let w= w, +w+w,+w. +ws=D.ds where Wy. V2. ,. ya. , respectively correspond with surfaces S, .S,, S,.S,.S, (in the figure below) respectively. For S, p= 2, ds = pdodza. w, = 2p(=+1) cosdS| =2(2) [ (=+ 1)d= [ cosodds = 8(12)(1) =96 For S, .= = 0. ds = pded p(-a;) "2 =[p coso pdodp= -]pap ] cosodd For S, = = 4,dS = pded pa. . 4, = For S,. $= 1/ 2, ds = d pdza. ". =![ p(=+ 1) sing d pd= 12 -(sin 2) " [papj ( = + 1)d= 2 2 . (12) = =(2)(12) =-24 E 3 For S,, $=0.dS = dpd=(-a,).w, = [[ p(=+1)sing apd=]._ =0 4 =96-4+4-24+0=72/C This is exactly the answer obtained in part (b).In a certain region, the electric field is given by D = 2p(2 + 1)cos ba - p(z + 1)sinbad + p cos ba, uC/m (a) Find the charge density. (b) Calculate the total charge enclosed by the volume 0